Question

Решите неравенство. 5(u+2)+14<6-u;4(u+3)<3(u+2);3/5 (5u-2/3)<u+7,6.срочно на сегодня надо

Answer 1

Solving the Inequalities

Let's solve the given inequalities step by step:

1. 5(u + 2) + 14 < 6 - u

To solve this inequality, we will simplify the expression and isolate the variable u.

Expanding the left side of the inequality: 5u + 10 + 14 < 6 - u

Combining like terms: 5u + 24 < 6 - u

Adding u to both sides: 5u + u + 24 < 6

Combining like terms: 6u + 24 < 6

Subtracting 24 from both sides: 6u < -18

Dividing both sides by 6: u < -3

Therefore, the solution to the first inequality is u < -3.

2. 4(u + 3) < 3(u + 2)

To solve this inequality, we will simplify the expression and isolate the variable u.

Expanding the left side of the inequality: 4u + 12 < 3u + 6

Subtracting 3u from both sides: 4u - 3u + 12 < 6

Combining like terms: u + 12 < 6

Subtracting 12 from both sides: u < -6

Therefore, the solution to the second inequality is u < -6.

3. 3/5 + (5u - 2/3) < u + 7

To solve this inequality, we will simplify the expression and isolate the variable u.

Combining the fractions on the left side: 3/5 + (15u/3 - 2/3) < u + 7

Combining the numerators of the fractions: 3/5 + (15u - 2)/3 < u + 7

To simplify further, we will find a common denominator for the fractions: Multiplying the first fraction by 3/3 and the second fraction by 5/5: (9/15) + (5(15u - 2)/15) < (3u + 21)/3

Combining the fractions: (9 + 5(15u - 2))/15 < (3u + 21)/3

Multiplying both sides by 15 to eliminate the denominators: 9 + 5(15u - 2) < 5(3u + 21)

Expanding and simplifying: 9 + 75u - 10 < 15u + 105

Combining like terms: 75u - 15u < 105 - 9 + 10

Simplifying further: 60u < 106

Dividing both sides by 60: u < 106/60

Simplifying the fraction: u < 53/30

Therefore, the solution to the third inequality is u < 53/30.

4. Summary of Solutions:

The solutions to the given inequalities are as follows: - u < -3 - u < -6 - u < 53/30

Please note that these solutions are based on the given inequalities and the steps taken to solve them.