Question

Đã trả lời bởi chuyên gia

Answer 1

\[\begin{array}{l}
\sqrt {{x^4} - 8{x^2} + 16} = 2 - x\\
dk:2 - x \ge 0 \Leftrightarrow x \le 2\\
\Leftrightarrow \sqrt {{x^4} - 2.{x^2} + {4^2}} = 2 - x\\
\Leftrightarrow \sqrt {{{\left( {{x^2} - 4} \right)}^2}} = 2 - x\\
\Leftrightarrow |{x^2} - 4| = 2 - x\\
\Leftrightarrow \left[ \begin{array}{l}
{x^2} - 4 = 2 - x\\
{x^2} - 4 = x - 2
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
{x^2} + x - 6 = 0\\
{x^2} - x - 2 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
{x^2} + 3x - 2x - 6 = 0\\
{x^2} + x - 2x - 2 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x\left( {x + 3} \right) - 2\left( {x + 3} \right) = 0\\
x\left( {x + 1} \right) - 2\left( {x + 1} \right) = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\left( {x + 3} \right)\left( {x - 2} \right) = 0\\
\left( {x + 1} \right)\left( {x - 2} \right) = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x + 3 = 0\\
x - 2 = 0\\
x + 1 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = - 3\\
x = 2\\
x = - 1
\end{array} \right.\left( {tm} \right)\\
Vay:S = \{ - 3;2; - 1\} \\
\sqrt {{x^2} + 2x + 1} = \sqrt {x + 1} \\
dk:x + 1 \ge 0 \Leftrightarrow x \ge - 1\\
\Leftrightarrow {x^2} + 2x + 1 = x + 1\\
\Leftrightarrow {x^2} + x = 0\\
\Leftrightarrow x\left( {x + 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x + 1 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = - 1
\end{array} \right.\left( {tm} \right)\\
vay:S = \{ 0; - 1\}
\end{array}\]

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Answer 2

\[\begin{array}{l}
\sqrt {{x^4} - 8{x^2} + 16} = 2 - x\\
dk:2 - x \ge 0 \Leftrightarrow x \le 2\\
\Leftrightarrow \sqrt {{x^4} - 2.{x^2} + {4^2}} = 2 - x\\
\Leftrightarrow \sqrt {{{\left( {{x^2} - 4} \right)}^2}} = 2 - x\\
\Leftrightarrow |{x^2} - 4| = 2 - x\\
\Leftrightarrow \left[ \begin{array}{l}
{x^2} - 4 = 2 - x\\
{x^2} - 4 = x - 2
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
{x^2} + x - 6 = 0\\
{x^2} - x - 2 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
{x^2} + 3x - 2x - 6 = 0\\
{x^2} + x - 2x - 2 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x\left( {x + 3} \right) - 2\left( {x + 3} \right) = 0\\
x\left( {x + 1} \right) - 2\left( {x + 1} \right) = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\left( {x + 3} \right)\left( {x - 2} \right) = 0\\
\left( {x + 1} \right)\left( {x - 2} \right) = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x + 3 = 0\\
x - 2 = 0\\
x + 1 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = - 3\\
x = 2\\
x = - 1
\end{array} \right.\left( {tm} \right)\\
Vay:S = \{ - 3;2; - 1\} \\
\sqrt {{x^2} + 2x + 1} = \sqrt {x + 1} \\
dk:x + 1 \ge 0 \Leftrightarrow x \ge - 1\\
\Leftrightarrow {x^2} + 2x + 1 = x + 1\\
\Leftrightarrow {x^2} + x = 0\\
\Leftrightarrow x\left( {x + 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x + 1 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = - 1
\end{array} \right.\left( {tm} \right)\\
vay:S = \{ 0; - 1\}
\end{array}\]

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