Toán Lớp 6: 3 x (1-3x) 5 x (2+x) 2 x (2-5x) 5 x (1-x)
_________ – _____________ = __________ ____________
4 6 9 12
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Toán Lớp 6: 3 x (1-3x) 5 x (2+x) 2 x (2-5x) 5 x (1-x)
_________ – _____________ = __________ ____________
4 6 9 12
Comments ( 1 )
Giải đáp: $x = 0;x = – \dfrac{{17}}{{43}}$
Lời giải và giải thích chi tiết:
$\begin{array}{l}
\dfrac{{3x\left( {1 – 3x} \right)}}{4} – \dfrac{{5x\left( {2 + x} \right)}}{6} = \dfrac{{2x\left( {2 – 5x} \right)}}{9} – \dfrac{{5x\left( {1 – x} \right)}}{{12}}\\
\Leftrightarrow \dfrac{{9.\left( {3x – 9{x^2}} \right) – 6.\left( {10x + 5{x^2}} \right)}}{{36}}\\
= \dfrac{{4.\left( {4x – 10{x^2}} \right) – 3.\left( {5x – 5{x^2}} \right)}}{{36}}\\
\Leftrightarrow 27x – 81{x^2} – 60x – 30{x^2} = 16x – 40{x^2} – 15x + 15{x^2}\\
\Leftrightarrow – 111{x^2} – 33x = – 25{x^2} + x\\
\Leftrightarrow 86{x^2} + 34x = 0\\
\Leftrightarrow 2x\left( {43x + 17} \right) = 0\\
\Leftrightarrow x = 0;x = – \dfrac{{17}}{{43}}\\
Vậy\,x = 0;x = – \dfrac{{17}}{{43}}
\end{array}$