Question

tìm x biết 

\(\dfrac{x-2}{x-3}\)=\(\dfrac{x+4}{x-2}\) 

(với x ≠ 2 , x ≠ 3

Answer 1

\(\dfrac{x-2}{x-3}=\dfrac{x+4}{x-2}\left(x\ne3;x\ne2\right)\)

suy ra

`(x-2)^2 =(x+4)(x-3)`

`<=> x^2 -4x+4=x^2 -3x+4x-12`

`<=> x^2 -x^2 -4x+3x-4x+4+12=0`

`<=> -5x +16=0`

`<=> -5x=-16`

`<=> x=16/5(tm)`

Answer 2

\(\dfrac{x-2}{x-3}=\dfrac{x+4}{x-2}\\ \left(x-2\right)\left(x-2\right)=\left(x-3\right)\left(x+4\right)\\ x^2-2x-2x+4=x^2-3x+4x-12\\ -4x+4=x-12\\ -4x-x=-4-12\\ -5x=-16\\ x=\dfrac{16}{5}\left(t/m\right)\)