Question

Giải phương trình \(\frac{1}{x^2}+\frac{3}{x+1}-\frac{2}{\left(x+1\right)^2}=2\)

Answer 1

@Nguyễn Việt Lâm

@Trần Thanh Phương

Answer 2

ĐKXĐ: ...

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=0\\\frac{1}{x^2}=x\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=-\frac{1}{2}\end{matrix}\right.\)

Answer 3

Đkxđ: x \(\ne\) -1.

pt <=>\(\frac{\left(x+1\right)^2}{x^2\left(x+1\right)^2}\) + \(\frac{3x^2\left(x+1\right)}{x^2\left(x+1\right)^2}\) \(-\) \(\frac{2x^2}{x^2\left(x+1\right)^2}\) = 2

<=> \(\frac{x^2+2x+1}{x^2\left(x+1\right)^2}\) + \(\frac{3x^3+3x^2}{x^2\left(x+1\right)^2}\) \(-\) \(\frac{2x^2}{x^2\left(x+1\right)^2}\) = 2

<=> \(\frac{3x^3+2x^2+2x+1}{x^2\left(x+1\right)^2}\) = 2

<=> \(3x^3+2x^2+2x+1=2x^2\left(x+1\right)^2\)

<=> \(3x^3+2x^2+2x+1=2x^2\left(x^2+2x+1\right)\)

<=> \(3x^3+2x^2+2x+1=2x^4+2x^3+2x^2\)

<=> \(2x^4-x^3-2x-1=0\)

<=> ....