Question

[hiền phạm]

làm giúp mình bài 10 và bài 11

Answer 1

[Thanh Hoàng Thanh]

Bài 1:

a) 7x + 3 = 24                                      b) 7x + 21 = 0

<=> 7x = 21                                         <=> 7x = -21

<=> x = 3                                             <=> x = -3

Vậy x \(\in\left\{3\right\}\)                                           Vậy x \(\in\left\{-3\right\}\)

c) 5 - 2x = 7

<=> 2x = -2

<=> x = -1

Vậy x \(\in\left\{-1\right\}\)

Bài 2:

a) \(\dfrac{2}{3}x-\dfrac{3}{2}x=\dfrac{5}{12}\)

<=> \(\dfrac{-5}{6}x=\dfrac{5}{12}< =>x=\dfrac{-1}{2}\)

Vậy x \(\in\left\{\dfrac{-1}{2}\right\}\)

b) \(\dfrac{-2}{3}x+\dfrac{1}{5}=\dfrac{3}{10}< =>\dfrac{-2}{3}x=\dfrac{1}{10}\)

<=> \(x=\dfrac{-3}{20}\)

Vậy x \(\in\left\{\dfrac{-3}{20}\right\}\)

c) \(\dfrac{2}{3}x-\dfrac{1}{2}=\dfrac{1}{10}< =>\dfrac{2}{3}x=\dfrac{3}{5}< =>x=\dfrac{9}{10}\)

Vậy x \(\in\left\{\dfrac{9}{10}\right\}\)

d) \(5\dfrac{4}{7}:x=13< =>\dfrac{39}{7}:x=13\)

<=> x = \(\dfrac{3}{7}\)

Vậy x \(\in\left\{\dfrac{3}{7}\right\}\)