Question

Đã trả lời bởi chuyên gia

Answer 1

\[\begin{array}{l}
a){\left( {x - 1} \right)^3} + \left( {2 - x} \right)\left( {4 + 2x + {x^2}} \right) + 3x\left( {x + 2} \right) = 17{\rm{ }}\\
\Leftrightarrow {x^3} - 3{x^2} + 3x - 1 + 8 - {x^3} + 3{x^2} + 6x - 17 = 0\\
\Leftrightarrow 9x - 10 = 0\\
\Leftrightarrow 9x = 10\\
\Leftrightarrow x = \frac{{10}}{9}\\
Vay:x = \frac{{10}}{9}\\
b)\left( {x + 2} \right)\left( {{x^2} - 2 + 4} \right) - x\left( {{x^2} - 2} \right) = 15{\rm{ }}\\
\Leftrightarrow (x + 2)({x^2} + 2) - x({x^2} - 2) = 15\\
\Leftrightarrow {x^3} + 2x + 2{x^2} + 4 - {x^3} + 2x = 15\\
\Leftrightarrow 2{x^2} + 4x - 11 = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = \frac{{ - 2 + \sqrt {26} }}{2}\\
x = \frac{{ - 2 - \sqrt {26} }}{2}
\end{array} \right.\\
Vay:S = \{ \frac{{ - 2 + \sqrt {26} }}{2};\frac{{ - 2 - \sqrt {26} }}{2}\} \\
c){\left( {x - 3} \right)^3} - \left( {x - 3} \right)\left( {{x^2} + 3x + 9} \right) + 9{\left( {x + 1} \right)^2} = 15{\rm{ }}\\
\Leftrightarrow {x^3} - 9{x^2} + 27x - 27 - ({x^3} - 27) + 9({x^2} + 2x + 1) = 15\\
\Leftrightarrow {x^3} - 9{x^2} + 27x - 27 - {x^3} + 27 + 9{x^2} + 18x + 9 - 15 = 0\\
\Leftrightarrow 45x - 6 = 0\\
\Leftrightarrow 45x = 6\\
\Leftrightarrow x = \frac{2}{{15}}\\
vay:x = \frac{2}{{15}}\\
d)x\left( {x - 5} \right)\left( {x + 5} \right) - \left( {x + 2} \right)\left( {{x^2} - 2x + 4} \right) = 3\\
\Leftrightarrow x({x^2} - 25) - ({x^3} + {2^3}) = 3\\
\Leftrightarrow {x^3} - 25x - {x^3} - 8 = 3\\
\Leftrightarrow - 25x = 11\\
\Leftrightarrow x = \frac{{ - 11}}{{25}}\\
vay:x = \frac{{ - 11}}{{25}}
\end{array}\]

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Answer 2

\[\begin{array}{l}
a){\left( {x - 1} \right)^3} + \left( {2 - x} \right)\left( {4 + 2x + {x^2}} \right) + 3x\left( {x + 2} \right) = 17{\rm{ }}\\
\Leftrightarrow {x^3} - 3{x^2} + 3x - 1 + 8 - {x^3} + 3{x^2} + 6x - 17 = 0\\
\Leftrightarrow 9x - 10 = 0\\
\Leftrightarrow 9x = 10\\
\Leftrightarrow x = \frac{{10}}{9}\\
Vay:x = \frac{{10}}{9}\\
b)\left( {x + 2} \right)\left( {{x^2} - 2 + 4} \right) - x\left( {{x^2} - 2} \right) = 15{\rm{ }}\\
\Leftrightarrow (x + 2)({x^2} + 2) - x({x^2} - 2) = 15\\
\Leftrightarrow {x^3} + 2x + 2{x^2} + 4 - {x^3} + 2x = 15\\
\Leftrightarrow 2{x^2} + 4x - 11 = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = \frac{{ - 2 + \sqrt {26} }}{2}\\
x = \frac{{ - 2 - \sqrt {26} }}{2}
\end{array} \right.\\
Vay:S = \{ \frac{{ - 2 + \sqrt {26} }}{2};\frac{{ - 2 - \sqrt {26} }}{2}\} \\
c){\left( {x - 3} \right)^3} - \left( {x - 3} \right)\left( {{x^2} + 3x + 9} \right) + 9{\left( {x + 1} \right)^2} = 15{\rm{ }}\\
\Leftrightarrow {x^3} - 9{x^2} + 27x - 27 - ({x^3} - 27) + 9({x^2} + 2x + 1) = 15\\
\Leftrightarrow {x^3} - 9{x^2} + 27x - 27 - {x^3} + 27 + 9{x^2} + 18x + 9 - 15 = 0\\
\Leftrightarrow 45x - 6 = 0\\
\Leftrightarrow 45x = 6\\
\Leftrightarrow x = \frac{2}{{15}}\\
vay:x = \frac{2}{{15}}\\
d)x\left( {x - 5} \right)\left( {x + 5} \right) - \left( {x + 2} \right)\left( {{x^2} - 2x + 4} \right) = 3\\
\Leftrightarrow x({x^2} - 25) - ({x^3} + {2^3}) = 3\\
\Leftrightarrow {x^3} - 25x - {x^3} - 8 = 3\\
\Leftrightarrow - 25x = 11\\
\Leftrightarrow x = \frac{{ - 11}}{{25}}\\
vay:x = \frac{{ - 11}}{{25}}
\end{array}\]

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