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\(B=\left(3-\sqrt{5}\right)\sqrt{3+\sqrt{5}}+\left(3+\sqrt{5}\right)\sqrt{3-\sqrt{5}}\)
\(\Leftrightarrow\sqrt{2B}=\left(3-\sqrt{5}\right)\sqrt{2}.\sqrt{3+\sqrt{5}}+\left(3+\sqrt{5}\right)\sqrt{2}.\sqrt{3-\sqrt{5}}\)
\(\Leftrightarrow\sqrt{2B}=\left(3-\sqrt{5}\right)\sqrt{2}.\sqrt{3+\sqrt{5}}+\left(3+\sqrt{5}\right)\sqrt{6-2\sqrt{5}}\)
\(\Leftrightarrow\sqrt{2B}=\left(3-\sqrt{5}\right)\sqrt{\left(\sqrt{5}+1\right)^2}+\left(3+\sqrt{5}\right)\sqrt{\left(\sqrt{5}-1^2\right)}\)
\(\Leftrightarrow\sqrt{2B}=\left(3-\sqrt{5}\right)\left(\sqrt{5}+1\right)+\left(3+\sqrt{5}\right)\left|\sqrt{5}-1\right|\)
\(=3\sqrt{5}+3-5-\sqrt{5}+\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)\)
\(=3\sqrt{5}+3-5-\sqrt{5}+3\sqrt{5}-3+5-\sqrt{5}\)
\(=6\sqrt{5}-2\sqrt{5}=4\sqrt{5}\)
\(\Rightarrow B=\frac{4\sqrt{5}}{\sqrt{2}}=2\sqrt{10}\)
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)
\(B=b^2\cdot a+a^2\cdot b=ab\left(a+b\right)=2\sqrt{10}\)
Sửa đề
\(A=\left(2-\sqrt{3}\right)\sqrt[3]{26+15\sqrt{3}}-\left(2+\sqrt{3}\right)\sqrt[3]{26-15\sqrt{3}}\)
\(=\left(2-\sqrt{3}\right)\sqrt[3]{8+12\sqrt{3}+18+3\sqrt{3}}-\left(2+\sqrt{3}\right)\sqrt[3]{8-12\sqrt{3}+18-3\sqrt{3}}\)
\(=\left(2-\sqrt{3}\right)\sqrt[3]{\left(2+\sqrt{3}\right)^3}-\left(2+\sqrt{3}\right)\sqrt[3]{\left(2-\sqrt{3}\right)^3}\)
\(=\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)-\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)=0\)
\(1.A=\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)=5-4=1\)
\(2.B=\left(\sqrt{45}+\sqrt{63}\right)\left(\sqrt{7}-\sqrt{5}\right)=\left(3\sqrt{5}+3\sqrt{7}\right)\left(\sqrt{7}-\sqrt{5}\right)=2\left(7-5\right)=4\) \(3.C=\left(\sqrt{5}+\sqrt{3}\right)\left(5-\sqrt{15}\right)=\sqrt{5}\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)=\sqrt{5}\left(5-3\right)=2\sqrt{5}\) \(4.\left(\sqrt{32}-\sqrt{50}+\sqrt{27}\right)\left(\sqrt{27}+\sqrt{50}-\sqrt{32}\right)=\left(4\sqrt{2}-5\sqrt{2}+3\sqrt{3}\right)\left(3\sqrt{3}+5\sqrt{2}-4\sqrt{2}\right)=\left(3\sqrt{3}-\sqrt{2}\right)\left(3\sqrt{3}+\sqrt{2}\right)=27-2=25\) \(5.E=\left(\sqrt{3}+1\right)^2-2\sqrt{3}+4=4+2\sqrt{3}-2\sqrt{3}+4=8\)
\(6.F=\left(\sqrt{15}-2\sqrt{3}\right)^2+12\sqrt{5}=27-12\sqrt{5}+12\sqrt{5}=27\)
a) \(\sqrt{\left(1-\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{2}+3\right)^2}=\left|1-\sqrt{2}\right|+\left|\sqrt{2}+3\right|=\sqrt{2}-1+\sqrt{2}+3=2\sqrt{2}+2\)b) \(\sqrt{\left(\sqrt{3}-2\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}=\left|\sqrt{3}-2\right|+\left|\sqrt{3}-1\right|=2-\sqrt{3}+\sqrt{3}-1=1\)
c) \(\sqrt{\left(\sqrt{5}-3\right)^2}+\sqrt{\left(\sqrt{5}-2\right)^2}=\left|\sqrt{5}-3\right|+\left|\sqrt{5}-2\right|=3-\sqrt{5}+\sqrt{5}-2=1\)
a) \(\sqrt{26+15\sqrt{3}}\)
\(=\frac{\sqrt{52+30\sqrt{3}}}{\sqrt{2}}\)
\(=\frac{\sqrt{\left(3\sqrt{3}\right)^2+2.3\sqrt{3}.5+5^2}}{\sqrt{2}}\)
\(=\frac{\sqrt{\left(3\sqrt{3}+5\right)^2}}{\sqrt{2}}=\frac{3\sqrt{3}+5}{\sqrt{2}}\)
b) \(\)\(\sqrt{2-\sqrt{3}}=\frac{\sqrt{4-2\sqrt{3}}}{\sqrt{2}}=\frac{\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{2}}\)
\(=\frac{\left|\sqrt{3}-1\right|}{\sqrt{2}}=\frac{\sqrt{3}-1}{\sqrt{2}}\)
c) \(\left(\sqrt{10}-\sqrt{2}\right).\left(\sqrt{3+5}\right)\)
\(=\sqrt{10}.\sqrt{8}-\sqrt{2}.\sqrt{8}\)
\(=\sqrt{80}-\sqrt{16}=4\sqrt{5}-4\)
d) \(\left(\sqrt{6}-2\right)\left(5+\sqrt{24}\right)\sqrt{5-\sqrt{24}}\)
\(=\left(\sqrt{6}-2\right)\left(\sqrt{5+\sqrt{24}}\right).\sqrt{5-\sqrt{24}}.\left(\sqrt{5+\sqrt{24}}\right)\)
\(=\left(\sqrt{6}-2\right)\left(\sqrt{5+\sqrt{24}}\right).1\)
\(=\left(\sqrt{6}-2\right).\left(\sqrt{5+\sqrt{24}}\right)\)
\(=\sqrt{2}.\left(\sqrt{3}-\sqrt{2}\right).\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}\)
\(=\sqrt{2}.\left(3-2\right)=\sqrt{2}\)
a: \(=6-\sqrt{15}+2\sqrt{15}=6+\sqrt{15}\)
b: \(=\left(\sqrt{7}-2\sqrt{3}\right)\cdot\sqrt{7}+2\sqrt{21}\)
\(=7-2\sqrt{21}+2\sqrt{21}=7\)
c: \(=10+5\sqrt{10}-5\sqrt{10}=10\)
d: \(=22-\sqrt{198}+\sqrt{198}=22\)
\(A=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{8-2\sqrt{15}}\)
\(A=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)
\(A=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)^2\)
\(A=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)
\(A=2\left(4+\sqrt{15}\right)\left(4-\sqrt{15}\right)=2\)
\(B=\sqrt{6-2\sqrt{5}}\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)\)
\(B=\sqrt{\left(\sqrt{5}-1\right)^2}\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)\)
\(B=\left(\sqrt{5}-1\right)^2\left(3+\sqrt{5}\right)\)
\(B=\left(6-2\sqrt{5}\right)\left(3+\sqrt{5}\right)=2\)
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