a) 7 + 2x = 22 – 3x b) (2x – 1)2 + (2 – x)(2x – 1) = 0 c) x+5/6 - x-2/4 = 2x+1/ d) x-1/x+2 - x/x-2 = 5x-8/x^2-4

Đáp án:

\(\begin{array}{l}
a)x = 3\\
c)x = 5\\
b)\left[ \begin{array}{l}
x = \dfrac{1}{2}\\
x =  - 1
\end{array} \right.\\
d)x = 1
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)7 + 2x = 22 - 3x\\
 \to 5x = 15\\
 \to x = 3\\
c)\dfrac{{x + 5}}{6} - \dfrac{{x - 2}}{4} = \dfrac{{2x + 1}}{{12}}\\
 \to 2\left( {x + 5} \right) - 3\left( {x - 2} \right) = 2x + 1\\
 \to 2x + 10 - 3x + 6 = 2x + 1\\
 \to  - 3x =  - 15\\
 \to x = 5\\
b){\left( {2x - 1} \right)^2} + \left( {2 - x} \right)\left( {2x - 1} \right) = 0\\
 \to \left( {2x - 1} \right)\left( {2x - 1 + 2 - x} \right) = 0\\
 \to \left[ \begin{array}{l}
2x - 1 = 0\\
x + 1 = 0
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = \dfrac{1}{2}\\
x =  - 1
\end{array} \right.\\
d)DK:x \ne  \pm 2\\
\dfrac{{x - 1}}{{x + 2}} - \dfrac{x}{{x - 2}} = \dfrac{{5x - 8}}{{{x^2} - 4}}\\
 \to \dfrac{{\left( {x - 1} \right)\left( {x - 2} \right) - x\left( {x + 2} \right) - 5x + 8}}{{\left( {x - 2} \right)\left( {x + 2} \right)}} = 0\\
 \to {x^2} - 3x + 2 - {x^2} - 2x - 5x + 8 = 0\\
 \to  - 10x + 10 = 0\\
 \to x = 1
\end{array}\)