Question

Giai pt

a/\(\dfrac{x+2}{x-2}-\dfrac{2}{x^2-2x}=\dfrac{1}{x}\)

b/\(\dfrac{1}{x-1}-\dfrac{3x^2}{x^3-1}=\dfrac{2x}{x^2+x+1}\)

Answer 1

a: \(\Leftrightarrow\dfrac{x\left(x+2\right)}{x\left(x-2\right)}-\dfrac{2}{x\left(x-2\right)}=\dfrac{x-2}{x\left(x-2\right)}\)

Suy ra: \(x^2+2x-2=x-2\)

\(\Leftrightarrow x^2+x=0\)

=>x(x+1)=0

=>x=0(loại) hoặc x=-1(nhận)

b: \(\Leftrightarrow x^2+x+1-3x^2=2x\left(x-1\right)\)

\(\Leftrightarrow-2x^2+x+1-2x^2+2x=0\)

\(\Leftrightarrow-4x^2+3x+1=0\)

\(\Leftrightarrow4x^2-3x-1=0\)

\(\Leftrightarrow4x^2-4x+x-1=0\)

\(\Leftrightarrow\left(x-1\right)\left(4x+1\right)=0\)

=>x=1(loại) hoặc x=-1/4(nhận)