Bài 100573

Question

Giải các phương trình :
$\begin{array}{l}
1)\,{\left[ {{{\left( {{2^{\sqrt x + 5}}} \right)}^{\frac{1}{{5\sqrt x + 1}}}}} \right]^{\frac{1}{{\sqrt x }}}} = \frac{1}{2}{.4^{\sqrt x }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(1)\\
2)\,\,{2^{{{\log }_8}\left( {{x^2} - 6x + 9} \right)}} = {3^{2{{\log }_x}\sqrt x - 1}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(2)
\end{array}$

score 0 · Answer 1

$1)$ ĐK $x > 0:\,(1)\, \Leftrightarrow {2^{\frac{{\sqrt x + 5}}{{\left( {5\sqrt x + 1} \right)\sqrt x }}}} = {2^{2\sqrt x - 1}} \Leftrightarrow \sqrt x + 5 = \left( {2\sqrt x - 1} \right)\left( {5x + \sqrt x } \right).$
Đặt $t = \sqrt x ,\,\,t \ge 0$ Ta có $\begin{array}{l}
\left( {2t - 1} \right)\left( {5{t^2} + 1} \right) = t + 5\\
\Leftrightarrow 10{t^3} - 3{t^2} - t = t + 5 \Leftrightarrow 10{t^3} - 3{t^2} - 2t - 5 = 0\\
\Leftrightarrow \left( {t - 1} \right)\left( {10{t^2} + 7t + 5} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
10{t^2} + 7t + 5 = 0\,\,\,\\
t - 1 = 0\,\,\,\,\,\, \Leftrightarrow t = 1
\end{array} \right.\\
t = 1 \Leftrightarrow \sqrt x = 1 \Leftrightarrow x = 1
\end{array}$
Phương trình $10{t^2} + 7t + 5 = 0$ vô nghiệm (vì $\Delta = 49 - 200 < 0$)
$2)$ ĐK: $x > 0,\,\,x \ne 1,\,x \ne 3$
$\begin{array}{l}
(2)\, \Leftrightarrow \,\,\,{2^{{{\log }_{{2^3}}}{{\left( {x - 3} \right)}^2}}} = {3^{2{{\log }_x}\sqrt x - 1}}\\
\Leftrightarrow {2^{\frac{2}{3}{{\log }_2}{{\left| {x - 3} \right|}^2}}} = {3^0} \Leftrightarrow {\log _2}\left| {x - 3} \right| = 0\\
\Leftrightarrow \left| {x - 3} \right| = {2^0} = 1 \Leftrightarrow x - 3 = \pm 1 \Leftrightarrow \left[ \begin{array}{l}
x = 4\\
x = 2
\end{array} \right.
\end{array}$

Bài 100573

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Bài 100573

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