Bài 1:
a: \(\sqrt{18}-\frac12\cdot\sqrt{72}+4\cdot\sqrt{\frac12}+\sqrt{72}\)
\(=3\sqrt2-\frac12\cdot6\sqrt2+2\sqrt2+6\sqrt2\)
\(=3\sqrt2-3\sqrt2+8\sqrt2=8\sqrt2\)
b: \(\frac{5\sqrt2-2\sqrt5}{\sqrt5-\sqrt2}+\frac{6}{2-\sqrt{10}}\)
\(=\frac{\sqrt{10}\left(\sqrt5-\sqrt2\right)}{\sqrt5-\sqrt2}+\frac{6\left(2+\sqrt{10}\right)}{\left(2-\sqrt{10}\right)\left(2+\sqrt{10}\right)}\)
\(=\sqrt{10}+\frac{6\left(2+\sqrt{10}\right)}{4-10}=\sqrt{10}-\left(2+\sqrt{10}\right)\)
=-2
c: \(\sqrt{3-2\sqrt2}+\sqrt{\left(2-\sqrt2\right)^2}\)
\(=\sqrt{\left(\sqrt2-1\right)^2}+2-\sqrt2\)
\(=\sqrt2-1+2-\sqrt2=2-1=1\)
d: \(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt6\right)\cdot\sqrt{4-\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\left(\sqrt5-\sqrt3\right)\cdot\sqrt{8-2\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\left(\sqrt5-\sqrt3\right)\left(\sqrt5-\sqrt3\right)\)
\(=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)=2\left(4+\sqrt{15}\right)\left(4-\sqrt{15}\right)\)
=2
Bài 2:
a: \(\sqrt{4x^2-12x+9}-7=0\)
=>\(\sqrt{\left(2x-3\right)^2}=7\)
=>|2x-3|=7
=>\(\left[\begin{array}{l}2x-3=7\\ 2x-3=-7\end{array}\right.\Rightarrow\left[\begin{array}{l}2x=10\\ 2x=-4\end{array}\right.\Rightarrow\left[\begin{array}{l}x=5\\ x=-2\end{array}\right.\)
b: ĐKXĐ: x>=5
\(\sqrt{x-5}+\frac13\cdot\sqrt{9x-45}=\frac15\cdot\sqrt{25x-125}+6\)
=>\(\sqrt{x-5}+\frac13\cdot3\sqrt{x-5}=\frac15\cdot5\sqrt{x-5}+6\)
=>\(\sqrt{x-5}=6\)
=>x-5=36
=>x=36+5=41(nhận)
