Question

Tìm x biết:

\(\dfrac{x-5}{x+3}=\dfrac{x-4}{x+2}\)

Answer 1

Ta có: (x-5)(x+2) = (x-4)(x+3)

=> \(x^2+2x-5x-10=x^2+3x-4x-12\)

=> \(x^2-x^2+2x-5x-3x+4x=-12+10\)

=> -2x = -2

=>x = 1

Answer 2

\(\dfrac{x-5}{x+3}=\dfrac{x-4}{x+2}\Rightarrow\left(x-5\right)\left(x+2\right)=\left(x-4\right)\left(x+3\right)\)

\(\Rightarrow x^2+2x-5x-10=x^2+3x-4x-12\)

\(\Rightarrow x^2-3x-10=x^2-x-12\Rightarrow x^2-3x-10-x^2+x+12=0\)

\(\Rightarrow-2x+2=0\Rightarrow-2x=-2\Rightarrow x=1\)

Vậy x=1