nhanh nhé bn

Bài 2
$\displaystyle  \begin{array}{{>{\displaystyle}l}}
A=\left(\frac{2x}{x^{3} +x-x^{2} -1} -\frac{1}{x-1}\right) :\left( 1+\frac{x}{x^{2} +1}\right) \ \ ( \ x\neq 1)\\
A=\left(\frac{2x}{\left( x^{2} +1\right)( x-1)} -\frac{1}{x-1}\right) :\left(\frac{x^{2} +1}{x^{2} +1} +\frac{x}{x^{2} +1}\right)\\
A=\left(\frac{2x}{\left( x^{2} +1\right)( x-1)} -\frac{x^{2} +1}{\left( x^{2} +1\right)( x-1)}\right) :\frac{x^{2} +x+1}{x^{2} +1}\\
A=\frac{-x^{2} +2x-1}{\left( x^{2} +1\right)( x-1)} .\frac{x^{2} +1}{x^{2} +x+1}\\
A=\frac{-( x-1)^{2}}{( x-1)\left( x^{2} +x+1\right)}\\
A=\frac{1-x}{x^{2} +x+1}
\end{array}$
$\displaystyle b,\ A=\frac{2}{7}$
$\displaystyle  \begin{array}{{>{\displaystyle}l}}
\Leftrightarrow \frac{1-x}{x^{2} +x+1} =\frac{2}{7}\\
\Leftrightarrow 2x^{2} +2x+2=-7x+7\\
\Leftrightarrow 2x^{2} +9x-5=0\\
\Leftrightarrow ( 2x-1)( x+5) =0\\
\Leftrightarrow \left[ \begin{array}{l l}
x=\frac{1}{2} & \\
x=-5 & 
\end{array} \right.
\end{array}$
$\displaystyle c,\ B=\frac{A}{1-x} =\frac{\frac{1-x}{x^{2} +x+1}}{1-x} =\frac{1}{x^{2} +x+1} =\frac{1}{\left( x+\frac{1}{2}\right)^{2} +\frac{3}{4}}$
Nhận xét 
$\displaystyle \left( x+\frac{1}{2}\right)^{2} \geqslant 0\ \forall x$
Suy ra
$\displaystyle \left( x+\frac{1}{2}\right)^{2} +\frac{3}{4} \geqslant \frac{3}{4}$
Suy ra $\displaystyle \frac{1}{\left( x+\frac{1}{2}\right)^{2} +\frac{3}{4}} \leqslant \frac{4}{3}$