c.
$\displaystyle \begin{array}{{>{\displaystyle}l}} c.\ |2x-5|=7-x\\ TH1:\ 2x-5=7-x\\ \Leftrightarrow 3x=12\\ \Leftrightarrow x=4\\ TH2:\ 2x-5=x-7\\ \Leftrightarrow x=-2 \end{array}$
Vậy x=4 hoặc x=-2
d.
$\displaystyle \begin{array}{{>{\displaystyle}l}} \frac{2x^{2} -1}{x^{3} +1} +\frac{1}{x+1} =2x\left( 1-\frac{x^{2} -x}{x^{2} -x+1}\right)\\ ĐK\ x\neq -1\\ \Leftrightarrow \frac{2x^{2} -1}{x^{3} +1} +\frac{x^{2} -x+1}{( x+1)\left( x^{2} -x+1\right)} =2x\left(\frac{x^{3} +1}{x^{3} +1} -\frac{\left( x^{2} -x\right)( x+1)}{\left( x^{2} -x+1\right)( x+1)}\right)\\ \Leftrightarrow 2x^{2} -1+x^{2} -x+1=2x\left[\left( x^{3} -1\right) -\left( x^{3} +x^{2} -x^{2} -x\right)\right]\\ \Leftrightarrow 3x^{2} -x=2x( x-1)\\ \Leftrightarrow x^{2} +x=0\\ \Leftrightarrow x( x+1) =0\\ \Leftrightarrow x=0( TM) \ or\ x=-1( loại) \end{array}$
Vậy x=0
