Question

giải pt sau

 

\(\frac{6}{x^2-9}=_{1-\frac{1}{3-x}}\)

Answer 1

Ta có : \(\frac{6}{x^2-9}=1-\frac{1}{3-x}\) (đk : x khác 3;-3)_

\(\Leftrightarrow\frac{6}{x^2-9}+\frac{1}{3-x}-1=0\)

\(\Leftrightarrow\frac{6}{x^2-9}-\frac{1}{x-3}-1=0\)

\(\Leftrightarrow\frac{6}{x^2-9}-\frac{x+3}{x^2-9}-\frac{x^2-9}{x^2-9}=0\)

\(\Leftrightarrow\frac{6-x-3-x^2+9}{x^2-9}=0\)

\(\Leftrightarrow\frac{-x^2-x+12}{x^2-9}=0\)

\(\Leftrightarrow\frac{-x^2+3x-4x+12}{x^2-9}=0\)

\(\Leftrightarrow\frac{x\left(x-3\right)-4\left(x-3\right)}{x^2-9}=0\)

\(\Leftrightarrow\frac{\left(x-4\right)\left(x-3\right)}{x^2-9}=0\)

\(\Leftrightarrow\frac{x-4}{x+3}=0\)

<=> x - 4 = 0

<=> x = 4 (t/m)