Question

Đã trả lời bởi chuyên gia

Answer 1

\[\begin{array}{l}
a)6{x^2} - (2x - 3)(3x + 2) = 1\\
\Leftrightarrow 6{x^2} - \left( {6{x^2} + 4x - 9x - 6} \right) = 1\\
\Leftrightarrow 6{x^2} - 6{x^2} + 5x + 6 = 1\\
\Leftrightarrow 5x = - 5\\
\Leftrightarrow x = - 1\\
vay:x = - 1\\
b){(x + 1)^3} - (x - 1)({x^2} + x + 1) - 2 = 0\\
\Leftrightarrow {x^3} + 3{x^2} + 3x + 1 - \left( {{x^3} - 1} \right) - 2 = 0\\
\Leftrightarrow {x^3} + 3{x^2} + 3x + 1 - {x^3} + 1 - 2 = 0\\
\Leftrightarrow 3{x^2} + 3x = 0\\
\Leftrightarrow 3x\left( {x + 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
3x = 0\\
x + 1 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = - 1
\end{array} \right.\\
vay:S = \{ 0; - 1\} \\
c)x(x + 4) - {x^2} - 6x = 10\\
\Leftrightarrow {x^2} + 4x - {x^2} - 6x = 10\\
\Leftrightarrow - 2x = 10\\
\Leftrightarrow x = - 5\\
vay:x = - 5\\
d){(3x - 1)^2} - {(x + 5)^2} = 0\\
\Leftrightarrow \left[ {\left( {3x - 1} \right) - \left( {x + 5} \right)} \right].\left[ {\left( {3x - 1} \right) + \left( {x + 5} \right)} \right] = 0\\
\Leftrightarrow \left( {3x - 1 - x - 5} \right)\left( {3x - 1 + x + 5} \right) = 0\\
\Leftrightarrow \left( {2x - 6} \right)\left( {4x + 4} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
2x - 6 = 0\\
4x + 4 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 3\\
x = - 1
\end{array} \right.\\
vay:S = \{ 3; - 1\}
\end{array}\]

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Answer 2

\[\begin{array}{l}
a)6{x^2} - (2x - 3)(3x + 2) = 1\\
\Leftrightarrow 6{x^2} - \left( {6{x^2} + 4x - 9x - 6} \right) = 1\\
\Leftrightarrow 6{x^2} - 6{x^2} + 5x + 6 = 1\\
\Leftrightarrow 5x = - 5\\
\Leftrightarrow x = - 1\\
vay:x = - 1\\
b){(x + 1)^3} - (x - 1)({x^2} + x + 1) - 2 = 0\\
\Leftrightarrow {x^3} + 3{x^2} + 3x + 1 - \left( {{x^3} - 1} \right) - 2 = 0\\
\Leftrightarrow {x^3} + 3{x^2} + 3x + 1 - {x^3} + 1 - 2 = 0\\
\Leftrightarrow 3{x^2} + 3x = 0\\
\Leftrightarrow 3x\left( {x + 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
3x = 0\\
x + 1 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = - 1
\end{array} \right.\\
vay:S = \{ 0; - 1\} \\
c)x(x + 4) - {x^2} - 6x = 10\\
\Leftrightarrow {x^2} + 4x - {x^2} - 6x = 10\\
\Leftrightarrow - 2x = 10\\
\Leftrightarrow x = - 5\\
vay:x = - 5\\
d){(3x - 1)^2} - {(x + 5)^2} = 0\\
\Leftrightarrow \left[ {\left( {3x - 1} \right) - \left( {x + 5} \right)} \right].\left[ {\left( {3x - 1} \right) + \left( {x + 5} \right)} \right] = 0\\
\Leftrightarrow \left( {3x - 1 - x - 5} \right)\left( {3x - 1 + x + 5} \right) = 0\\
\Leftrightarrow \left( {2x - 6} \right)\left( {4x + 4} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
2x - 6 = 0\\
4x + 4 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 3\\
x = - 1
\end{array} \right.\\
vay:S = \{ 3; - 1\}
\end{array}\]

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