Toán Lớp 10: Giúp em với ạ:
Giải các pt sau:
a) | 3x + 2 | = | 2x + 1 |
b) | 2$x^{2}$ – x – 1 | = | 2x + 1 |
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Toán Lớp 10: Giúp em với ạ:
Giải các pt sau:
a) | 3x + 2 | = | 2x + 1 |
b) | 2$x^{2}$ – x – 1 | = | 2x + 1 |
Comments ( 1 )
a)\left| {3x + 2} \right| = \left| {2x + 1} \right|\\
\Leftrightarrow \left[ \begin{array}{l}
3x + 2 = 2x + 1\\
3x + 2 = – 2x – 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
3x – 2x = 1 – 2\\
3x + 2x = – 1 – 2
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = – 1\\
5x = – 3
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = – 1\\
x = – \dfrac{3}{5}
\end{array} \right.\\
Vậy\,x = – 1;x = – \dfrac{3}{5}\\
b)\left| {2{x^2} – x – 1} \right| = \left| {2x + 1} \right|\\
\Leftrightarrow \left[ \begin{array}{l}
2{x^2} – x – 1 = 2x + 1\\
2{x^2} – x – 1 = – 2x – 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
2{x^2} – 3x – 2 = 0\\
2{x^2} + x = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\left( {2x + 1} \right)\left( {x – 2} \right) = 0\\
x\left( {2x + 1} \right) = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = – \dfrac{1}{2}\\
x = 2\\
x = 0
\end{array} \right.\\
Vậy\,x = – \dfrac{1}{2};x = 0;x = 2
\end{array}$