\(\dfrac{\left(x+1\right)^2}{x^2+2x+2}-\dfrac{x^2+2x}{\left(x+1\right)^2}=\dfrac{1}{90}\)
\(\Leftrightarrow\dfrac{\left(x+1\right)^2}{\left(x+1\right)^2+1}-\dfrac{\left(x+1\right)^2-1}{\left(x+1\right)^2}=\dfrac{1}{90}\)
Đặt \(\left(x+1\right)^2=a\left(a\ge0\right)\)thì ta có
\(\dfrac{a}{a+1}-\dfrac{a-1}{a}=\dfrac{1}{90}\)
\(\Leftrightarrow a^2+a-90=0\)
\(\Leftrightarrow\left(a-9\right)\left(a+10\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=9\\a=-10\left(l\right)\end{matrix}\right.\)
\(\Rightarrow a=9\)
\(\Leftrightarrow\left(x+1\right)^2=9\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=3\\x+1=-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\)
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