\(B=\dfrac{21}{2}\left(\sqrt{4+2\sqrt{3}}+\sqrt{6-2\sqrt{5}}\right)^2-3\left(\sqrt{4-2\sqrt{3}}+\sqrt{6+2\sqrt{5}}\right)^2-15\sqrt{15}\)

\(=\dfrac{21}{2}\left(\sqrt{3}+1+\sqrt{5}-1\right)^2-3\left(\sqrt{3}-1+\sqrt{5}+1\right)^2-15\sqrt{15}\)

\(=\dfrac{21}{2}\left(\sqrt{3}+\sqrt{5}\right)^2-3\left(\sqrt{3}+\sqrt{5}\right)^2-15\sqrt{15}\)

\(=\dfrac{15}{2}\left(8+2\sqrt{15}\right)-15\sqrt{15}\)

\(=60+15\sqrt{15}-15\sqrt{15}=60\)

Sửa đề

\(A=\left(2-\sqrt{3}\right)\sqrt[3]{26+15\sqrt{3}}-\left(2+\sqrt{3}\right)\sqrt[3]{26-15\sqrt{3}}\)

\(=\left(2-\sqrt{3}\right)\sqrt[3]{8+12\sqrt{3}+18+3\sqrt{3}}-\left(2+\sqrt{3}\right)\sqrt[3]{8-12\sqrt{3}+18-3\sqrt{3}}\)

\(=\left(2-\sqrt{3}\right)\sqrt[3]{\left(2+\sqrt{3}\right)^3}-\left(2+\sqrt{3}\right)\sqrt[3]{\left(2-\sqrt{3}\right)^3}\)

\(=\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)-\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)=0\)

Hơi mờ nha cậu

Làm luôn nhé

\(2B=21.2\left[\left(\sqrt{2+\sqrt{3}}+\sqrt{3-\sqrt{5}}\right)-6\left(\sqrt{2-\sqrt{3}}+\sqrt{3+\sqrt{5}}\right)\right]^2-2.15\sqrt{15}\)

\(2B=21\left(\sqrt{3}+1+\sqrt{5}-1\right)^2-6\left(\sqrt{3}-1+\sqrt{5}-1\right)^2-30\sqrt{15}\)

\(2B=21\left(\sqrt{3}+\sqrt{5}\right)^2-6\left(\sqrt{3}+\sqrt{5}\right)^2-30\sqrt{15}\)

\(2B=15\left(\sqrt{3}+\sqrt{5}\right)^2-30\sqrt{15}\)

\(2B=15\left(8+2\sqrt{15}\right)-30\sqrt{15}\)

\(2B=120+30\sqrt{15}-30\sqrt{5}\)

\(2B=120\)

\(B=60\)

\(=\frac{21}{2}\left(\sqrt{4+2\sqrt{3}}+\sqrt{6-2\sqrt{5}}\right)^2-3\left(\sqrt{4-2\sqrt{3}}+\sqrt{6+2\sqrt{5}}\right)^2-15\sqrt{15}\)

\(=\frac{21}{2}\left(\sqrt{3}+1+\sqrt{5}-1\right)^2-3\left(\sqrt{3}-1+\sqrt{5}+1\right)^2-15\sqrt{15}\)

\(=\frac{15}{2}\left(\sqrt{3}+\sqrt{5}\right)^2-15\sqrt{15}\)

\(=\frac{15}{2}\left(8+2\sqrt{15}\right)-15\sqrt{15}\)

\(=60+15\sqrt{15}-15\sqrt{15}=60\)

câu b trc nha

B = \(\dfrac{4+\sqrt{2}-\sqrt{3}-\sqrt{6}+\sqrt{8}}{2+\sqrt{2}-\sqrt{3}}\)

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

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

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

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

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

A = \(\dfrac{21}{2}\) . (\(\sqrt{4+2\sqrt{3}}\) + \(\sqrt{6-2\sqrt{5}}\) )² - 15\(\sqrt{15}\)

- 3(\(\sqrt{4-2\sqrt{3}}\) +\(\sqrt{6+2\sqrt{5}}\) )²

= \(\dfrac{21}{2}\).(\(\sqrt{\left(\sqrt{3}+1\right)^2}\) + \(\sqrt{\left(\sqrt{5}-1\right)^2}\))²-15\(\sqrt{15}\)

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

= \(\dfrac{21}{2}\).(\(\sqrt{3}\) +1+ \(\sqrt{5}\) - 1)² -3.(\(\sqrt{3}\) - 1 + \(\sqrt{5}\) +1)²

- 15\(\sqrt{15}\)

= \(\dfrac{21}{2}\).(8+2\(\sqrt{15}\) ) - 3(8 + 2\(\sqrt{15}\) ) -15\(\sqrt{15}\)

= \(\dfrac{15}{2}\) .2.(4+\(\sqrt{15}\) ) - 15\(\sqrt{15}\)

= 15.( 4 + \(\sqrt{15}\) ) - 15\(\sqrt{15}\)

= 15.(4+\(\sqrt{15}\) -\(\sqrt{15}\)) =15.4 = 60

Vậy A = 60.

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

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

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

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

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

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

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

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

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

\(B=31\)

_____________________________

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

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

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

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

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