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Làm luôn nhé
\(2B=21.2\left[\left(\sqrt{2+\sqrt{3}}+\sqrt{3-\sqrt{5}}\right)-6\left(\sqrt{2-\sqrt{3}}+\sqrt{3+\sqrt{5}}\right)\right]^2-2.15\sqrt{15}\)
\(2B=21\left(\sqrt{3}+1+\sqrt{5}-1\right)^2-6\left(\sqrt{3}-1+\sqrt{5}-1\right)^2-30\sqrt{15}\)
\(2B=21\left(\sqrt{3}+\sqrt{5}\right)^2-6\left(\sqrt{3}+\sqrt{5}\right)^2-30\sqrt{15}\)
\(2B=15\left(\sqrt{3}+\sqrt{5}\right)^2-30\sqrt{15}\)
\(2B=15\left(8+2\sqrt{15}\right)-30\sqrt{15}\)
\(2B=120+30\sqrt{15}-30\sqrt{5}\)
\(2B=120\)
\(B=60\)
\(=\sqrt{3-\sqrt{5}}.\sqrt{2}.\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)\)
\(=\sqrt{6-2\sqrt{5}}\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)\)
\(=\sqrt{\left(\sqrt{5}-1\right)^2}\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)\)
\(=\left(\sqrt{5}-1\right)\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)\)
\(=\left(\sqrt{5}-1\right)^2\left(3+\sqrt{5}\right)\)
\(=\left(6-2\sqrt{5}\right)\left(3+\sqrt{5}\right)\)
\(=2\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)\)
\(=2\left(9-\left(\sqrt{5}\right)^2\right)\)
\(=2.4=8\)
Chỉ vậy thôi nha bạn ^_^
\(C=\sqrt{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}.\sqrt{3+\sqrt{5}.}\sqrt{2}\left(\sqrt{5}-1\right)\)
\(C=\sqrt{4}.\sqrt{6+2\sqrt{5}}\left(\sqrt{5}-1\right)\)
\(C=2.\sqrt{\left(\sqrt{5}+1\right)^2}.\left(\sqrt{5}-1\right)\)
\(C=2.\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)=2.4=8\)
b) \(\sqrt{\left(7-\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(=7-\sqrt{3}+\sqrt{3}+1\)
\(=8\)
a) \(\left(2\sqrt{3}+\sqrt{5}\right)\sqrt{3}-\sqrt{60}\) = \(6+\sqrt{15}-2\sqrt{15}\)
= \(6-\sqrt{15}\)
b) \(\left(5\sqrt{2}+2\sqrt{5}\right)\sqrt{5}-\sqrt{250}\) = \(5\sqrt{10}+10-5\sqrt{10}\) = \(10\)
c) \(\left(\sqrt{28}-\sqrt{12}-\sqrt{7}\right)\sqrt{7}+2\sqrt{21}\) = \(14-2\sqrt{21}-7+2\sqrt{21}\)
= \(7\)
d) \(\left(\sqrt{99}-\sqrt{18}-\sqrt{11}\right)\sqrt{11}+3\sqrt{22}\)
= \(33-3\sqrt{22}-11+3\sqrt{22}\) = \(22\)
a)(2√3+√5)√3-√60
=6+√15-2√15
=6-√15
b)(5√2+2√5)√5-√250
=5√10+10-5√10
=10
c)(√28-√12-√7)√7+2√21
=14-2√21-7+2√21
=7
d)(√99-√18-√11)√11+3√22
=33-3√22-11+3√22
=22
\(B=\left(3-\sqrt{5}\right)\sqrt{3+\sqrt{5}}+\left(3+\sqrt{5}\right)\sqrt{3-\sqrt{5}}\)
\(\Leftrightarrow\sqrt{2B}=\left(3-\sqrt{5}\right)\sqrt{2}.\sqrt{3+\sqrt{5}}+\left(3+\sqrt{5}\right)\sqrt{2}.\sqrt{3-\sqrt{5}}\)
\(\Leftrightarrow\sqrt{2B}=\left(3-\sqrt{5}\right)\sqrt{2}.\sqrt{3+\sqrt{5}}+\left(3+\sqrt{5}\right)\sqrt{6-2\sqrt{5}}\)
\(\Leftrightarrow\sqrt{2B}=\left(3-\sqrt{5}\right)\sqrt{\left(\sqrt{5}+1\right)^2}+\left(3+\sqrt{5}\right)\sqrt{\left(\sqrt{5}-1^2\right)}\)
\(\Leftrightarrow\sqrt{2B}=\left(3-\sqrt{5}\right)\left(\sqrt{5}+1\right)+\left(3+\sqrt{5}\right)\left|\sqrt{5}-1\right|\)
\(=3\sqrt{5}+3-5-\sqrt{5}+\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)\)
\(=3\sqrt{5}+3-5-\sqrt{5}+3\sqrt{5}-3+5-\sqrt{5}\)
\(=6\sqrt{5}-2\sqrt{5}=4\sqrt{5}\)
\(\Rightarrow B=\frac{4\sqrt{5}}{\sqrt{2}}=2\sqrt{10}\)
Đặt \(\sqrt{3+\sqrt{5}}=a>0;\sqrt{3-\sqrt{5}}=b>0\Rightarrow ab=\sqrt{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}=\sqrt{3^2-5}=2\)
Và \(a^2+b^2=6 \Rightarrow\left(a+b\right)^2=a^2+b^2+2ab=6+4=10\Rightarrow a+b=\sqrt{10}\) (vì a + b > 0 do a > 0,b>0)
\(B=b^2\cdot a+a^2\cdot b=ab\left(a+b\right)=2\sqrt{10}\)