Bài 1:
a: \(\sqrt{11-6\sqrt2}-\sqrt{6-4\sqrt2}\)
\(=\sqrt{9-2\cdot3\cdot\sqrt2+2}-\sqrt{4-2\cdot2\cdot\sqrt2+2}\)
\(=\sqrt{\left.\left(3-\sqrt2\right)^2\right.}-\sqrt{\left(2-\sqrt2\right)^2}\)
\(=3-\sqrt2-2+\sqrt2=1\)
b: \(\frac{\sqrt{10}-\sqrt2}{\sqrt5-1}-\frac{7}{\sqrt7}+\frac{5}{\sqrt7-\sqrt2}\)
\(=\frac{\sqrt2\left(\sqrt5-1\right)}{\sqrt5-1}-\sqrt7+\frac{5\left(\sqrt7+\sqrt2\right)}{\left(\sqrt7-\sqrt2\right)\left(\sqrt7+\sqrt2\right)}\)
\(=\sqrt2-\sqrt7+\sqrt7+\sqrt2=2\sqrt2\)
Bài 2:
a: ĐKXĐ: x>=-1
\(\sqrt{4x+4}-\sqrt{9x+9}+\sqrt{x+1}=-\sqrt{x+7}\)
=>\(2\sqrt{x+1}-3\sqrt{x+1}+\sqrt{x+1}=-\sqrt{x+7}\)
=>\(-\sqrt{x+7}=0\)
=>x+7=0
=>x=-7(loại)
b: \(\sqrt{x^2-4x+4}-5=8\)
=>\(\sqrt{\left(x-2\right)^2}=5+8\)
=>|x-2|=13
=>\(\left[\begin{array}{l}x-2=13\\ x-2=-13\end{array}\right.\Rightarrow\left[\begin{array}{l}x=15\left(nhận\right)\\ x=-11\left(nhận\right)\end{array}\right.\)
