Question

Giải phương trình sau:

d) \(\dfrac{x^2}{\left(x+1\right)^2}+\dfrac{x}{x+1}-2=0\)

Answer 1

ĐKXĐ: x<>-1

\(\dfrac{x^2}{\left(x+1\right)^2}+\dfrac{x}{x+1}-2=0\)

\(\Leftrightarrow\left(\dfrac{x}{x+1}\right)^2+\left(\dfrac{x}{x+1}\right)-2=0\)

=>\(\left(\dfrac{x}{x+1}\right)^2+2\left(\dfrac{x}{x+1}\right)-\dfrac{x}{x+1}-2=0\)

=>\(\dfrac{x}{x+1}\left(\dfrac{x}{x+1}+2\right)-\left(\dfrac{x}{x+1}+2\right)=0\)

=>\(\left(\dfrac{x}{x+1}+2\right)\left(\dfrac{x}{x+1}-1\right)=0\)

=>\(\dfrac{x+2x+2}{x+1}\cdot\dfrac{x-x-1}{x+1}=0\)

=>\(\dfrac{3x+2}{x+1}\cdot\dfrac{-1}{x+1}=0\)

=>3x+2=0

=>x=-2/3(nhận)