Question

tìm x sao cho giá trị của 2 biểu thức \(\dfrac{6x-1}{3x+2}\) và \(\dfrac{2x+5}{x-3}\) bằng nhau

Answer 1

ĐK: \(x\ne-\dfrac{2}{3};x\ne3\)

\(\dfrac{6x-1}{3x+2}=\dfrac{2x+5}{x-3}\Rightarrow\left(6x-1\right)\left(x-3\right)=\left(2x+5\right)\left(3x+2\right)\)

\(\Leftrightarrow6x^2-19x+3=6x^2+19x+10\Leftrightarrow38x=-7\Leftrightarrow x=-\dfrac{7}{38}\).

Answer 2

ĐKXĐ : x ≠ -2/3 ; x ≠ 3

\(\dfrac{6x-1}{3x+2}=\dfrac{2x+5}{x-3}\Rightarrow\left(6x-1\right)\left(x-3\right)=\left(3x+2\right)\left(2x+5\right)\)

\(\Leftrightarrow6x^2-19x+3=6x^2+19x+10\)

\(\Leftrightarrow-38x=7\Leftrightarrow x=-\dfrac{7}{38}\)(tm)

Vậy ...

Answer 3

ĐK: `x \ne -2/3 ; x \ne 3`

`(6x-1)/(3x+2)=(2x+5)/(x-3)`

`<=> (6x-1)(x-3)=(2x+5)(3x+2)`

`<=> 6x^2-19x+3=6x^2+19x+10`

`<=>-19x+3=19x+10`

`<=>-38x=7`

`<=>x=-7/38` (TM)