Đáp án:
\(a.x = \dfrac{9}{{11}}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.2x - 2 - 15 + 20x = 1\\
\to 22x = 18\\
\to x = \dfrac{9}{{11}}\\
b.{\left( {x + 2} \right)^2}\left( {x - 1} \right)\left( {3x - 1} \right) = 0\\
\to \left[ \begin{array}{l}
x + 2 = 0\\
x - 1 = 0\\
3x - 1 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = - 2\\
x = 1\\
x = \dfrac{1}{3}
\end{array} \right.\\
c.\left( {x - 3} \right)\left( {2x + 5} \right) = 0\\
\to \left[ \begin{array}{l}
x - 3 = 0\\
2x + 5 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 3\\
x = - \dfrac{5}{2}
\end{array} \right.\\
d.DK:x \ne \left\{ { - 1;0} \right\}\\
Pt \to \dfrac{{\left( {x + 3} \right)x + \left( {x - 2} \right)\left( {x + 1} \right)}}{{x\left( {x + 1} \right)}} = \dfrac{{2\left( {{x^2} + x} \right)}}{{x\left( {x + 1} \right)}}\\
\to {x^2} + 3x + {x^2} - x - 2 = 2{x^2} + 2x\\
\to 0x = 2\left( {vô lý} \right)
\end{array}\)
⇒ Phương trình vô nghiệm
