Operations with decimal fractions. Decimal fractions. Solving equations (presentation)

Lesson-tale DECIMAL FRACTIONS. SOLVING EQUATIONS

Denisova Svetlana Ivanovna

mathematic teacher

MOU " high school No. 1"

Kimry, Tver region

And he had three sisters

Ivan Tsarevich gave his sisters in marriage to the kings

copper kingdom

silver kingdom

golden kingdom

He lived without his sisters for a whole year, and he became bored. He decided to visit his sisters

and hit the road

They went out to the river, and there a huge stone blocked the road to the bridge

(y - 0.371)+ 5.44= 27.7

(0.127 + m) – 9.8= 3.2

(x + 0.379) – 1.97=1.83

If you solve them correctly, the stone will turn and clear the way

2.4 – 3x = 0.21 (2)

2.5x + 0.8x = 99 (2)

5x – 7.35 = 0.3 (3)

7.2y – 0.3y = 27.6 (3)

She had been at enmity with Koshchei for a long time and agreed to help Ivan Tsarevich, but only if his warriors solved six equations

5.8y – 2.7y = 62 (1)

0.65 + 2x = 5.9 (1)

Saying goodbye to Tsarevich Ivan, Baba Yaga told him about the power of the equation.

If you need a lock to unlock or close tightly, say out loud the roots of the equation. It will be fulfilled in an instant.

Koschey waylaid Ivan the Tsarevich and his soldiers, grabbed them and threw them into a deep dungeon. Locked with six locks.

3.5:x – 2 = 1.5 (1)

(x – 0.5) * 5 =0.4 * 2 – 0.3 * 2 (1)

y: 0.2 + 0.35 = 3.6 (2)

(0.3 + x) * 4 = 0.3 * 3 + 0.7 * 3 (2)

m: 0.12 * 0.2 = 7.2 (3)

(0.7 + x) * 5 = 0.8 * 5 + 0.6 * 5 (3)

Ivan Tsarevich uttered “magic words” and named the roots of all equations. The dungeon doors opened. The soldiers stood in front of the gates of the Koshcheev Palace

y + 0.0015: 0.001 = 1.5

After that, Tsarevich Ivan and the beautiful Elena visited his sisters, came home and began to live, live and make good

Fairy tale “The Magic Word” In a certain kingdom, in a certain state, there lived Ivan Tsarevich. Ivan Tsarevich once met Elena the Beautiful. They fell in love with each other. But the evil Koschey the Immortal kidnapped Elena the Beautiful. Ivan Tsarevich went to help out his beloved. So he drove up to the river, and there a huge stone blocked the road to the bridge. There are 3 equations written on the stone: 1) (y – 3.71) - 5.46 = 12.77 2) (12.7 +x) – 9.8 = 3.2 3) (y +3.79) – 1.79 = 1.83. If they are solved correctly, the stone will turn and clear the way. Help Ivan Tsarevich

Kingdom of Baba Yaga: -Ivan the prince rode through the forest for a long time until the road led him to Baba Yaga’s hut. She had long been at enmity with Kosh the Immortal and agreed to help Ivan, the prince, but only on the condition that he solved the equations written on the walls of the hut

Solve the equations 1) 6.5 + 2x = 14.5 2) 12.4 – 3x = 3.4 3) 7.5 + 5x - 1.5 = 16 -When saying goodbye to Ivan, the prince, Baba Yaga told him about the power roots of the equation: “If you need a lock to unlock or close tightly, say out loud the roots of the equation. It will be fulfilled in an instant.”

Black Raven: Black Raven overheard this conversation and told Koshchei about everything. He waylaid Ivan, the prince, grabbed him and threw him into a deep dungeon. Locked with 3 locks. Help Ivan - Tsarevich 1) 35: x – 1.2 = 3.8 2) y: = 7.7 3)(x – 5.4) - 2.3 = 5.2

« Magic words“: Ivan Tsarevich uttered “magic words” and named the roots of all the equations. The dungeon doors opened. And Ivan, the prince, stood in front of the gates of Koshcheev’s kingdom. And on the gate there is an equation written: (y + 2.84) -1.84 = 6.4 – Ivan, the prince, solved it orally. The gates opened. Ivan - Tsarevich freed Elena the Beautiful and on the same day they played a wedding. Can you solve this equation orally?

From the history of mathematics. –The rules of calculations with decimal fractions were described by the famous scientist al-Kashi Jemshid Ibn Masud at the beginning of the 15th century. He wrote fractions in the same way as is customary now, but did not use a comma: he wrote the fractional part in red ink or separated it with a vertical line. But in Europe they did not find out about this, and only 150 years later the scientist Simon Stephen wrote down decimal fractions in a rather complex way: in place of the decimal point, a zero in a circle. A comma or period to separate a whole part has been used since the 17th century. In Russia, L. F. Magnitsky outlined decimal fractions in 1703 in the first mathematics textbook “Arithmetic, that is, the science of numerals.”

Complete task 1).2.01 = 2) 105.11 – 8.7 = 3) Solve the equation: 1 – x = 0.89 4) Solve the equation: x + 15.35 = 19.4 5) On the first day They sold 12.52 m of fabric, and on the second day another 19.7 m. How much fabric were sold in two days? 6). The mass of two heads of cabbage is 10.67 kg, and one of them is 5.29 kg. What is the mass of the other head of cabbage?

Interesting page: p/pKSCHTIYA 12,4463,22455,1554,215,20,110,151,0510,830,75 57,1830,229,4332,2115,9614,2713,44,08

Knowledge test Option 1 Mandatory part. 1). Calculate: a) 28..7 + 1.53 b) 75.4 – 4.23 2). Find the value of the expression: 8.3 + 4, – 1.25. Additional part: 3). From a piece of wire 20 m long, 4 pieces were cut: the first one was 1.7 m long, and each next one was half a meter longer than the previous one. Determine the length of the remaining piece of wire. Option 2 Mandatory part. 1). Calculate: a) 32.9 + 3.61 b) 10 -4.26. 2). Find the meaning of the expression: , – Additional part 3). The route consists of 3 sections. The first section is 4.2 km long, the second is one and a half kilometers longer, and the third is one and a half kilometers longer. less than the first. What is the length of the entire route?

Solving equations with fractions Let's look at examples. The examples are simple and illustrative. With their help, you will be able to understand in the most understandable way.
For example, you need to solve the simple equation x/b + c = d.

An equation of this type is called linear, because The denominator contains only numbers.

The solution is performed by multiplying both sides of the equation by b, then the equation takes the form x = b*(d – c), i.e. the denominator of the fraction on the left side cancels.

For example, how to solve a fractional equation:
x/5+4=9
We multiply both sides by 5. We get:
x+20=45
x=45-20=25

Another example when the unknown is in the denominator:

Equations of this type are called fractional-rational or simply fractional.

We would solve a fractional equation by getting rid of fractions, after which this equation, most often, turns into a linear or quadratic equation, which can be solved in the usual way. You just need to consider the following points:

• the value of a variable that turns the denominator to 0 cannot be a root;
• You cannot divide or multiply an equation by the expression =0.

This is where the concept of the region of permissible values ​​(ADV) comes into force - these are the values ​​of the roots of the equation for which the equation makes sense.

Thus, when solving the equation, it is necessary to find the roots, and then check them for compliance with the ODZ. Those roots that do not correspond to our ODZ are excluded from the answer.

For example, you need to solve a fractional equation:

Based the above rule x cannot be = 0, i.e. ODZ in this case: x – any value other than zero.

We get rid of the denominator by multiplying all terms of the equation by x

And we solve the usual equation

5x – 2x = 1
3x = 1
x = 1/3

Answer: x = 1/3

Let's solve a more complicated equation:

ODZ is also present here: x -2.

When solving this equation, we will not move everything to one side and bring the fractions to a common denominator. We will immediately multiply both sides of the equation by an expression that will cancel out all the denominators at once.

To reduce the denominators you need left side multiply by x+2, and the right hand by 2. This means that both sides of the equation must be multiplied by 2(x+2):

This is the most common multiplication of fractions, which we have already discussed above.

Let's write the same equation, but slightly differently

The left side is reduced by (x+2), and the right by 2. After the reduction, we obtain the usual linear equation:

x = 4 – 2 = 2, which corresponds to our ODZ

Answer: x = 2.

Solving equations with fractions not as difficult as it might seem. In this article we have shown this with examples. If you have any difficulties with how to solve equations with fractions, then unsubscribe in the comments.

Equations with fractions themselves are not difficult and are very interesting. Let's look at the types of fractional equations and how to solve them.

How to solve equations with fractions - x in the numerator

If a fractional equation is given, where the unknown is in the numerator, the solution does not require additional conditions and is solved without unnecessary hassle. General form such an equation is x/a + b = c, where x is the unknown, a, b and c are ordinary numbers.

Find x: x/5 + 10 = 70.

In order to solve the equation, you need to get rid of fractions. Multiply each term in the equation by 5: 5x/5 + 5x10 = 70x5. 5x and 5 are cancelled, 10 and 70 are multiplied by 5 and we get: x + 50 = 350 => x = 350 – 50 = 300.

Find x: x/5 + x/10 = 90.

This example is a slightly more complicated version of the first one. There are two possible solutions here.

• Option 1: We get rid of fractions by multiplying all terms of the equation by a larger denominator, that is, by 10: 10x/5 + 10x/10 = 90×10 => 2x + x = 900 => 3x = 900 => x=300.
• Option 2: Add the left side of the equation. x/5 + x/10 = 90. The common denominator is 10. Divide 10 by 5, multiply by x, we get 2x. Divide 10 by 10, multiply by x, we get x: 2x+x/10 = 90. Hence 2x+x = 90×10 = 900 => 3x = 900 => x = 300.

We often encounter fractional equations in which the x's are on opposite sides of the equal sign. In such situations, it is necessary to move all the fractions with X's to one side, and the numbers to the other.

• Find x: 3x/5 = 130 – 2x/5.
• Move 2x/5 to the right with opposite sign: 3x/5 + 2x/5 = 130 => 5x/5 = 130.
• We reduce 5x/5 and get: x = 130.

How to solve an equation with fractions - x in the denominator

This type of fractional equations requires writing additional conditions. The indication of these conditions is a mandatory and integral part of the right decision. By not adding them, you run the risk, since the answer (even if it is correct) may simply not be counted.

The general form of fractional equations, where x is in the denominator, is: a/x + b = c, where x is the unknown, a, b, c are ordinary numbers. Please note that x may not be any number. For example, x cannot equal zero, since it cannot be divided by 0. This is exactly what it is additional condition, which we must specify. This is called the range of permissible values, abbreviated as VA.

Find x: 15/x + 18 = 21.

We immediately write the ODZ for x: x ≠ 0. Now that the ODZ is indicated, we solve the equation using standard scheme, getting rid of fractions. We multiply all terms of the equation by x. 15x/x+18x = 21x => 15+18x = 21x => 15 = 3x => x = 15/3 = 5.

Often there are equations where the denominator contains not only x, but also some other operation with it, for example, addition or subtraction.

Find x: 15/(x-3) + 18 = 21.

We already know that the denominator cannot be equal to zero, which means x-3 ≠ 0. We transfer -3 to right side, changing the “-” sign to “+” and we get that x ≠ 3. The ODZ is indicated.

We solve the equation, multiply everything by x-3: 15 + 18×(x – 3) = 21×(x – 3) => 15 + 18x – 54 = 21x – 63.

Move the X's to the right, numbers to the left: 24 = 3x => x = 8.

Back forward

Attention! Slide previews are for informational purposes only and may not represent all the features of the presentation. If you are interested this work, please download the full version.

Lesson objectives:

• test the ability to perform operations with decimal fractions orally and in writing; consolidate and test the ability to solve equations and problems involving decimals;
• develop fast work thoughts, ingenuity and attentiveness; develop interest in mathematics.
• cultivate friendships in the classroom and a sense of empathy for each other; develop the ability to speak up.

Lesson type: generalization and systematization of knowledge.

Lesson type: Olympiad lesson using presentation.

Equipment: a table with decimal fractions written on it, cards with equations, cards with tasks, a table with a task for ingenuity, a table with examples for mental calculation.

DURING THE CLASSES

1. Organizing time (3 min.)

Calm down and seat the children.

Teacher: We already had the “Olympiad in Natural Numbers”. Now we have learned about decimals. It's time for the Decimal Olympiad (Slide 1). Some things will be similar to the previous Olympiad, but many tasks will be new. And the most important thing is that all actions, assignments and tasks will only be in decimal fractions. Therefore, how well you perform depends on your knowledge on the topic. The teams, like last time, will be in rows. The result of some tasks will directly depend on the concentration of the entire team.

2. Warm-up – oral work(3 min) (Slide 2)

Teacher: Any competition begins with a warm-up. Our warm-up will be mental counting. But this time the warm-up will not affect the result of the competition, and the tasks will be given randomly. Therefore, the most important thing now is not to answer correctly, but to tune in to the lesson.

Examples are given randomly in order to involve as many students as possible from all rows.

3. “Who is faster?”(5 min) (Slide 3)

Teacher: Well, now let's move on to the competition. The first competition will be speed. A table of numbers will now open on our board. Decimal fractions are scattered on it. Your task will be this: find a fraction that matches the condition as quickly as possible. This task is not addressed to any particular series, so everyone will be looking. Whoever finds the fraction raises his hand and reads it, saying in which row and in which column it is located. The rest will have time to correct themselves, in case someone else finds a fraction that satisfies the condition. Each find is awarded a point for the team.
A table is posted or opened.

The conditions are given one by one. Find:

– fraction, more than 2.5, but less than 3;
– the smallest fraction, located in the range from 2 to 3;
– the largest fraction in the range from 1 to 2;
– a fraction in which one digit is repeated several times.

It should be taken into account that the first and fourth tasks have several answers, this must be played out. You can score more points for these tasks. The second and third tasks have only one answer. But he may not be found. Perhaps an answer will be offered that satisfies the conditions, but is inaccurate, and no one will be able to interrupt it. The point is added to the piggy bank by the one whose result remains the last. Finally, the team scores are calculated.

4. “Who is more accurate?”(4 + 3 min) (Slide 4)

Teacher: Our next competition will let you find out whose row is more accurate. Cards with equations are distributed. Everyone has their own card, their own equation. It needs to be solved not for speed, but for accuracy. The one who solves it faster will not receive any points. He will still wait for the others. But still, time is limited, 4-5 minutes are given to solve. After this, starting from the first, the answers to the equations will be read and checked. If the equation is solved correctly, then a point is added; if the answer is incorrect, then there will be no point.

Cards are distributed. The first card is the simplest, so it is given to weak students. On command, students begin solving. After 5 minutes a check is carried out. Each equation is for three participants from different rows. One reads the answer, the other speaks out loud, right or wrong, if wrong, offers his result. And the teacher checks the third and says which of the participants has the correct answer and which does not. To check, of course, you need to make a template. After checking all equations, scores are calculated. If no one could solve an equation, it must be analyzed on the board. If one, or maybe two students made a mistake, they come up after the lesson, or in the next lesson the equation is analyzed on the board.

5. “Who is taller?”(10 min) (Slide 5)

Teacher: Now it's time to find out who can jump higher. In order to jump as high as possible, you need to solve a task using ingenuity. In these examples, the occupied positions must be arranged in such a way that the equalities are true. There are 9 examples in total, 3 for each row. To jump high, you need to solve all three examples. Solving less means jumping lower. Everyone answers in turn: first the student from the first row, then from the second, and then from the third. No more than two attempts are given for each jump. This means that if two options are proposed, and neither is correct, then the height is not taken.

–Examples are written on the board in three columns:

Whoever is the first in the row to raise his hand answers. If you answered correctly, then the first height has been passed. The second row answers, then the third. If the answer is incorrect, then the height is not taken, one more attempt remains. You cannot return to the same example three times. If an example is not solved in class, it is written down to be solved at home. For all three examples, 5 points are given as the highest height. If one example is not solved, then 3 points are given. If only one example is solved, then 1 point is given. At the end, the results are summed up this type work and for everything together.

6. “Who is stronger?”(10 min) (Slide 6)

Teacher: Now it's time to find out who is stronger. In this, as at the last Olympiad, problem solving will help us, and it will take place in this way. The problems will be on decimal fractions. There are 5 tasks of varying difficulty per row. You will choose the complexity of the problem yourself. Each task is a stage. If someone from the group solves this problem, then the stage is considered completed.

Stages go from one to five. The first, second and third stages are worth three points each.

The fourth stage is worth 4 points, and the fifth stage is worth 5 points.

First, everyone is given cards with tasks. It is necessary to check that each task is solved by at least one person. After all the cards have been dealt, you have 7 minutes to solve. After this time, the answers are checked. After checking the answers of all rows, points are calculated.

1) Two types of sweets were placed in a vase. Find the mass of a mixture of candies if it contains 3.8 kg of candies of the first type, and 1.5 kg more candies of the second type.

2) Three vehicles carry 14.5 tons of cargo. On the first car it is 5.2t, and on the second it is 0.8t less than on the first. How many tons of cargo is on the third vehicle?

3) A load of 11.2 tons was distributed among two vehicles so that one of them carried 0.84 tons more than the other. How many tons of cargo were on each vehicle?

4) Two motorcyclists are moving in opposite directions. The speed of one of them is 22 km/h, and the other is 4 km/h more. What will be the distance between them in 0.25 hours if there is now 0.8 km between them?

5) Sewing a coat took 4 times more fabric than a skirt. How many meters of fabric did it take to sew a coat if the skirt took 2.55 meters less fabric than the coat?

7. “The most dexterous?”(4 min) (Slide 7)

Teacher: To find out who is the most dexterous, let's complete a task of ingenuity. There is a poster hanging on the board with a web on it connecting circles with decimals. The task is this: you need to connect numbers from one corner to the other using arithmetic signs so that from 0.1 you get 1. Whoever thought of such a combination raises his hand and shows his solution on the board. If the decision is correct, then the team earns 3 points.

8. Summing up(3 min) (Slad 8)

Count the points and praise the winning team. Give everyone good marks for activity and friendship. Praise the active guys in each row. Discuss with the children what they can already solve well and what needs to be strengthened. Give homework. Collect notebooks for checking. Equations and problems are checked in the notebooks, which can also be graded later. But the main thing is that the notebooks will show which equations and problems the children have coped with, and what type of tasks still need to be consolidated before the test. It will immediately be seen whether the children can cope with the design of equations and problems.

9. Homework: (Slide 8) page 138, “Infinite division” (for those who are interested).