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Câu 1,2 bạn đã đăng và có lời giải rồi
Câu 3:
\(=\frac{(\sqrt{3})^2+(2\sqrt{5})^2-2.\sqrt{3}.2\sqrt{5}}{\sqrt{2}(\sqrt{3}-2\sqrt{5})}=\frac{(\sqrt{3}-2\sqrt{5})^2}{\sqrt{2}(\sqrt{3}-2\sqrt{5})}=\frac{\sqrt{3}-2\sqrt{5}}{\sqrt{2}}\)
a: \(=\dfrac{\sqrt{3}\left(\sqrt{5}-2\right)}{\sqrt{5}-2}-3\sqrt{3}+\dfrac{\sqrt{3}\left(\sqrt{3}+\sqrt{2}\right)}{\sqrt{3}+\sqrt{2}}\)
\(=\sqrt{3}-3\sqrt{3}+\sqrt{3}=-\sqrt{3}\)
b: \(=\left(\left(2-2\sqrt{5}\right)\left(\sqrt{5}+2\right)+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\left(2\sqrt{5}+4-10-4\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\left(-2\sqrt{5}+\sqrt{3}-6\right)\left(\sqrt{5}-\sqrt{3}\right)\)
\(=-20+2\sqrt{15}+\sqrt{15}-3-6\sqrt{5}+6\sqrt{3}\)
\(=-23+3\sqrt{15}-6\sqrt{5}+6\sqrt{3}\)
a) \(\sqrt{\dfrac{3-\sqrt{5}}{3+\sqrt{5}}}\)
\(=\sqrt{\dfrac{\left(3-\sqrt{5}\right)^2}{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}}\)
\(=\dfrac{\sqrt{\left(3-\sqrt{5}\right)^2}}{\sqrt{3^2-\left(\sqrt{5}\right)^2}}\)
\(=\dfrac{\left|3-\sqrt{5}\right|}{\sqrt{9-5}}\)
\(=\dfrac{3-\sqrt{5}}{2}\)
b) \(\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}\)
\(=\sqrt{\dfrac{\left(2-\sqrt{3}\right)^2}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}}\)
\(=\dfrac{\sqrt{\left(2-\sqrt{3}\right)^2}}{\sqrt{2^2-\left(\sqrt{3}\right)^2}}\)
\(=\dfrac{\left|2-\sqrt{3}\right|}{\sqrt{4-3}}\)
\(=\dfrac{2-\sqrt{3}}{1}\)
\(=2-\sqrt{3}\)
a: \(=\sqrt{\dfrac{\left(3-\sqrt{5}\right)\left(3-\sqrt{5}\right)}{4}}=\dfrac{3-\sqrt{5}}{2}\)
b: \(=\sqrt{\dfrac{\left(2-\sqrt{3}\right)^2}{1}}=2-\sqrt{3}\)
d: \(=\left(-3+3\sqrt{6}+4+2\sqrt{6}-12-4\sqrt{6}\right)\left(\sqrt{6}+11\right)\)
=(căn 6-11)(căn 6+11)
=6-121=-115
\(A=\left(2+\dfrac{5-2\sqrt{5}}{2-\sqrt{5}}\right)\left(2+\dfrac{5+3\sqrt{5}}{3+\sqrt{5}}\right)\)
\(A=\left[2-\dfrac{\sqrt{5}\left(\sqrt{5}-2\right)}{\sqrt{5}-2}\right]\left[2+\dfrac{\sqrt{5}\left(\sqrt{5}+3\right)}{\sqrt{5}+3}\right]\)
\(A=\left(2-\sqrt{5}\right)\left(2+\sqrt{5}\right)\)
\(A=2^2-\left(\sqrt{5}\right)^2\)
\(A=4-5\)
\(A=-1\)
____
\(B=\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}+11\right)\)
\(B=\left[\dfrac{15\left(\sqrt{6}-1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}+\dfrac{4\left(\sqrt{6}+2\right)}{\left(\sqrt{6}-2\right)\left(\sqrt{6}+2\right)}-\dfrac{12\left(3+\sqrt{6}\right)}{\left(3+\sqrt{6}\right)\left(3-\sqrt{6}\right)}\right]\left(\sqrt{6}+11\right)\)
\(B=\left[\dfrac{15\left(\sqrt{6}-1\right)}{5}+\dfrac{4\left(\sqrt{6}+2\right)}{2}-\dfrac{12\left(3+\sqrt{6}\right)}{3}\right]\left(\sqrt{6}+11\right)\)
\(B=\left(3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}\right)\left(\sqrt{6}+11\right)\)
\(B=\left(\sqrt{6}-11\right)\left(\sqrt{6}+11\right)\)
\(B=6-121\)
\(B=-115\)
\(a,=\dfrac{\sqrt{5}+1+\sqrt{5}-1}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}=\dfrac{2\sqrt{5}}{4}=\dfrac{\sqrt{5}}{2}\\ b,=\sqrt{\left(3-\sqrt{5}\right)^2}+\left|2-\sqrt{5}\right|=3-\sqrt{5}+\sqrt{5}-2=1\\ c,=\dfrac{2\left(\sqrt{5}-\sqrt{3}\right)}{2}-\dfrac{-\sqrt{3}\left(\sqrt{5}-\sqrt{3}\right)}{\sqrt{5}-\sqrt{3}}=\sqrt{5}-\sqrt{3}+\sqrt{3}=\sqrt{5}\)
Làm luôn nhé
\(2B=21.2\left[\left(\sqrt{2+\sqrt{3}}+\sqrt{3-\sqrt{5}}\right)-6\left(\sqrt{2-\sqrt{3}}+\sqrt{3+\sqrt{5}}\right)\right]^2-2.15\sqrt{15}\)
\(2B=21\left(\sqrt{3}+1+\sqrt{5}-1\right)^2-6\left(\sqrt{3}-1+\sqrt{5}-1\right)^2-30\sqrt{15}\)
\(2B=21\left(\sqrt{3}+\sqrt{5}\right)^2-6\left(\sqrt{3}+\sqrt{5}\right)^2-30\sqrt{15}\)
\(2B=15\left(\sqrt{3}+\sqrt{5}\right)^2-30\sqrt{15}\)
\(2B=15\left(8+2\sqrt{15}\right)-30\sqrt{15}\)
\(2B=120+30\sqrt{15}-30\sqrt{5}\)
\(2B=120\)
\(B=60\)
\(B=\dfrac{21}{2}\left(\sqrt{4+2\sqrt{3}}+\sqrt{6-2\sqrt{5}}\right)^2-3\left(\sqrt{4-2\sqrt{3}}+\sqrt{6+2\sqrt{5}}\right)^2-15\sqrt{15}\)
\(=\dfrac{21}{2}\left(\sqrt{3}+1+\sqrt{5}-1\right)^2-3\left(\sqrt{3}-1+\sqrt{5}+1\right)^2-15\sqrt{15}\)
\(=\dfrac{21}{2}\left(\sqrt{3}+\sqrt{5}\right)^2-3\left(\sqrt{3}+\sqrt{5}\right)^2-15\sqrt{15}\)
\(=\dfrac{15}{2}\left(8+2\sqrt{15}\right)-15\sqrt{15}\)
\(=60+15\sqrt{15}-15\sqrt{15}=60\)
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.