Đáp án:
$\begin{array}{l}
a)\left( {x - 2} \right)\left( {3 + x} \right) - {\left( {x + 1} \right)^2} = 2 - 4x\\
\Leftrightarrow 3x + {x^2} - 6 - 2x\\
- \left( {{x^2} + 2x + 1} \right) - 2 + 4x = 0\\
\Leftrightarrow {x^2} + x - 6 - {x^2} - 2x - 1 - 2 + 4x = 0\\
\Leftrightarrow 3x - 9 = 0\\
\Leftrightarrow x = 3\\
Vậy\,x = 3\\
b)Dkxd:x\# 3\\
\dfrac{{{x^2} - 3x}}{{x - 3}} + 2x = 7\\
\Leftrightarrow \dfrac{{x\left( {x - 3} \right)}}{{x - 3}} + 2x = 7\\
\Leftrightarrow x + 2x = 7\\
\Leftrightarrow 3x = 7\\
\Leftrightarrow x = \dfrac{7}{3}\\
Vậy\,x = \dfrac{7}{3}\\
c)\dfrac{{x + 2}}{2} = \dfrac{{2x - 1}}{3} + 1\\
\Leftrightarrow \dfrac{{3\left( {x + 2} \right) - 2\left( {2x - 1} \right) - 6}}{6} = 0\\
\Leftrightarrow 3x + 6 - 4x + 2 - 6 = 0\\
\Leftrightarrow x = 2\\
Vậy\,x = 2\\
d){x^2} + {\left( {\dfrac{x}{{x + 1}}} \right)^2} = \dfrac{5}{4}\\
\Leftrightarrow \dfrac{{{x^2}{{\left( {x + 1} \right)}^2} + {x^2}}}{{{{\left( {x + 1} \right)}^2}}} = \dfrac{5}{4}\\
\Leftrightarrow 5\left( {{x^2} + 2x + 1} \right) = 4\left( {{x^4} + 2{x^3} + {x^2} + {x^2}} \right)\\
\Leftrightarrow 4{x^4} + 8{x^3} + 3{x^2} - 10x - 5 = 0
\end{array}$
(Em kiểm tra lại đề bài câu cuối)
