Hết rồi nhé !

\(\dfrac{8}{\sqrt{5}-1}-\dfrac{22}{4+\sqrt{5}}+\dfrac{\sqrt{15}+2\sqrt{5}}{2+\sqrt{3}}\)
\(=\dfrac{8\left(\sqrt{5}+1\right)}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}-\dfrac{22\left(4-\sqrt{5}\right)}{\left(\sqrt{5}+4\right)\left(4-\sqrt{5}\right)}+\dfrac{\sqrt{5}\left(\sqrt{3}+2\right)}{2+\sqrt{3}}\)
\(=\dfrac{8\sqrt{5}+8}{5-1}-\dfrac{88-22\sqrt{5}}{16-5}+\sqrt{5}\)
\(=\dfrac{8\sqrt{5}+8}{4}-\dfrac{88-22\sqrt{5}}{11}+\sqrt{5}\)
\(=2\sqrt{5}+2-8+2\sqrt{5}+\sqrt{5}=5\sqrt{5}-6\)

\(\sqrt{13-\sqrt{160}}-\sqrt{53+4\sqrt{60}}=\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}-\sqrt{\left(4\sqrt{3}+\sqrt{5}\right)^2}\)

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

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

\(\sqrt{\left(4+\sqrt{3}\right)\sqrt{19-8\sqrt{3}}+3}=\sqrt{\left(4+\sqrt{3}\right)\sqrt{\left(4-\sqrt{3}\right)^2}+3}\)

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

a) Ta có: \(\sqrt{13-\sqrt{160}}-\sqrt{53+4\sqrt{60}}\)

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

\(=2\sqrt{2}-4\sqrt{3}\)

b) Ta có: \(\sqrt{\left(4+\sqrt{3}\right)\cdot\sqrt{19-8\sqrt{3}+3}}\)

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

=4

Ta có \(P=\left(\frac{\sqrt{14}-\sqrt{7}}{\sqrt{8}-2}-\frac{\sqrt{15}-\sqrt{3}}{2-2\sqrt{5}}\right):\frac{1}{\sqrt{7}-\sqrt{3}}\)

\(=\left(\frac{\sqrt{7}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}-\frac{\sqrt{3}\left(\sqrt{5}-1\right)}{2\left(1-\sqrt{5}\right)}\right).\left(\sqrt{7}-\sqrt{3}\right)\)

\(=\left(\frac{\sqrt{7}}{2}+\frac{\sqrt{3}}{2}\right).\left(\sqrt{7}-\sqrt{3}\right)=\frac{\sqrt{7}+\sqrt{3}}{2}.\left(\sqrt{7}-\sqrt{3}\right)\)

\(=\frac{7-3}{2}=2\)

Vậy \(P=2\)

a: Ta có: \(\sqrt{8+2\sqrt{15}}-\sqrt{6+2\sqrt{5}}\)

\(=\sqrt{5}+\sqrt{3}-\sqrt{5}-1\)

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

b: Ta có: \(\sqrt{17-2\sqrt{72}}+\sqrt{19+2\sqrt{18}}\)

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

\(=4+\sqrt{2}\)

c: Ta có: \(\sqrt{12-2\sqrt{32}}+\sqrt{9+4\sqrt{2}}\)

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

\(=4\sqrt{2}-1\)

a)

\(\sqrt{8+2\sqrt{15}}-\sqrt{6+2\sqrt{5}}\\ =\sqrt{5+2\sqrt{5}\cdot\sqrt{3}+3}-\sqrt{5+2\sqrt{5}\cdot\sqrt{1}+1}\\ =\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{5}+\sqrt{1}\right)^2}\\ =\sqrt{5}+\sqrt{3}-\sqrt{5}-\sqrt{1}\\ =\sqrt{3}-\sqrt{1}\)

b)

\(\sqrt{17-2\sqrt{72}}+\sqrt{19+2\sqrt{18}}\\ =\sqrt{9-2\sqrt{9}\cdot\sqrt{8}+8}+\sqrt{18+2\sqrt{18}\cdot\sqrt{1}+1}\\ =\sqrt{\left(3-2\sqrt{2}\right)^2}+\sqrt{\left(3\sqrt{2}+1\right)^2}\\ =3-2\sqrt{2}+3\sqrt{2}+1\\ =4+\sqrt{2}\)

c)

\(\sqrt{12-2\sqrt{32}}+\sqrt{9+4\sqrt{2}}\\ =\sqrt{8-2\sqrt{8}\cdot\sqrt{4}+4}+\sqrt{8+2\sqrt{8}\cdot\sqrt{1}+1}\\ =\sqrt{\left(2\sqrt{2}-2\right)^2}+\sqrt{\left(2\sqrt{2}+1\right)^2}\\ =2\sqrt{2}-2+2\sqrt{2}+1\\ =4\sqrt{2}-1\)

\(\sqrt{19-8\sqrt{3}}\)

\(=\sqrt{16-2.4.\sqrt{3}+3}\)

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

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

\(\sqrt{19-8\sqrt{3}}\)

\(=\sqrt{\left(\sqrt{16}\right)^2-2.\sqrt{48}+\left(\sqrt{3}\right)^2}\)

\(=\sqrt{\left(\sqrt{16}-\sqrt{3}\right)^2}\)

\(=\sqrt{16}-\sqrt{3}\)

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

 \(A=\left(\sqrt{5}+2\right)\frac{5+4\sqrt{5}+4-8\sqrt{5}}{\sqrt{5}-2}\\ A=\left(\sqrt{5}+2\right)\frac{5-4\sqrt{5}+4}{\sqrt{5}-2}\\ A=\left(\sqrt{5}+2\right)\frac{\left(\sqrt{5}-2\right)^2}{\sqrt{5}-2}\\ A=\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)\\ A=1\)

\(B=\sqrt{11+2\sqrt{24}}-4\sqrt{2}-\sqrt{3}\)

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

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

\(\dfrac{24-9\sqrt{3}}{3-\sqrt{3}}\)

\(=\dfrac{\sqrt{3}\cdot\left(8\sqrt{3}-9\right)}{\sqrt{3}\cdot\left(\sqrt{3}-1\right)}\)

\(=\dfrac{8\sqrt{3}-9}{\sqrt{3}-1}\)

\(=\dfrac{\left(8\sqrt{3}-9\right)\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)

\(=\dfrac{8\cdot3+8\sqrt{3}-9\sqrt{3}-9}{2}\)

\(=\dfrac{15-\sqrt{3}}{2}\)

Tham khảo:

\(\dfrac{15-\sqrt{3}}{2}\)