Question

\(\dfrac{5}{x-3}\)- \(\dfrac{4}{x+3}\)= \(\dfrac{x-5}{x^2-9}\)

Answer 1

Đkxđ : x \(\ne-3\) , x\(\ne3\)

<=> 5(x+3)-4(x-3)=x-5

<=> 5x+15+4x-12=x-5

<=>5x+4x-x=-15+12-5

<=> 8x = -8

<=> x= -1 (nhận)

S=\(\left\{-1\right\}\)

Answer 2

\(\Leftrightarrow\dfrac{\left(5x+15\right)\left(4x-12\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{x-5}{x^2-9}\)

\(\Leftrightarrow\dfrac{\left(5x+15\right)\left(4x-12\right)}{x^2+3x-3x-9}=\dfrac{x-5}{x^2-9}\)

\(\Leftrightarrow\dfrac{\left(5x+15\right)\left(4x-12\right)}{x^2-9}=\dfrac{x-5}{x^2-9}\)

\(\Leftrightarrow\left(5x+15\right)\left(4x-12\right)=x-5\)