Question

Đã trả lời bởi chuyên gia

Answer 1

\[\begin{array}{l}
a)\sqrt {{x^2} - 4x + 4} = x - 3\\
\Leftrightarrow \sqrt {{{\left( {x - 2} \right)}^2}} = x - 3\\
\Leftrightarrow |x - 2| = x - 3\\
\Leftrightarrow \left[ \begin{array}{l}
x - 2 = x - 3\\
x - 2 = 3 - x
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
0x = - 1\\
2x = 5
\end{array} \right.\\
\Leftrightarrow x = \frac{5}{2}\\
vay:x = \frac{5}{2}\\
b)\sqrt {{x^2} + 8x + 16} = 5\\
\Leftrightarrow \sqrt {{x^2} + 2.x.4 + {4^2}} = 5\\
\Leftrightarrow \sqrt {{{\left( {x + 4} \right)}^2}} = 5\\
\Leftrightarrow |x + 4| = 5\\
\Leftrightarrow \left[ \begin{array}{l}
x + 4 = 5\\
x + 4 = - 5
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = - 9
\end{array} \right.\\
Vay:S = \{ 1; - 9\} \\
c)\sqrt {{x^2} + 4} = \sqrt {2x + 3} \\
dk:x \ge \frac{{ - 3}}{2}\\
\Leftrightarrow {x^2} + 4 = 2x + 3\\
\Leftrightarrow {x^2} - 2x + 1 = 0\\
\Leftrightarrow {\left( {x - 1} \right)^2} = 0\\
\Leftrightarrow x - 1 = 0\\
\Leftrightarrow x = 1\left( {tm} \right)\\
vay:x = 1\\
d)\frac{{\sqrt {2x - 3} }}{{\sqrt {x - 1} }} = 2\\
dk:x \ge \frac{3}{2}\\
\Leftrightarrow \sqrt {2x - 3} = 2\sqrt {x - 1} \\
\Leftrightarrow 2x - 3 = 4\left( {x - 1} \right)\\
\Leftrightarrow 2x - 3 = 4x - 4\\
\Leftrightarrow 2x = 1\\
\Leftrightarrow x = \frac{1}{2}\left( {ktm} \right)
\end{array}\]

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Answer 2

\[\begin{array}{l}
a)\sqrt {{x^2} - 4x + 4} = x - 3\\
\Leftrightarrow \sqrt {{{\left( {x - 2} \right)}^2}} = x - 3\\
\Leftrightarrow |x - 2| = x - 3\\
\Leftrightarrow \left[ \begin{array}{l}
x - 2 = x - 3\\
x - 2 = 3 - x
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
0x = - 1\\
2x = 5
\end{array} \right.\\
\Leftrightarrow x = \frac{5}{2}\\
vay:x = \frac{5}{2}\\
b)\sqrt {{x^2} + 8x + 16} = 5\\
\Leftrightarrow \sqrt {{x^2} + 2.x.4 + {4^2}} = 5\\
\Leftrightarrow \sqrt {{{\left( {x + 4} \right)}^2}} = 5\\
\Leftrightarrow |x + 4| = 5\\
\Leftrightarrow \left[ \begin{array}{l}
x + 4 = 5\\
x + 4 = - 5
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = - 9
\end{array} \right.\\
Vay:S = \{ 1; - 9\} \\
c)\sqrt {{x^2} + 4} = \sqrt {2x + 3} \\
dk:x \ge \frac{{ - 3}}{2}\\
\Leftrightarrow {x^2} + 4 = 2x + 3\\
\Leftrightarrow {x^2} - 2x + 1 = 0\\
\Leftrightarrow {\left( {x - 1} \right)^2} = 0\\
\Leftrightarrow x - 1 = 0\\
\Leftrightarrow x = 1\left( {tm} \right)\\
vay:x = 1\\
d)\frac{{\sqrt {2x - 3} }}{{\sqrt {x - 1} }} = 2\\
dk:x \ge \frac{3}{2}\\
\Leftrightarrow \sqrt {2x - 3} = 2\sqrt {x - 1} \\
\Leftrightarrow 2x - 3 = 4\left( {x - 1} \right)\\
\Leftrightarrow 2x - 3 = 4x - 4\\
\Leftrightarrow 2x = 1\\
\Leftrightarrow x = \frac{1}{2}\left( {ktm} \right)
\end{array}\]

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