Question

a)
\(\dfrac{x-1}{x+2}-\dfrac{x}{x-2}=\dfrac{5x-2}{4-x^2}\)
b)
\(\dfrac{x-3}{x+2}+\dfrac{x-2}{x-4}=3\dfrac{1}{5}\)

Answer 1

a)dkxd:\(x\ne2;x\ne-2\)

\(\frac{x-1}{x+2}-\frac{x}{x-2}=\frac{5x-2}{4-x^2}\\ \Leftrightarrow\frac{x-1}{x+2}-\frac{x}{x-2}+\frac{5x-2}{x^2-4}=0\\ \Leftrightarrow\frac{\left(x-1\right)\left(x-2\right)-x\left(x+2\right)+5x-2}{\left(x-2\right)\left(x+2\right)}=0\\ \Leftrightarrow x^2-3x+2-x^2-2x+5x-2=0\\ \Leftrightarrow6x=0\Leftrightarrow x=0\left(tmdkxd\right)\)

Vậy phương trình có tập nghiệm là S={0}