Bài 3:
a: \(Q=\frac{3}{\sqrt{x}-1}+\frac{2}{\sqrt{x}+1}-\frac{4\sqrt{x}+2}{x-1}\)
\(=\frac{3\left(\sqrt{x}+1\right)+2\left(\sqrt{x}-1\right)-4\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{3\sqrt{x}+3+2\sqrt{x}-2-4\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{1}{\sqrt{x}+1}\)
b: \(Q=\frac12\)
=>\(\frac{1}{\sqrt{x}+1}=\frac12\)
=>\(\sqrt{x}+1=2\)
=>\(\sqrt{x}=1\)
=>x=1(loại)
Bài 2:
a: ĐKXĐ: 9x-9>=0
=>9x>=9
=>x>=1
\(\sqrt{9x-9}+1=7\)
=>\(3\sqrt{x-1}=7-1=6\)
=>\(\sqrt{x-1}=2\)
=>x-1=4
=>x=4+1
=>x=5(nhận)
b: ĐKXĐ: x>=-3
\(\sqrt{9x+27}-\frac14\cdot\sqrt{16x+48}+\sqrt{x+3}=9\)
=>\(3\sqrt{x+3}-\frac14\cdot4\sqrt{x+3}+\sqrt{x+3}=9\)
=>\(3\cdot\sqrt{x+3}=9\)
=>\(\sqrt{x+3}=3\)
=>x+3=9
=>x=6(nhận)
