\(=5\sqrt{2}-9\sqrt{5}-6\sqrt{2}+10\sqrt{5}=\sqrt{5}-\sqrt{2}\)

√2–9√5–6√2+10√5

=√5–√2

Lời giải:
a.

$=2\sqrt{5}-9\sqrt{5}-2\sqrt{5}=(2-9-2)\sqrt{5}=-9\sqrt{5}$

b.

$=36\sqrt{6}-2\sqrt{6}+6\sqrt{6}=(36-2+6)\sqrt{6}=40\sqrt{6}$

a) \(5\sqrt{\frac{1}{5}}+\frac{1}{2}\sqrt{20}+\sqrt{5}\)

\(=\sqrt{25.\frac{1}{5}}+\sqrt{\frac{1}{4}.20}+\sqrt{5}\)

\(=\sqrt{5}+\sqrt{5}+\sqrt{5}\)

\(=3\sqrt{5}\)

b) \(\sqrt{20}-\sqrt{45}+3\sqrt{18}+\sqrt{72}\)

\(=\sqrt{4.5}-\sqrt{5.9}+3\sqrt{18}+\sqrt{8.9}\)

\(=2\sqrt{5}-3\sqrt{5}+3\sqrt{18}+3\sqrt{8}\)

\(=2\sqrt{5}-3\sqrt{5}+3\sqrt{18}+3\sqrt{8}\)

\(=-\sqrt{5}+3.\left(\sqrt{18}+\sqrt{8}\right)\) (Tới đây không biết làm gì nữa)

\(5\sqrt{\dfrac{1}{5}}+\dfrac{1}{3}\sqrt{45}+\sqrt{\left(2-\sqrt{5}\right)^2}\)

\(=\dfrac{5}{\sqrt{5}}+\dfrac{1}{3}\cdot3\sqrt{5}+\left|2-\sqrt{5}\right|\)

\(=\sqrt{5}+\sqrt{5}+\sqrt{5}-2\)

\(=3\sqrt{5}-2\)

\(B=\left(5+2\sqrt{6}\right)\left(49-20\sqrt{6}\right)\cdot\sqrt{5-2\sqrt{6}}\)

\(=\left(5+2\sqrt{6}\right)\left(\sqrt{3}-\sqrt{2}\right)\cdot\left(5-2\sqrt{6}\right)\)

\(=\sqrt{3}-\sqrt{2}\)

\(B=\left(\dfrac{4}{1-\sqrt{5}}+\dfrac{1}{2+\sqrt{5}}-\dfrac{4}{3-\sqrt{5}}\right)\left(\sqrt{5}-6\right)\)

\(B=\left[\dfrac{4\left(1+\sqrt{5}\right)}{\left(1-\sqrt{5}\right)\left(1+\sqrt{5}\right)}+\dfrac{2-\sqrt{5}}{\left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right)}-\dfrac{4\left(3+\sqrt{5}\right)}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}\right]\left(\sqrt{5}-6\right)\)

\(B=\left[\dfrac{4\left(1+\sqrt{5}\right)}{1-5}+\dfrac{2-\sqrt{5}}{4-5}-\dfrac{4\left(3+\sqrt{5}\right)}{9-5}\right]\left(\sqrt{5}-6\right)\)

\(B=\left[-\dfrac{4\left(1+\sqrt{5}\right)}{4}-\dfrac{2-\sqrt{5}}{1}-\dfrac{4\left(3+\sqrt{5}\right)}{4}\right]\left(\sqrt{5}-6\right)\)

\(B=\left(-1-\sqrt{5}-2+\sqrt{5}-3-\sqrt{5}\right)\left(\sqrt{5}-6\right)\)

\(B=\left(-\sqrt{5}-6\right)\left(\sqrt{5}-6\right)\)

\(B=-\left(\sqrt{5}+6\right)\left(\sqrt{5}-6\right)\)

\(B=-\left(5-36\right)\)

\(B=-\left(-31\right)\)

\(B=31\)

_____________________________

\(\sqrt{48}-\dfrac{\sqrt{21}-\sqrt{15}}{\sqrt{7}-\sqrt{5}}+\dfrac{2}{\sqrt{3}+1}\)

\(=4\sqrt{3}-\dfrac{\sqrt{3}\left(\sqrt{7}-\sqrt{5}\right)}{\sqrt{7}-\sqrt{5}}+\dfrac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)

\(=4\sqrt{3}-\sqrt{3}-\dfrac{2\left(\sqrt{3}-1\right)}{2}\)

\(=3\sqrt{3}-\sqrt{3}+1\)

\(=2\sqrt{3}+1\)

\(\left(20.\sqrt{0.03}+12.\sqrt{3}-\frac{1}{5}.\sqrt{75}\right).\sqrt{6}\)

\(=20.\sqrt{0,03.6}+12.\sqrt{3.6}-\frac{1}{5}.\sqrt{75.6}\)

\(=20.\sqrt{\frac{9}{50}}+12.\sqrt{3^2.2}-\frac{1}{5}.\sqrt{15^2.2}\)

\(=6\sqrt{2}+36\sqrt{2}-3\sqrt{2}\)

\(=39\sqrt{2}\)

\(=\sqrt{6}\left(2\sqrt{3}+12\sqrt{3}-\sqrt{3}\right)\)

\(=\sqrt{6}.13\sqrt{3}=13\sqrt{18}=39\sqrt{2}\)

\(a,=\sqrt{5}\left(2\sqrt{5}-3\right)+3\sqrt{5}=10-3\sqrt{5}+3\sqrt{5}=10\\ b,=5-\sqrt{3}-\left(2-\sqrt{3}\right)=3\\ c,=\dfrac{2\left(\sqrt{5}-1\right)}{4}-\dfrac{2\left(3+\sqrt{5}\right)}{4}=\dfrac{2\sqrt{5}-2-6-2\sqrt{5}}{4}=\dfrac{-8}{4}=-2\)