Question

Đã trả lời bởi chuyên gia

Answer 1

\[\begin{array}{l}
1)\\
A)\;2x.{\left( {2x - 1} \right)^2} - 3x.\left( {x + 3} \right).\left( {x - 3} \right) - 4x.{\left( {x + 1} \right)^2}\\
= 2x\left( {4{x^2} - 4x + 1} \right) - 3x\left( {{x^2} - \;{\bf{9}}} \right) - 4x\left( {{x^2} + 2x + 1} \right)\\
= 8{x^3} - 8{x^2} + 2x - 3{x^3} + 27x - 4{x^3} - 8{x^2} - 4x\\
= {x^3} - 16{x^2} + 25x\\
2)\\
\begin{array}{*{20}{l}}
\begin{array}{l}
A){\rm{ }}\frac{{{{\left( {x + 1} \right)}^2}}}{3}{\rm{ }} - \frac{{{{\left( {x - 2} \right)}^2}}}{3} = \frac{{2x + 1}}{2} - \frac{{{{\left( {x - 3} \right)}^2}}}{6}\\
\Leftrightarrow 2{\left( {x + 1} \right)^2} - 2{\left( {x - 2} \right)^2} = 3\left( {2x + 1} \right) - {\left( {x - 3} \right)^2}\\
\Leftrightarrow 2\left( {{x^2} + 2x + 1} \right) - 2\left( {{x^2} - 4x + 4} \right) = 6x + 3 - \left( {{x^2} - 6x + {\bf{9}}} \right)\\
\Leftrightarrow 2{x^2} + 4x + 2 - 2{x^2} + 8x - 8 = 6x + 3 - {x^2} + 6x - {\bf{9}}\\
\Leftrightarrow 12x - 6 = - {x^2} + 12x - 6\\
\Leftrightarrow {x^2} = 0\\
\Leftrightarrow x = 0\\
Vay:x = 0
\end{array}\\
{B){\rm{ }}8{x^3} - 12{x^2} + 6x - 1 = 0}
\end{array}\\
\Leftrightarrow {\left( {2x} \right)^3} - 3.{\left( {2x} \right)^2}.1 + 3.2x{.1^2} - {1^3} = 0\\
\Leftrightarrow {\left( {2x - 1} \right)^3} = 0\\
\Leftrightarrow 2x - 1 = 0\\
\Leftrightarrow x = \frac{1}{2}\\
Vay:x = \frac{1}{2}
\end{array}\]

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Answer 2

\[\begin{array}{l}
1)\\
A)\;2x.{\left( {2x - 1} \right)^2} - 3x.\left( {x + 3} \right).\left( {x - 3} \right) - 4x.{\left( {x + 1} \right)^2}\\
= 2x\left( {4{x^2} - 4x + 1} \right) - 3x\left( {{x^2} - \;{\bf{9}}} \right) - 4x\left( {{x^2} + 2x + 1} \right)\\
= 8{x^3} - 8{x^2} + 2x - 3{x^3} + 27x - 4{x^3} - 8{x^2} - 4x\\
= {x^3} - 16{x^2} + 25x\\
2)\\
\begin{array}{*{20}{l}}
\begin{array}{l}
A){\rm{ }}\frac{{{{\left( {x + 1} \right)}^2}}}{3}{\rm{ }} - \frac{{{{\left( {x - 2} \right)}^2}}}{3} = \frac{{2x + 1}}{2} - \frac{{{{\left( {x - 3} \right)}^2}}}{6}\\
\Leftrightarrow 2{\left( {x + 1} \right)^2} - 2{\left( {x - 2} \right)^2} = 3\left( {2x + 1} \right) - {\left( {x - 3} \right)^2}\\
\Leftrightarrow 2\left( {{x^2} + 2x + 1} \right) - 2\left( {{x^2} - 4x + 4} \right) = 6x + 3 - \left( {{x^2} - 6x + {\bf{9}}} \right)\\
\Leftrightarrow 2{x^2} + 4x + 2 - 2{x^2} + 8x - 8 = 6x + 3 - {x^2} + 6x - {\bf{9}}\\
\Leftrightarrow 12x - 6 = - {x^2} + 12x - 6\\
\Leftrightarrow {x^2} = 0\\
\Leftrightarrow x = 0\\
Vay:x = 0
\end{array}\\
{B){\rm{ }}8{x^3} - 12{x^2} + 6x - 1 = 0}
\end{array}\\
\Leftrightarrow {\left( {2x} \right)^3} - 3.{\left( {2x} \right)^2}.1 + 3.2x{.1^2} - {1^3} = 0\\
\Leftrightarrow {\left( {2x - 1} \right)^3} = 0\\
\Leftrightarrow 2x - 1 = 0\\
\Leftrightarrow x = \frac{1}{2}\\
Vay:x = \frac{1}{2}
\end{array}\]

Answer 3

Tải quanda

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