a) \(\sqrt{\dfrac{25}{81}.\dfrac{16}{49}.\dfrac{196}{9}}=\sqrt{\dfrac{25}{81}}.\sqrt{\dfrac{16}{49}}.\sqrt{\dfrac{196}{9}}=\dfrac{5}{9}.\dfrac{4}{7}.\dfrac{14}{3}=\dfrac{40}{27}\)

b) \(\sqrt{3\dfrac{1}{16}.2\dfrac{14}{25}.2\dfrac{34}{81}}=\sqrt{\dfrac{49}{16}.\dfrac{64}{25}.\dfrac{196}{81}}=\sqrt{\dfrac{49}{16}}.\sqrt{\dfrac{64}{25}}.\sqrt{\dfrac{196}{81}}=\dfrac{7}{4}.\dfrac{8}{5}.\dfrac{14}{9}=\dfrac{196}{45}\)

c) \(\dfrac{\sqrt{640}.\sqrt{34,3}}{\sqrt{567}}=\sqrt{\dfrac{640.34,3}{567}}=\sqrt{\dfrac{64.49}{81}}=\dfrac{\sqrt{64}.\sqrt{49}}{\sqrt{81}}=\dfrac{8.7}{9}=\dfrac{56}{9}\)

d) \(\sqrt{21,6}.\sqrt{810}.\sqrt{11^2-5^2}=\sqrt{21,6.810.\left(11^2-5^2\right)}=\sqrt{216.81.\left(11+5\right)\left(11-5\right)}=\sqrt{36^2.9^2.4^2}=36.9.4=1296\)

a) Ta có: \(\sqrt{27\left(\sqrt{5}-\sqrt{3}\right)^2}\)

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

\(=3\sqrt{15}-9\)

c) Ta có: \(\dfrac{1+x\sqrt{x}}{1+\sqrt{x}}=\dfrac{\left(1+\sqrt{x}\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}+1}=x-\sqrt{x}+1\)

\(=\sqrt{\left(2-\sqrt{3}\right)^2\left(26+15\sqrt{3}\right)}-\sqrt{\left(2+\sqrt{3}\right)^2\left(26-15\sqrt{3}\right)}=\)

\(=\sqrt{\left(7-4\sqrt{3}\right)\left(26+15\sqrt{3}\right)}-\sqrt{\left(7+4\sqrt{3}\right)\left(26-15\sqrt{3}\right)=}\)

\(=\sqrt{7.26+7.15\sqrt{3}-4.26\sqrt{3}-180}-\sqrt{7.26-7.15\sqrt{3}+4.26\sqrt{3}-180}=\)

\(=\sqrt{4+\sqrt{3}}-\sqrt{4-\sqrt{3}}\)

1:

\(A=\sqrt{x^2+\dfrac{2x^2}{3}}=\sqrt{\dfrac{5x^2}{3}}=\left|\sqrt{\dfrac{5}{3}}x\right|=-x\sqrt{\dfrac{5}{3}}\)

2: \(=\left(\dfrac{\sqrt{100}+\sqrt{40}}{\sqrt{5}+\sqrt{2}}+\sqrt{6}\right)\cdot\dfrac{2\sqrt{5}-\sqrt{6}}{2}\)

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

\(=\dfrac{20-6}{2}=7\)

a: \(A=6\sqrt{3}+10\sqrt{3}-12\sqrt{3}=4\sqrt{3}\)

b: \(B=7\sqrt{3}+5\sqrt{3}-12\sqrt{3}=0\)

c: \(=12\sqrt{2}-6+3\left(9-4\sqrt{2}\right)=12\sqrt{2}-6+27-12\sqrt{2}=21\)

d: \(=2\sqrt{5}-5\sqrt{5}-4\sqrt{5}+11\sqrt{5}=4\sqrt{5}\)

1.

 Ta có: \(A=\sqrt{31-2\sqrt{30}}=\sqrt{\left(\sqrt{30}-1\right)^2}=\left|\sqrt{30}-1\right|=\sqrt{30}-1\)

\(B=\sqrt{11-2\sqrt{30}}=\sqrt{\left(\sqrt{6}-\sqrt{5}\right)^2}=\left|\sqrt{6}-\sqrt{5}\right|=\sqrt{6}-\sqrt{5}\)

\(C=\sqrt{13-2\sqrt{30}}=\sqrt{\left(\sqrt{10}-\sqrt{3}\right)^2}=\left|\sqrt{10}-\sqrt{3}\right|=\sqrt{10}-\sqrt{3}\)

\(D=\sqrt{39-6\sqrt{30}}=\sqrt{\left(\sqrt{30}-3\right)^2}=\left|\sqrt{30}-3\right|=\sqrt{30}-3\)

\(A=\sqrt{31-2\sqrt{30}}=\sqrt{30}-1\)

\(B=\sqrt{11-2\sqrt{30}}=\sqrt{6}-\sqrt{5}\)

\(C=\sqrt{13-2\sqrt{30}}=\sqrt{10}-\sqrt{3}\)

\(D=\sqrt{39-6\sqrt{30}}=\sqrt{30}-3\)