a)√(4x^2-4x+1)-5=0 b)(√x/√(x)+3+x+9/9-x)/3√x+3/x-3√x

Đáp án:

$\begin{array}{l}
a)\sqrt {4{x^2} - 4x + 1}  - 5 = 0\\
 \Leftrightarrow \sqrt {{{\left( {2x - 1} \right)}^2}}  = 5\\
 \Leftrightarrow \left| {2x - 1} \right| = 5\\
 \Leftrightarrow \left[ \begin{array}{l}
2x - 1 = 5\\
2x - 1 =  - 5
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
2x = 6\\
2x =  - 4
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 3\\
x =  - 2
\end{array} \right.\\
Vay\,x = 3;x =  - 2\\
b)\\
\left( {\dfrac{{\sqrt x }}{{\sqrt x  + 3}} + \dfrac{{x + 9}}{{9 - x}}} \right):\dfrac{{3\sqrt x  + 3}}{{x - 3\sqrt x }}\\
 = \left( {\dfrac{{\sqrt x }}{{\sqrt x  + 3}} - \dfrac{{x + 9}}{{x - 9}}} \right).\dfrac{{\sqrt x \left( {\sqrt x  - 3} \right)}}{{3\left( {\sqrt x  + 1} \right)}}\\
 = \dfrac{{\sqrt x \left( {\sqrt x  - 3} \right) - x - 9}}{{\left( {\sqrt x  + 3} \right)\left( {\sqrt x  - 3} \right)}}.\dfrac{{\sqrt x \left( {\sqrt x  - 3} \right)}}{{3\left( {\sqrt x  + 1} \right)}}\\
 = \dfrac{{x - 3\sqrt x  - x - 9}}{{\sqrt x  + 3}}.\dfrac{{\sqrt x }}{{3\left( {\sqrt x  + 1} \right)}}\\
 = \dfrac{{ - 3\left( {\sqrt x  + 3} \right)}}{{\sqrt x  + 3}}.\dfrac{{\sqrt x }}{{3\left( {\sqrt x  + 1} \right)}}\\
 =  - \sqrt x 
\end{array}$