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\(\left(x+1\right)^2-3\left(x+1\right)=\left(x+1\right)\left(x+1-3\right)=\left(x+1\right)\left(x-2\right)\)
\(2x\left(x-2\right)-\left(x-2\right)^2=\left(x-2\right)\left[2x-\left(x-2\right)\right]=\left(x-2\right)\left(2x-x+2\right)=\left(x-2\right)\left(x+2\right)\)
\(4x^2-20xy+25y^2=\left(2x\right)^2-2.2x.5y+\left(5y\right)^2=\left(2x-5y\right)^2\)
\(x^2+3x-x-3=x\left(x+3\right)-\left(x+3\right)=\left(x-1\right)\left(x+3\right)\)
\(x^2-xy+x-y=x\left(x-y\right)+\left(x-y\right)=\left(x-y\right)\left(x+1\right)\)
\(2y\left(x+2\right)-3x-6=2y\left(x+2\right)-3\left(x+2\right)=\left(x+2\right)\left(2y-3\right)\)
ĐKXĐ:\(\left\{{}\begin{matrix}x-2\ne0\\x-5\ne0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ne2\\x\ne5\end{matrix}\right.\)
\(\frac{3x}{x-2}-\frac{x}{x-5}=\frac{3x}{\left(x-2\right)\left(5-x\right)}\)
\(\frac{3x\left(5-x\right)}{\left(x-2\right)\left(5-x\right)}+\frac{x\left(x-2\right)}{\left(x-2\right)\left(5-x\right)}-\frac{3x}{\left(x-2\right)\left(5-x\right)}=0\)
\(15x-3x^2+x^2-2x-3x=0\)
\(10x-2x^2=0\)
\(2x.\left(5-x\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}2x=0\\5-x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\left(chọn\right)\\x=5\left(loại\right)\end{matrix}\right.\)
\(\Leftrightarrow\left(2x+7\right)^2-\left(3x+6\right)^2=0\)
\(\Leftrightarrow\left(2x+7-3x-6\right)\left(2x+7+3x+6\right)=0\)
=>(1-x)(5x+13)=0
=>x=1 hoặc x=-13/5