Question

[Nguyễn Đa Vít]

Giải phương trình:

\(\frac{1}{x^2-2x+2}+\frac{1}{x^2-2x+3}=\frac{9}{2\left(x^2-2x+4\right)}\)

Help me ai nhanh mình tick nha!!!!

Answer 1

[Bui Huyen]

\(x^2-2x+3=t\left(t\ge0\right)\)

\(pt\Leftrightarrow\frac{1}{t-1}+\frac{1}{t}=\frac{9}{2\left(t+1\right)}\)

\(\Leftrightarrow\frac{2t\left(t+1\right)}{2t\left(t^2-1\right)}+\frac{2\left(t^2-1\right)}{2t\left(t^2-1\right)}-\frac{9t\left(t-1\right)}{2t\left(t^2-1\right)}=0\)

\(\Leftrightarrow-5t^2+11t-2=0\)

\(\Leftrightarrow\left(5x-1\right)\left(x-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=2\end{cases}}\)