\(VT=\left[\dfrac{\sqrt{7}\left(\sqrt{2}-1\right)}{1-\sqrt{2}}+\dfrac{\sqrt{5}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}\right]\cdot\left(\sqrt{7}-\sqrt{5}\right)\\ =\left(-\sqrt{7}-\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)\\ =-\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)=-\left(7-5\right)=-2=VP\)

\(\left(\dfrac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\dfrac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right):\dfrac{1}{\sqrt{7}-\sqrt{5}}=\left(-\sqrt{7}-\sqrt{5}\right).\left(\sqrt{7}-\sqrt{5}\right)=-\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)=-\left(7-5\right)=-2\)

a) \(=2\sqrt{5}-3\sqrt{5}+\sqrt{5}-1=-1\)

b) \(=\left[\sqrt{14}+\dfrac{\sqrt{6}\left(\sqrt{2}+\sqrt{5}\right)}{\sqrt{2}+\sqrt{5}}\right].\sqrt{\left(\sqrt{\dfrac{7}{2}}-\sqrt{\dfrac{3}{2}}\right)^2}\)

\(=\left(\sqrt{14}+\sqrt{6}\right)\left(\sqrt{\dfrac{7}{2}}-\sqrt{\dfrac{3}{2}}\right)\)

\(=\sqrt{49}-\sqrt{21}+\sqrt{21}-\sqrt{9}\)

\(=7-3=4\)

a) \(\sqrt{20}-15\sqrt{\dfrac{1}{5}}+\sqrt{\left(1-\sqrt{5}\right)^2}=2\sqrt{5}-3\sqrt{5}+\sqrt{5}-1=-1\)

b) \(\left(\dfrac{14}{\sqrt{14}}+\dfrac{\sqrt{12}+\sqrt{30}}{\sqrt{2}+\sqrt{5}}\right).\sqrt{5-\sqrt{21}}=\left(\sqrt{14}+\sqrt{6}\right)\sqrt{5-\sqrt{21}}=\left(\sqrt{7}+\sqrt{3}\right)\sqrt{10-2\sqrt{21}}=\left(\sqrt{7}+\sqrt{3}\right)\left(\sqrt{7}-\sqrt{3}\right)=4\)

\(=\sqrt{25-\sqrt{21}}\left(\sqrt{14}+\sqrt{6}\right)\)

Ta có: \(\dfrac{2\sqrt{3-\sqrt{5}}\left(3+\sqrt{5}\right)}{\sqrt{10}-\sqrt{2}}-\dfrac{\sqrt{15}+\sqrt{5}}{\sqrt{12}+\sqrt{2}}\)

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

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

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

\(=\dfrac{\sqrt{2}\left(3\sqrt{6}+3+\sqrt{30}+\sqrt{5}\right)-\sqrt{15}-\sqrt{5}}{\sqrt{2}\left(\sqrt{6}+1\right)}\)

\(=\dfrac{6\sqrt{3}+3\sqrt{2}+2\sqrt{15}+\sqrt{10}-\sqrt{15}-\sqrt{5}}{\sqrt{2}\left(\sqrt{6}+1\right)}\)

\(=\dfrac{6\sqrt{3}+3\sqrt{2}+\sqrt{15}+\sqrt{10}-\sqrt{5}}{ }\)

Đề sai rồi bạn

Ta có: \(\dfrac{2\sqrt{3-\sqrt{5}}\left(3+\sqrt{5}\right)}{\sqrt{10}-\sqrt{2}}-\dfrac{\sqrt{15}+\sqrt{5}}{\sqrt{12}+2}\)

\(=\dfrac{\left(6+2\sqrt{5}\right)\sqrt{6-2\sqrt{5}}}{\sqrt{20}-2}-\dfrac{\sqrt{5}\left(\sqrt{3}+1\right)}{2\left(\sqrt{3}+1\right)}\)

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

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

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

\(VT=\left(\dfrac{\sqrt{14}.\sqrt{14}}{\sqrt{14}}+\dfrac{\sqrt{6}\left(\sqrt{2}+\sqrt{5}\right)}{\sqrt{2}+\sqrt{5}}\right)\sqrt{5-\sqrt{21}}\\ =\left(\sqrt{14}+\sqrt{6}\right)\sqrt{5-\sqrt{21}}\\ =\sqrt{14}.\sqrt{5-\sqrt{2}1}+\sqrt{6}.\sqrt{5-\sqrt{21}}\\ =\sqrt{70-14\sqrt{21}}+\sqrt{30-6\sqrt{21}}\\ =\sqrt{49-2.7.\sqrt{21}+21}+\sqrt{9-2.3.\sqrt{21}+21}\\ =\sqrt{\left(7-\sqrt{21}\right)^2}+\sqrt{\left(3-\sqrt{21}\right)^2}\\ =\left|7-\sqrt{21}\right|+\left|3-\sqrt{21}\right|\\ =7-\sqrt{21}+\sqrt{21}-3\\ =7-3=4=VP\)

help me now

\(VT=\left(\sqrt{14}+\sqrt{6}\right)\cdot\sqrt{5-\sqrt{21}}\)

\(=\left(\sqrt{7}+\sqrt{3}\right)\cdot\sqrt{10-2\sqrt{21}}\)

=(căn 7+căn 3)(căn 7-căn 3)

=7-3

=4=VP

Bài 1

a) Đặt VT = A

<=> \(2\sqrt{2}A=\left(8+2\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{8-2\sqrt{15}}\)

<=> \(2\sqrt{2}A=\left(\sqrt{5}+\sqrt{3}\right)^2.\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right).\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)

<=> \(2A=\left(\sqrt{5}+\sqrt{3}\right)^2.\left(\sqrt{5}-\sqrt{3}\right)^2\)

<=> 2A = \(\left(5-3\right)^2=4\)

<=> A = 2

b) Đặt VT = B

<=> \(2\sqrt{2}B=\left(10+2\sqrt{21}\right).\left(\sqrt{14}-\sqrt{6}\right)\sqrt{10-2\sqrt{21}}\)

<=> \(2\sqrt{2}B=\left(\sqrt{7}+\sqrt{3}\right)^2.\sqrt{2}\left(\sqrt{7}-\sqrt{3}\right).\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}\)

<=> \(2B=\left(\sqrt{7}+\sqrt{3}\right)^2.\left(\sqrt{7}-\sqrt{3}\right)^2=\left(7-3\right)^2=16\)

<=> B = 8 

Bài 2

Đặt VT = A

<=> A² = \(\dfrac{\sqrt{5}+2+\sqrt{5}-2+2\sqrt{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}}{2}\)

<=> A² = \(\dfrac{2\sqrt{5}+2\sqrt{5-4}}{2}=\dfrac{2\sqrt{5}+2}{2}=\sqrt{5}+1\)

<=> \(A=\sqrt{\sqrt{5}+1}\)

\(\left(4-\sqrt{7}\right)^2=4^2-2\cdot4\cdot\sqrt{7}+7\)

\(=16-8\sqrt{7}+7=23-8\sqrt{7}\)

\(\sqrt{9-4\sqrt{5}}-\sqrt{5}\)

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

\(=\sqrt{\left(\sqrt{5}-2\right)^2}-\sqrt{5}\)

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

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

\(\dfrac{\sqrt{4-2\sqrt{3}}}{1+\sqrt{2}}:\dfrac{\sqrt{2}-1}{\sqrt{3}+1}\)

\(=\dfrac{\sqrt{3-2\cdot\sqrt{3}\cdot1+1}}{\sqrt{2}+1}\cdot\dfrac{\sqrt{3}+1}{\sqrt{2}-1}\)

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

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

\(\left(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\dfrac{\sqrt{216}}{3}\right)\cdot\dfrac{1}{\sqrt{6}}\)

\(=\left(\dfrac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}-\dfrac{6\sqrt{6}}{3}\right)\cdot\dfrac{1}{\sqrt{6}}\)

\(=\left(\dfrac{1}{2}\sqrt{6}-2\sqrt{6}\right)\cdot\dfrac{1}{\sqrt{6}}\)

\(=\dfrac{1}{2}-2=-\dfrac{3}{2}=-1,5\)