10: ĐKXĐ: x>=2
\(\sqrt{x-1+2\sqrt{x-2}}-\sqrt{x-1-2\sqrt{x-2}}=1\)
=>\(\sqrt{x-2+2\cdot\sqrt{x-2}+1}-\sqrt{x-2-2\cdot\sqrt{x-2}+1}=1\)
=>\(\sqrt{\left(\sqrt{x-2}+1\right)^2}-\sqrt{\left(\sqrt{x-2}-1\right)^2}=1\)
=>\(\sqrt{x-2}+1-\left|\sqrt{x-2}-1\right|=1\)
=>\(\sqrt{x-2}-\left|\sqrt{x-2}-1\right|=0\) (1)
TH1: x>=3
=>x-2>=1
=>\(\sqrt{x-2}-1\ge0\)
(1) trở thành: \(\sqrt{x-2}-\left(\sqrt{x-2}-1\right)=0\)
=>1=0(vô lý)
TH2: 2<=x<=3
=>\(\sqrt{x-2}-1\le0\)
(1) sẽ trở thành: \(\sqrt{x-2}+\left(\sqrt{x-2}-1\right)=0\)
=>\(2\sqrt{x-2}=1\)
=>\(\sqrt{x-2}=\frac12\)
=>\(x-2=\frac14\)
=>\(x=2+\frac14=\frac94\) (nhận)
Bài 11:
\(\sqrt{2+\sqrt3}-\sqrt{2-\sqrt3}\)
\(=\frac{1}{\sqrt2}\left(\sqrt{4+2\sqrt3}-\sqrt{4-2\sqrt3}\right)\)
\(=\frac{1}{\sqrt2}\left(\sqrt{\left(\sqrt3+1\right)^2}-\sqrt{\left(\sqrt3-1\right)^2}\right)\)
\(=\frac{1}{\sqrt2}\left(\sqrt3+1-\sqrt3+1\right)=\frac{2}{\sqrt2}=\sqrt2\)
