a)\(2\sqrt{\dfrac{16}{3}}-3\sqrt{\dfrac{1}{27}}-6\sqrt{\dfrac{4}{75}}\)

\(=2.\sqrt{\dfrac{4^2}{3}}-3.\sqrt{\dfrac{1}{3.3^2}}-6\sqrt{\dfrac{2^2}{3.5^2}}\)

\(=2.\dfrac{4}{\sqrt{3}}-3.\dfrac{1}{3\sqrt{3}}-6.\dfrac{2}{5\sqrt{3}}=\dfrac{8}{\sqrt{3}}-\dfrac{1}{\sqrt{3}}-\dfrac{12}{5\sqrt{3}}\)\(=\dfrac{23}{5\sqrt{3}}=\dfrac{23\sqrt{3}}{15}\)

b)\(\left(6\sqrt{\dfrac{8}{9}}-5\sqrt{\dfrac{32}{25}}+14\sqrt{\dfrac{18}{49}}\right).\sqrt{\dfrac{1}{2}}\)

\(=6\sqrt{\dfrac{8}{9}.\dfrac{1}{2}}-5\sqrt{\dfrac{32}{25}.\dfrac{1}{2}}+14\sqrt{\dfrac{18}{49}.\dfrac{1}{2}}\)

\(=6\sqrt{\dfrac{4}{9}}-5\sqrt{\dfrac{16}{25}}+14\sqrt{\dfrac{9}{49}}\)\(=6.\dfrac{2}{3}-5.\dfrac{4}{5}+14.\dfrac{3}{7}=6\)

c)\(\sqrt{\left(\sqrt{2}-2\right)^2}-\sqrt{6+4\sqrt{2}}=\left|\sqrt{2}-2\right|-\sqrt{4+2.2\sqrt{2}+2}=2-\sqrt{2}-\sqrt{\left(2+\sqrt{2}\right)^2}\)

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

1:

a: \(\sqrt{36}-\sqrt{100}=6-10=-4\)

b: Để \(\sqrt{\dfrac{2}{2x-1}}\) có nghĩa thì \(\dfrac{2}{2x-1}>=0\)

=>2x-1>0

=>x>1/2

2:

a: \(A=\dfrac{\left(15\sqrt{180}-5\sqrt{200}-3\sqrt{450}\right)}{\sqrt{10}}\)

\(=15\sqrt{\dfrac{180}{10}}-5\sqrt{\dfrac{200}{10}}-3\sqrt{\dfrac{450}{10}}\)

\(=15\sqrt{18}-5\sqrt{20}-3\sqrt{45}\)

\(=45\sqrt{2}-10\sqrt{5}-9\sqrt{5}\)

\(=45\sqrt{2}-19\sqrt{5}\)

b: \(B=\sqrt{32}-\sqrt{50}-16\sqrt{\dfrac{1}{8}}\)

\(=4\sqrt{2}-5\sqrt{2}-\dfrac{16}{\sqrt{8}}\)

\(=-\sqrt{2}-2\sqrt{8}=-\sqrt{2}-4\sqrt{2}=-5\sqrt{2}\)

Bài 1:

a: \(\sqrt{252}-\sqrt{700}+\sqrt{1008}-\sqrt{448}\)

\(=6\sqrt{7}-10\sqrt{7}+12\sqrt{7}-8\sqrt{7}\)

\(=8\sqrt{7}\)

Bài 3: 

a: \(\sqrt{27^2-23^2}=10\sqrt{2}\)

b: \(\sqrt{37^2-35^2}=12\)

c: \(\sqrt{65^2-63^2}=16\)

d: \(\sqrt{117^2-108^2}=45\)

a: \(\dfrac{-4}{3}\cdot\sqrt{\left(-0.4\right)^2}=\dfrac{-4}{3}\cdot\dfrac{2}{5}=\dfrac{-8}{15}\)

b: \(\sqrt[3]{\dfrac{3}{4}}\cdot\sqrt[3]{\dfrac{9}{16}}=\dfrac{3}{4}\)

b, \(\dfrac{2}{\sqrt{5}+2}+\dfrac{2}{2-\sqrt{5}}\)

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

\(=2\sqrt{5}-4-2\sqrt{5}-4=-8\)

a, \(\sqrt{2}\left(\sqrt{8}+\sqrt{32}-\sqrt{98}\right)\)

\(=\sqrt{2}\left(2\sqrt{2}+4\sqrt{2}-7\sqrt{2}\right)\)

\(=\sqrt{2}.\left(-\sqrt{2}\right)=-2\)

\(a,=\sqrt{17}-4-\sqrt{17}-2=-6\\ b,=7\left(\sqrt{3}+\sqrt{2}\right)-7\sqrt{3}-6\sqrt{2}\\ =7\sqrt{3}+7\sqrt{2}-7\sqrt{3}-6\sqrt{2}=\sqrt{2}\\ c,=\dfrac{6\sqrt{5}+12-6\sqrt{5}+12}{3}+2\sqrt{2}-\dfrac{4\sqrt{7}}{7}\\ =8+2\sqrt{2}-\dfrac{4\sqrt{7}}{7}=\dfrac{56+14\sqrt{2}-4\sqrt{7}}{7}\\ d,=\left(\dfrac{\sqrt{2}}{4}-\dfrac{6\sqrt{2}}{4}+8\sqrt{2}\right):\dfrac{1}{8}=\dfrac{-5\sqrt{2}+32\sqrt{2}}{4}\cdot8=54\sqrt{2}\)

a.\(\sqrt{\left(\sqrt{7}-1\right)^2}=\left|\sqrt{7}-1\right|=\sqrt{7}-1\)

b.\(\sqrt{\left(2-\sqrt{3}\right)^2}=\left|2-\sqrt{3}\right|=2-\sqrt{3}\)

c.\(\sqrt{\left(\sqrt{2}+5\right)^2}-\sqrt{2}=\left|\sqrt{2}+5\right|-\sqrt{2}=\sqrt{2}+5-\sqrt{2}=5\)

d.\(\sqrt{\left(3+\sqrt{5}\right)^2}+\sqrt{\left(\sqrt{5}-6\right)^2}=\left|3+\sqrt{5}\right|+\left|\sqrt{5}-6\right|=3+\sqrt{5}+6-\sqrt{5}=9\)

a)\(=\sqrt{7}-1\)

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

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

d)\(=3+\sqrt{5}+\sqrt{5}+6=9\)

\(A=\left|2-\sqrt{3}\right|+\left|1+\sqrt{3}\right|=2-\sqrt{3}+1+\sqrt{3}=3\)

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

\(C=\left|1-\sqrt{5}\right|-\left|2-\sqrt{5}\right|=\sqrt{5}-1-\sqrt{5}+2=1\)

\(A=\left|2-\sqrt{3}\right|+\left|1+\sqrt{3}\right|=2-\sqrt{3}+1+\sqrt{3}=3\)

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

C=\(\left|1-\sqrt{5}\right|-\left|2-\sqrt{5}\right|=\sqrt{5}-1-\sqrt{5}+2=1\)

a: \(\frac{\sqrt{3-\sqrt5}\cdot\left(3+\sqrt5\right)}{\sqrt{10}+\sqrt2}\)

\(=\frac{\sqrt{6-2\sqrt5}\cdot\left(3+\sqrt5\right)}{\sqrt{20}+\sqrt4}\)

\(=\frac{\sqrt{\left(\sqrt5-1\right)^2}\cdot\left(3+\sqrt5\right)}{2\left(\sqrt5+\sqrt1\right)}\)

\(=\frac{\left(\sqrt5-1\right)^{}\cdot\left(3+\sqrt5\right)}{2\left(\sqrt5+\sqrt1\right)}=\frac{3\sqrt5+5-3-\sqrt5}{2\left(\sqrt5+1\right)}=\frac{2\sqrt5+2}{2\sqrt5+2}=1\)

b: \(\sqrt{8\sqrt3}-\sqrt{25\sqrt{12}}+4\sqrt{\sqrt{192}}\)

\(=2\sqrt{2\sqrt3}-5\sqrt{2\sqrt3}+4\sqrt{\sqrt{64}\cdot\sqrt3}\)

\(=-3\sqrt{2\sqrt3}+4\sqrt{8\sqrt3}\)

\(=-3\sqrt{2\sqrt3}+4\cdot2\sqrt{2\sqrt3}=5\sqrt{2\sqrt3}\)

c: \(\sqrt{2-\sqrt3}\cdot\left(\sqrt5+\sqrt2\right)\)

\(=\frac{\sqrt{4-2\sqrt3}\cdot\left(\sqrt5+\sqrt2\right)}{\sqrt2}=\frac{\left(\sqrt3-1\right)\left(\sqrt5+\sqrt2\right)}{\sqrt2}=\frac{\left(\sqrt6-\sqrt2\right)\left(\sqrt5+\sqrt2\right)}{2}\)

\(=\frac{\sqrt{30}+2\sqrt3-\sqrt{10}-2}{2}\)

d: \(\sqrt{3-\sqrt5}+\sqrt{3+\sqrt5}\)

\(=\frac{\sqrt{6-2\sqrt5}+\sqrt{6+2\sqrt5}}{\sqrt2}=\frac{\sqrt{\left(\sqrt5-1\right)^2}+\sqrt{\left(\sqrt5+1\right)^2}}{\sqrt2}\)

\(=\frac{\sqrt5-1+\sqrt5+1}{\sqrt2}=\frac{2\sqrt5}{\sqrt2}=\sqrt{10}\)

e: Đặt \(A=\sqrt{4+\sqrt{10+2\sqrt5}}+\sqrt{4-\sqrt{10+2\sqrt5}}\)

=>\(A^2=4+\sqrt{10+2\sqrt5}+4-\sqrt{10+2\sqrt5}+2\cdot\sqrt{16-\left(10+2\sqrt5\right)}\)

=>\(A^2=8+2\cdot\sqrt{6-2\sqrt5}\)

=>\(A^2=8+2\cdot\sqrt{\left(\sqrt5-1\right)^2}=8+2\left(\sqrt5-1\right)=6+2\sqrt5=\left(\sqrt5+1\right)^2\)

=>\(A=\sqrt5+1\)

f: \(\left(5+2\sqrt6\right)\left(49-20\sqrt6\right)\cdot\sqrt{5-2\sqrt6}\)

\(=\left(245-100\sqrt6+98\sqrt6-240\right)\cdot\sqrt{\left(\sqrt3-\sqrt2\right)^2}\)

\(=\left(5-2\sqrt6\right)\left(\sqrt3-\sqrt2\right)=\left(\sqrt3-\sqrt2\right)^2\)

g: \(\frac{1}{\sqrt2+\sqrt{2+\sqrt3}}+\frac{1}{\sqrt2-\sqrt{2-\sqrt3}}\)

\(=\frac{\sqrt2}{2+\sqrt{4+2\sqrt3}}+\frac{\sqrt2}{2-\sqrt{4-2\sqrt3}}\)

\(=\frac{\sqrt2}{2+\sqrt{\left(\sqrt3+1\right)^2}}+\frac{\sqrt2}{2-\sqrt{\left(\sqrt3-1\right)^2}}\)

\(=\frac{\sqrt2}{2+\left(\sqrt3+1\right)^{}}+\frac{\sqrt2}{2-\left(\sqrt3-1\right)}\)

\(=\frac{\sqrt2}{2+\sqrt3+1^{}}+\frac{\sqrt2}{2-\sqrt3+1}=\frac{\sqrt2}{3+\sqrt3}+\frac{\sqrt2}{3-\sqrt3}=\frac{\sqrt2\left(3-\sqrt3\right)+\sqrt2\left(3+\sqrt3\right)}{9-3}\)

\(=\frac{3\sqrt2-\sqrt6+3\sqrt2+\sqrt6}{6}=\frac{6\sqrt2}{6}=\sqrt2\)

i: \(\frac{\left(\sqrt5+2\right)^2-8\sqrt5}{2\sqrt5-4}\)

\(=\frac{9+4\sqrt5-8\sqrt5}{2\left(\sqrt5-2\right)}\)

\(=\frac{9-4\sqrt5}{2\left(\sqrt5-2\right)}=\frac{\left(\sqrt5-2\right)^2}{2\left(\sqrt5-2\right)}=\frac{\sqrt5-2}{2}\)

k: \(\sqrt{14-8\sqrt3}-\sqrt{24-12\sqrt3}\)
\(=\sqrt{8-2\cdot2\sqrt2\cdot\sqrt6+6}-\sqrt{6\left(4-2\sqrt3\right)}\)
\(=\sqrt{\left(2\sqrt2-\sqrt6\right)^2}-\sqrt6\left(\sqrt3-1\right)=2\sqrt2-\sqrt6-\sqrt{18}+\sqrt6=2\sqrt2-3\sqrt2=-\sqrt2\)

l: \(\frac{4}{\sqrt3+1}+\frac{1}{\sqrt3-2}+\frac{6}{\sqrt3-3}\)

\(=\frac{4\left(\sqrt3-1\right)}{3-1}-\frac{1\left(2+\sqrt3\right)}{\left(2-\sqrt3\right)\left(2+\sqrt3\right)}-\frac{6\left(3+\sqrt3\right)}{9-3}\)

\(=2\left(\sqrt3-1\right)-\left(2+\sqrt3\right)-\left(3+\sqrt3\right)=2\sqrt3-2-2-\sqrt3-3-\sqrt3\)

=-7

m: \(\left(\sqrt2+1\right)^3-\left(\sqrt2-1\right)^3\)

\(=\left(2\sqrt2+3\cdot2\cdot1+3\cdot\sqrt2\cdot1+1\right)-\left(2\sqrt2-3\cdot2\cdot1+3\cdot\sqrt2\cdot1-1\right)\)

\(=\left(5\sqrt2+7\right)-\left(5\sqrt2-7\right)=14\)