d: ĐKXĐ: \(x\ne0\)
\(\left(x+1+\dfrac{1}{x}\right)^2=\left(x-1-\dfrac{1}{x}\right)^2\)
=>\(\left(x+1+\dfrac{1}{x}-x+1+\dfrac{1}{x}\right)\left(x+1+\dfrac{1}{x}+x-1-\dfrac{1}{x}\right)=0\)
=>\(2x\left(\dfrac{2}{x}+2\right)=0\)
=>\(\left[{}\begin{matrix}x=0\left(loại\right)\\\dfrac{2}{x}+2=0\end{matrix}\right.\Leftrightarrow\dfrac{2}{x}+2=0\)
=>\(\dfrac{2}{x}=-2\)
=>x=-1(nhận)
e: ĐKXĐ: \(x\ne1\)
\(\dfrac{1}{x-1}-\dfrac{3x^2}{x^3-1}=\dfrac{2x}{x^2+x+1}\)
=>\(\dfrac{x^2+x+1-3x^2}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
=>\(-2x^2+x+1=2x^2-2x\)
=>\(-4x^2+3x+1=0\)
=>\(4x^2-3x-1=0\)
=>\(4x^2-4x+x-1=0\)
=>(x-1)(4x+1)=0
=>\(\left[{}\begin{matrix}x=1\left(loại\right)\\x=-\dfrac{1}{4}\left(nhận\right)\end{matrix}\right.\)
f: ĐKXĐ: \(x\notin\left\{3;-3;-\dfrac{7}{2}\right\}\)
\(\dfrac{13}{\left(x-3\right)\left(2x+7\right)}+\dfrac{1}{2x+7}=\dfrac{6}{\left(x-3\right)\left(x+3\right)}\)
=>\(\dfrac{13\left(x+3\right)}{\left(x-3\right)\left(x+3\right)\left(2x+7\right)}+\dfrac{\left(x-3\right)\left(x+3\right)}{\left(2x+7\right)\left(x-3\right)\left(x+3\right)}=\dfrac{6\left(2x+7\right)}{\left(x-3\right)\left(x+3\right)\left(2x+7\right)}\)
=>\(13\left(x+3\right)+\left(x-3\right)\left(x+3\right)=6\left(2x+7\right)\)
=>\(x^2-9+13x+39-12x-42=0\)
=>\(x^2+x-12=0\)
=>(x+4)(x-3)=0
=>\(\left[{}\begin{matrix}x=-4\left(nhận\right)\\x=3\left(loại\right)\end{matrix}\right.\)
