Toán Lớp 8: (4x + 3)(x – 3)(x + 3) – (x + 3)(4x – 3)(4x + 3)=0
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Toán Lớp 8: (4x + 3)(x – 3)(x + 3) – (x + 3)(4x – 3)(4x + 3)=0
Comments ( 1 )
Giải đáp: $\,x \in \left\{ { – 3; – \dfrac{3}{4};0} \right\}$
Lời giải và giải thích chi tiết:
$\begin{array}{l}
\left( {4x + 3} \right).\left( {x – 3} \right)\left( {x + 3} \right) – \left( {x + 3} \right)\left( {4x – 3} \right).\left( {4x + 3} \right) = 0\\
\Leftrightarrow \left( {4x + 3} \right).\left( {x + 3} \right).\left( {x – 3 – \left( {4x – 3} \right)} \right) = 0\\
\Leftrightarrow \left( {4x + 3} \right).\left( {x + 3} \right).\left( {x – 3 – 4x + 3} \right) = 0\\
\Leftrightarrow \left( {4x + 3} \right).\left( {x + 3} \right).\left( { – 3x} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
4x + 3 = 0\\
x + 3 = 0\\
– 3x = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = – \dfrac{3}{4}\\
x = – 3\\
x = 0
\end{array} \right.\\
Vậy\,x \in \left\{ { – 3; – \dfrac{3}{4};0} \right\}
\end{array}$