Toán Lớp 8: tìm x
a/ 4(2x – 1) + (x + 2)2 = 0
b/ x2(3x – 2) + 15x = 10
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Toán Lớp 8: tìm x
a/ 4(2x – 1) + (x + 2)2 = 0
b/ x2(3x – 2) + 15x = 10
Comments ( 2 )
a)4\left( {2x–1} \right) + {\left( {x + 2} \right)^2} = 0\\
\Leftrightarrow 8x – 4 + {x^2} + 4x + 4 = 0\\
\Leftrightarrow 12x + {x^2} = 0\\
\Leftrightarrow x\left( {12 + x} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
x = 0\\
12 + x = 0
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
x = 0\\
x = – 12
\end{array} \right.\\
b){x^2}\left( {3x–2} \right) + 15x = 10\\
\Leftrightarrow {x^2}\left( {3x–2} \right) + 15x–10 = 0\\
\Leftrightarrow {x^2}\left( {3x–2} \right) + 5\left( {3x–2} \right) = 0\\
\Leftrightarrow \left( {{x^2}{\rm{ }} + {\rm{ }}5} \right)\left( {3x{\rm{ }}–{\rm{ }}2} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
{x^2} + 5 = 0\\
3x – 2 = 0
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
x = \sqrt { – 5} \\
x = \frac{2}{3}
\end{array} \right.
\end{array}\]