Toán Lớp 8: h. (4x -3)2 – 3x(3 – 4x) = 0
i. 3x (x – 4) – x(5+3x) = -34
j. x3 – 9x – 5×2 + 45 = 0
Giúp nhanh vs ak đag cần gấp
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Toán Lớp 8: h. (4x -3)2 – 3x(3 – 4x) = 0
i. 3x (x – 4) – x(5+3x) = -34
j. x3 – 9x – 5×2 + 45 = 0
Giúp nhanh vs ak đag cần gấp
Comments ( 1 )
h)\left( {4x{\rm{ }} – 3} \right)2{\rm{ }}–{\rm{ }}3x\left( {3{\rm{ }}–{\rm{ }}4x} \right){\rm{ }} = {\rm{ }}0\\
\Leftrightarrow \left( {4x{\rm{ }} – 3} \right)2{\rm{ }} + 3x\left( {4x – 3} \right) = 0\\
\Leftrightarrow \left( {4x – 3} \right)\left( {2 + 3x} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
4x – 3 = 0\\
2 + 3x = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \frac{3}{4}\\
x = \frac{{ – 2}}{3}
\end{array} \right.\\
i)3x{\rm{ }}\left( {x{\rm{ }}–{\rm{ }}4} \right){\rm{ }}–{\rm{ }}x\left( {5 + 3x} \right){\rm{ }} = {\rm{ }} – 34\\
\Leftrightarrow 3{x^2} – 12x – 5x – 3{x^2} = 34\\
\Leftrightarrow – 17x = 34\\
\Leftrightarrow – x = 2\\
\Leftrightarrow x = – 2\\
j){x^3}{\rm{ }}–{\rm{ }}9x{\rm{ }}–{\rm{ }}5{x^2}{\rm{ }} + {\rm{ }}45{\rm{ }} = {\rm{ }}0\\
\Leftrightarrow {x^2}\left( {x + 5} \right) + 9\left( {x + 5} \right) = 0\\
\Leftrightarrow \left( {{x^2} + 9} \right)\left( {x + 5} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
{x^2} + 9 = 0\\
x + 5 = 0
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
x = \emptyset \\
x = – 5
\end{array} \right.
\end{array}\]