Bài 3:
\(H=\left(\frac{2\sqrt{a}}{\sqrt{a}+3}+\frac{\sqrt{a}}{\sqrt{a}-3}-\frac{3a+3}{a-9}\right):\left(\frac{2\sqrt{a}-2}{\sqrt{a}-3}-1\right)\)
\(=\frac{2\sqrt{a}\left(\sqrt{a}-3\right)+\sqrt{a}\left(\sqrt{a}+3\right)-3a-3}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}:\frac{2\sqrt{a}-2-\sqrt{a}+3}{\sqrt{a}-3}\)
\(=\frac{2a-6\sqrt{a}+a+3\sqrt{a}-3a-3}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}\cdot\frac{\sqrt{a}-3}{\sqrt{a}+1}\)
\(=\frac{-3\sqrt{a}-3}{\left(\sqrt{a}+1\right)\left(\sqrt{a}+3\right)}=\frac{-3}{\sqrt{a}+3}\)
Bài 2:
a: ĐKXĐ: x>=2
TA có: \(\sqrt{16x-32}-\sqrt{4x-8}+\sqrt{9x-18}=1\)
=>\(4\sqrt{x-2}-2\sqrt{x-2}+3\sqrt{x-2}=1\)
=>\(5\sqrt{x-2}=1\)
=>\(\sqrt{x-2}=\frac15\)
=>\(x-2=\frac{1}{25}\)
=>\(x=2+\frac{1}{25}=\frac{51}{25}\) (nhận)
b: ĐKXĐ:x>=1/2
Ta có: \(\sqrt{8x-4}-12\cdot\sqrt{\frac{2x-1}{9}}+\sqrt{18x-9}=3\)
=>\(2\sqrt{2x-1}-12\cdot\frac{\sqrt{2x-1}}{3}+3\sqrt{2x-1}=3\)
=>\(5\sqrt{2x-1}-4\sqrt{2x-1}=3\)
=>\(\sqrt{2x-1}=3\)
=>2x-1=9
=>2x=10
=>x=5(nhận)
c: \(\sqrt{9x^2-12x+4}=2\)
=>\(\sqrt{\left(3x-2\right)^2}=2\)
=>|3x-2|=2
=>\(\left[\begin{array}{l}3x-2=2\\ 3x-2=-2\end{array}\right.\Rightarrow\left[\begin{array}{l}3x=4\\ 3x=0\end{array}\right.=>\left[\begin{array}{l}x=\frac43\\ x=0\end{array}\right.\)
d: \(\sqrt{x^2+12x+36}=\left|-7\right|\)
=>\(\sqrt{\left(x+6\right)^2}=7\)
=>|x+6|=7
=>x+6=7 hoặc x+6=-7
=>x=1 hoặc x=-13
