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Làm luôn nhé
\(2B=21.2\left[\left(\sqrt{2+\sqrt{3}}+\sqrt{3-\sqrt{5}}\right)-6\left(\sqrt{2-\sqrt{3}}+\sqrt{3+\sqrt{5}}\right)\right]^2-2.15\sqrt{15}\)
\(2B=21\left(\sqrt{3}+1+\sqrt{5}-1\right)^2-6\left(\sqrt{3}-1+\sqrt{5}-1\right)^2-30\sqrt{15}\)
\(2B=21\left(\sqrt{3}+\sqrt{5}\right)^2-6\left(\sqrt{3}+\sqrt{5}\right)^2-30\sqrt{15}\)
\(2B=15\left(\sqrt{3}+\sqrt{5}\right)^2-30\sqrt{15}\)
\(2B=15\left(8+2\sqrt{15}\right)-30\sqrt{15}\)
\(2B=120+30\sqrt{15}-30\sqrt{5}\)
\(2B=120\)
\(B=60\)
Mình ghi nhầm. \(x=\frac{\sqrt{4+2\sqrt{3}}.\left(\sqrt{3}-1\right)}{\sqrt{6+2\sqrt{5}}-\sqrt{5}}\)nhé
a,A.√2= √(4+2√3)-√(4-2√3)
= √(1+√3)2 -√( √3 -1)2
= 1+√3-√3+1= 2
=> A= 2/√2=√2
B2= (4+√15)2.(4-√15).(√10-√6)2
= (4+√15).1.(16-4√15)
= (4+√15).(4-√15).4
= 4
=> B = √4 = 2
\(a,\sqrt{3+2\sqrt{2}}\)\(+\sqrt{6-4\sqrt{2}}\)
\(=\sqrt{\left(\sqrt{2}+1\right)^2}\)\(+\sqrt{\left(2-\sqrt{2}\right)^2}\)
\(=\sqrt{2}\)\(+1+2-\sqrt{2}\)
\(=\left(\sqrt{2}-\sqrt{2}\right)+1+2\)
\(=0+1+2=3\)
a) \(\sqrt{3+2\sqrt{2}}\)+\(\sqrt{6-4\sqrt{2}}\)
=\(\sqrt{\left(\sqrt{2}+1\right)^2}\)+\(\sqrt{\left(2-\sqrt{2}\right)^2}\)=\(\sqrt{2} +1+2-\sqrt{2}\)=3
b)\(\left(\sqrt{5}-2\sqrt{6}+\sqrt{2}\right)\sqrt{3}\)
=\((\sqrt{\left(\sqrt{3})-\sqrt{2}\right)^2}+\sqrt{2})\sqrt{3}\)
=\(\left(\sqrt{3}-\sqrt{2}+\sqrt{2}\right)\sqrt{3}\)
=\(\sqrt{3}.\sqrt{3}=3\)