a,\(\left(x+1\right)\left(x-5\right)-x\left(x-6\right)=3x+7\)
\(\Leftrightarrow x^2-5x+x-5-x^2+6x-3x-7=0\)
\(\Leftrightarrow x-12=0\)
\(\Leftrightarrow x=12\)
b,\(\dfrac{x-2}{x+3}-\dfrac{2x-1}{x}=\dfrac{11-2x^2}{x^2+3x}\)
ĐKXĐ:\(x^2+3x\ne0\Leftrightarrow x\left(x+3\right)\ne0\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\x+3\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\x\ne-3\end{matrix}\right.\)
Ta có : MTC : \(x^2+3x\)
\(\Leftrightarrow x\left(x-2\right)-\left(x+3\right)\left(2x-1\right)=11-2x^2\)
\(\Leftrightarrow x^2-2x-2x^2+x-6x+3-11+2x^2=0\)
\(\Leftrightarrow x^2-7x-8=0\)
\(\Leftrightarrow x^2+x-8x-8=0\)
\(\Leftrightarrow x\left(x+1\right)-8\left(x+1\right)=0\)
\(\Leftrightarrow\left(x-8\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+1=0\\x-8=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\x=8\end{matrix}\right.\)(Thỏa mãn điều kiện xác định)