Question

Tìm x

a)\(\dfrac{6}{x-3}=\dfrac{9}{2x-7}\)

b)\(\dfrac{-7}{x+1}=\dfrac{6}{x+27}\)

Answer 1

a, \(\dfrac{6}{x-3}=\dfrac{9}{2x-7}\)

=> 6(2x-7) = 9(x-3)

=> 12x - 42 = 9x - 27

=> 12x - 9x = -27 + 42

=> 3x = 15 

=> x = 5

Vậy x = 5

b, \(\dfrac{-7}{x+1}=\dfrac{6}{x+27}\)

=> -7(x + 27) = 6(x + 1)

=> -7x - 189 = 6x + 6

=> -7x - 6x = 6 + 189

=> -13x = 195

=> x = -15

Vậy x = -15

Answer 2

a) Ta có: \(\dfrac{6}{x-3}=\dfrac{9}{2x-7}\)

\(\Leftrightarrow6\left(2x-7\right)=9\left(x-3\right)\)

\(\Leftrightarrow12x-42=9x-27\)

\(\Leftrightarrow12x-9x=-27+42\)

\(\Leftrightarrow3x=15\)

hay x=5

Vậy: x=5

b) Ta có: \(\dfrac{-7}{x+1}=\dfrac{6}{x+27}\)

\(\Leftrightarrow6\left(x+1\right)=-7\left(x+27\right)\)

\(\Leftrightarrow6x+6=-7x+189\)

\(\Leftrightarrow6x+7x=189-6\)

\(\Leftrightarrow13x=183\)

hay \(x=\dfrac{183}{13}\)

Vậy: \(x=\dfrac{183}{13}\)