Toán Lớp 8: Tìm x:
a) 8×3 – 10x = 0
b) x2 + 16 = 10x.
c) x3 + 27 = (x + 3)(9 – x)
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Toán Lớp 8: Tìm x:
a) 8×3 – 10x = 0
b) x2 + 16 = 10x.
c) x3 + 27 = (x + 3)(9 – x)
Comments ( 1 )
a)\left[ \begin{array}{l}
x = 0\\
x = \dfrac{{\sqrt 5 }}{2}\\
x = – \dfrac{{\sqrt 5 }}{2}
\end{array} \right.\\
b)\left[ \begin{array}{l}
x = 8\\
x = 2
\end{array} \right.\\
c)\left[ \begin{array}{l}
x = 0\\
x = 2\\
x = – 3
\end{array} \right.
\end{array}\)
a)8{x^3} – 10x = 0\\
\to 2x\left( {4{x^2} – 5} \right) = 0\\
\to 2x\left( {2x – \sqrt 5 } \right)\left( {2x + \sqrt 5 } \right) = 0\\
\to \left[ \begin{array}{l}
x = 0\\
x = \dfrac{{\sqrt 5 }}{2}\\
x = – \dfrac{{\sqrt 5 }}{2}
\end{array} \right.\\
b){x^2} + 16 = 10x\\
\to {x^2} – 10x = – 16\\
\to {x^2} – 10x + 25 = 9\\
\to {\left( {x – 5} \right)^2} = 9\\
\to \left| {x – 5} \right| = 3\\
\to \left[ \begin{array}{l}
x – 5 = 3\\
x – 5 = – 3
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 8\\
x = 2
\end{array} \right.\\
c){x^3} + 27 = \left( {x + 3} \right)\left( {9 – x} \right)\\
\to \left( {x + 3} \right)\left( {{x^2} – 3x + 9} \right) – \left( {x + 3} \right)\left( {9 – x} \right) = 0\\
\to \left( {x + 3} \right)\left( {{x^2} – 3x + 9 – 9 + x} \right) = 0\\
\to \left( {x + 3} \right)\left( {{x^2} – 2x} \right) = 0\\
\to x\left( {x – 2} \right)\left( {x + 3} \right) = 0\\
\to \left[ \begin{array}{l}
x = 0\\
x = 2\\
x = – 3
\end{array} \right.
\end{array}\)