Question

Đã trả lời bởi chuyên gia

Answer 1

\[\begin{array}{l}
4)5 - \sqrt {{{\left( {x + 1} \right)}^2}} = 2x + 5\\
\Leftrightarrow 5 - |x + 1| = 2x + 5\\
= > |x + 1| = 5 - 2x - 5\\
\Leftrightarrow |x + 1| = - 2x\\
\Leftrightarrow \left[ \begin{array}{l}
x + 1 = - 2x\\
x + 1 = 2x
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x + 2x = - 1\\
2x - x = 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
3x = - 1\\
x = 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \frac{{ - 1}}{3}\\
x = 1
\end{array} \right.\\
vay:S = \{ \frac{{ - 1}}{3};1\} \\
13)\sqrt {\frac{{x + 1}}{{2x - 1}}} = 2\\
dk:\frac{{x + 1}}{{2x - 1}} \ge 0 \Leftrightarrow \left[ {\begin{array}{*{20}{c}}
{\left\{ \begin{array}{l}
x + 1 \ge 0\\
2x - 1 > 0
\end{array} \right.}\\
{\left\{ \begin{array}{l}
x + 1 \le 0\\
2x - 1 < 0
\end{array} \right.}
\end{array}} \right. \Leftrightarrow \left[ {\begin{array}{*{20}{c}}
{\left\{ \begin{array}{l}
x \ge - 1\\
x > \frac{1}{2}
\end{array} \right.}\\
{\left\{ \begin{array}{l}
x \le - 1\\
x < \frac{1}{2}
\end{array} \right.}
\end{array}} \right. \Leftrightarrow \left[ \begin{array}{l}
x > \frac{1}{2}\\
x \le - 1
\end{array} \right.\\
\Leftrightarrow \frac{{x + 1}}{{2x - 1}} = {2^2}\\
\Leftrightarrow \frac{{x + 1}}{{2x - 1}} = 4\\
\Leftrightarrow x + 1 = 4\left( {2x - 1} \right)\\
\Leftrightarrow x + 1 = 8x - 4\\
\Leftrightarrow 8x - x = 1 + 4\\
\Leftrightarrow 7x = 5\\
\Leftrightarrow x = \frac{5}{7}\left( {tm} \right)\\
Vay:x = \frac{5}{7}
\end{array}\]

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Answer 2

\[\begin{array}{l}
4)5 - \sqrt {{{\left( {x + 1} \right)}^2}} = 2x + 5\\
\Leftrightarrow 5 - |x + 1| = 2x + 5\\
= > |x + 1| = 5 - 2x - 5\\
\Leftrightarrow |x + 1| = - 2x\\
\Leftrightarrow \left[ \begin{array}{l}
x + 1 = - 2x\\
x + 1 = 2x
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x + 2x = - 1\\
2x - x = 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
3x = - 1\\
x = 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \frac{{ - 1}}{3}\\
x = 1
\end{array} \right.\\
vay:S = \{ \frac{{ - 1}}{3};1\} \\
13)\sqrt {\frac{{x + 1}}{{2x - 1}}} = 2\\
dk:\frac{{x + 1}}{{2x - 1}} \ge 0 \Leftrightarrow \left[ {\begin{array}{*{20}{c}}
{\left\{ \begin{array}{l}
x + 1 \ge 0\\
2x - 1 > 0
\end{array} \right.}\\
{\left\{ \begin{array}{l}
x + 1 \le 0\\
2x - 1 < 0
\end{array} \right.}
\end{array}} \right. \Leftrightarrow \left[ {\begin{array}{*{20}{c}}
{\left\{ \begin{array}{l}
x \ge - 1\\
x > \frac{1}{2}
\end{array} \right.}\\
{\left\{ \begin{array}{l}
x \le - 1\\
x < \frac{1}{2}
\end{array} \right.}
\end{array}} \right. \Leftrightarrow \left[ \begin{array}{l}
x > \frac{1}{2}\\
x \le - 1
\end{array} \right.\\
\Leftrightarrow \frac{{x + 1}}{{2x - 1}} = {2^2}\\
\Leftrightarrow \frac{{x + 1}}{{2x - 1}} = 4\\
\Leftrightarrow x + 1 = 4\left( {2x - 1} \right)\\
\Leftrightarrow x + 1 = 8x - 4\\
\Leftrightarrow 8x - x = 1 + 4\\
\Leftrightarrow 7x = 5\\
\Leftrightarrow x = \frac{5}{7}\left( {tm} \right)\\
Vay:x = \frac{5}{7}
\end{array}\]

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