1: Ta có: \(\frac{\sqrt{x}+1}{\sqrt{x}-1}-\frac{\sqrt{x}-1}{\sqrt{x}+1}+\frac{8\sqrt{x}}{1-x}\)
\(=\frac{\left(\sqrt{x}+1\right)^2-\left(\sqrt{x}-1\right)^2-8\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{x+2\sqrt{x}+1-x+2\sqrt{x}-1-8\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{-4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
Ta có: \(\frac{-x+\sqrt{x}-3}{x-1}-\frac{1}{\sqrt{x}-1}\)
\(=\frac{-x+\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{1}{\sqrt{x}-1}\)
\(=\frac{-x+\sqrt{x}-3-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{-x-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
Ta có: \(K=\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}-\frac{\sqrt{x}-1}{\sqrt{x}+1}+\frac{8\sqrt{x}}{1-x}\right):\left(\frac{-x+\sqrt{x}-3}{x-1}-\frac{1}{\sqrt{x}-1}\right)\)
\(=\frac{-4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\frac{-x-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{4\sqrt{x}}{x+4}\)
2: \(K=\frac35\)
=>\(\frac{4\sqrt{x}}{x+4}=\frac35\)
=>\(3\left(x+4\right)=4\sqrt{x}\cdot5=20\sqrt{x}\)
=>\(3x-20\sqrt{x}+12=0\)
=>\(3x-18\sqrt{x}-2\sqrt{x}+12=0\)
=>\(\left(\sqrt{x}-6\right)\left(3\sqrt{x}-2\right)=0\)
TH1: \(\sqrt{x}-6=0\)
=>\(\sqrt{x}=6\)
=>x=36(nhận)
TH2: \(3\sqrt{x}-2=0\)
=>\(3\sqrt{x}=2\)
=>\(\sqrt{x}=\frac23\)
=>\(x=\frac49\) (nhận)
3: Thay \(x=14-6\sqrt5=\left(3-\sqrt5\right)^2\) vào K, ta được:
\(K=\frac{4\cdot\sqrt{\left(3-\sqrt5\right)^2}}{14-6\sqrt5+4}=\frac{4\left(3-\sqrt5\right)}{18-6\sqrt5}=\frac{4\left(3-\sqrt5\right)}{6\left(3-\sqrt5\right)}=\frac46=\frac23\)
4: \(K-1=\frac{4\sqrt{x}}{x+4}-1=\frac{4\sqrt{x}-x-4}{x+4}=\frac{-\left(\sqrt{x}-2\right)^2}{x+4}<=0\) ∀x thỏa mãn ĐKXĐ
=>K<=1∀x thỏa mãn ĐKXĐ
