mình chưa biết cách làm mn jups vs

Đáp án:

\(\begin{array}{l}
a)\dfrac{{ - x\sqrt x  - 3x - \sqrt x  + 2}}{{ - x\sqrt x  + \sqrt x }}\\
b)\left[ \begin{array}{l}
x = \dfrac{1}{4}\\
x = 14 + 4\sqrt {10} 
\end{array} \right.
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
A = \left( {\dfrac{1}{{x + 2\sqrt x }} - \dfrac{1}{{\sqrt x  + 1}}} \right):\dfrac{{1 - \sqrt x }}{{x + 4\sqrt x  + 4}}\\
 = \dfrac{{\sqrt x  + 1 - x - 2\sqrt x }}{{\sqrt x \left( {\sqrt x  + 1} \right)\left( {\sqrt x  + 2} \right)}}.\dfrac{{{{\left( {\sqrt x  + 2} \right)}^2}}}{{1 - \sqrt x }}\\
 = \dfrac{{ - x - \sqrt x  + 1}}{{\sqrt x \left( {\sqrt x  + 1} \right)\left( {\sqrt x  + 2} \right)}}.\dfrac{{{{\left( {\sqrt x  + 2} \right)}^2}}}{{1 - \sqrt x }}\\
 = \dfrac{{\left( { - x - \sqrt x  + 1} \right)\left( {\sqrt x  + 2} \right)}}{{\sqrt x \left( {\sqrt x  + 1} \right)\left( {1 - \sqrt x } \right)}}\\
 = \dfrac{{ - x\sqrt x  - 2x - x - 2\sqrt x  + \sqrt x  + 2}}{{\sqrt x \left( {1 - x} \right)}}\\
 = \dfrac{{ - x\sqrt x  - 3x - \sqrt x  + 2}}{{ - x\sqrt x  + \sqrt x }}\\
b)A = \dfrac{5}{3}\\
 \to \dfrac{{ - x\sqrt x  - 3x - \sqrt x  + 2}}{{ - x\sqrt x  + \sqrt x }} = \dfrac{5}{3}\\
 \to  - 3x\sqrt x  - 9x - 3\sqrt x  + 6 =  - 5x\sqrt x  + 5\sqrt x \\
 \to 2x\sqrt x  - 9x - 8\sqrt x  + 6 = 0\\
 \to 2x\sqrt x  - x - 8x + 4\sqrt x  - 12\sqrt x  + 6 = 0\\
 \to x\left( {2\sqrt x  - 1} \right) - 4\sqrt x \left( {2\sqrt x  - 1} \right) - 6\left( {2\sqrt x  - 1} \right) = 0\\
 \to \left( {2\sqrt x  - 1} \right)\left( {x - 4\sqrt x  - 6} \right) = 0\\
 \to \left[ \begin{array}{l}
\sqrt x  = \dfrac{1}{2}\\
\sqrt x  = 2 + \sqrt {10} \\
\sqrt x  = 2 - \sqrt {10} \left( l \right)
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = \dfrac{1}{4}\\
x = 14 + 4\sqrt {10} 
\end{array} \right.
\end{array}\)