Toán Lớp 10: Giải các phương trình sau:
a. $x^{3}$ – 3x – 2 = 0
b. $x^{3}$ + 64 = 0
c. 3$x^{3}$ + 6$x^{2}$ – 4x = 0
d. $x^{3}$ – 2x – 4 = 0
e. $x^{3}$ – 125 = 0
f. 1,2$x^{3}$ – $x^{2}$ – 0,2x = 0
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Toán Lớp 10: Giải các phương trình sau:
a. $x^{3}$ – 3x – 2 = 0
b. $x^{3}$ + 64 = 0
c. 3$x^{3}$ + 6$x^{2}$ – 4x = 0
d. $x^{3}$ – 2x – 4 = 0
e. $x^{3}$ – 125 = 0
f. 1,2$x^{3}$ – $x^{2}$ – 0,2x = 0
Comments ( 1 )
a){x^3} – 3x – 2 = 0\\
\Leftrightarrow {x^3} – 2{x^2} + 2{x^2} – 4x + x – 2 = 0\\
\Leftrightarrow \left( {x – 2} \right)\left( {{x^2} + 2x + 1} \right) = 0\\
\Leftrightarrow \left( {x – 2} \right){\left( {x + 1} \right)^2} = 0\\
\Leftrightarrow x = 2;x = – 1\\
Vậy\,x = 2;x = – 1\\
b){x^3} + 64 = 0\\
\Leftrightarrow {x^3} = – 64\\
\Leftrightarrow {x^3} = {\left( { – 4} \right)^3}\\
\Leftrightarrow x = – 4\\
Vậy\,x = – 4\\
c)3{x^3} + 6{x^2} – 4x = 0\\
\Leftrightarrow x\left( {3{x^2} + 6x – 4} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
3{x^2} + 6x – 4 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = \dfrac{{ – 3 \pm \sqrt {21} }}{3}
\end{array} \right.\\
Vậy\,x = 0;x = \dfrac{{ – 3 \pm \sqrt {21} }}{3}\\
d){x^3} – 2x – 4 = 0\\
\Leftrightarrow {x^3} – 2{x^2} + 2{x^2} – 4x + 2x – 4 = 0\\
\Leftrightarrow \left( {x – 2} \right)\left( {{x^2} + 2x + 2} \right) = 0\\
\Leftrightarrow x = 2\\
Vậy\,x = 2\\
e){x^3} – 125 = 0\\
\Leftrightarrow {x^3} = 125\\
\Leftrightarrow x = 5\\
Vậy\,x = 5\\
f)1,2{x^3} – {x^2} – 0,2x = 0\\
\Leftrightarrow x.\left( {12{x^2} – 10x – 2} \right) = 0\\
\Leftrightarrow x\left( {6{x^2} – 5x – 1} \right) = 0\\
\Leftrightarrow x\left( {6x + 1} \right)\left( {x – 1} \right) = 0\\
\Leftrightarrow x = 0;x = – \dfrac{1}{6};x = 1\\
Vậy\,x = – \dfrac{1}{6};x = 0;x = 1
\end{array}$