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a: \(A=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{8-2\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)
\(=32-8\sqrt{15}+8\sqrt{15}-30=2\)
b: \(\sqrt{2}\cdot B=\left(3-\sqrt{5}\right)\left(\sqrt{5}+1\right)+\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)\)
\(\Leftrightarrow B\sqrt{2}=3\sqrt{5}+3-5-\sqrt{5}+3\sqrt{5}-3+5-\sqrt{5}\)
\(\Leftrightarrow B\sqrt{2}=4\sqrt{5}\)
hay \(B=2\sqrt{10}\)
d: \(D\sqrt{2}=\sqrt{5}+\sqrt{3}+\sqrt{5}-\sqrt{3}-2\cdot\left(\sqrt{5}-1\right)\)
\(=2\sqrt{5}-2\sqrt{5}+2=2\)
hay \(D=\sqrt{2}\)
a) \(\sqrt{6-\sqrt{11}}\cdot\sqrt{6+\sqrt{11}}\)
\(=\sqrt{\left(6-\sqrt{11}\right)\left(6+\sqrt{11}\right)}\)
\(=\sqrt{6^2-\left(\sqrt{11}\right)^2}\)
\(=\sqrt{36-11}\)
\(=\sqrt{25}\)
\(=\sqrt{5^2}\)
\(=5\)
b) \(\sqrt{8+\sqrt{15}}\cdot\sqrt{8-\sqrt{15}}\)
\(=\sqrt{\left(8+\sqrt{15}\right)\left(8-\sqrt{15}\right)}\)
\(=\sqrt{8^2-\left(\sqrt{15}\right)^2}\)
\(=\sqrt{64-15}\)
\(=\sqrt{49}\)
\(=\sqrt{7^2}\)
\(=7\)
a: \(=\sqrt{6^2-11}=\sqrt{25}=5\)
b: \(=\sqrt{8^2-15}=\sqrt{49}=7\)
Ta có: \(\left(4+\sqrt{15}\right).\left(\sqrt{10}-\sqrt{6}\right).\sqrt{4-\sqrt{15}}\)
\(=\left(\sqrt{2}.\sqrt{4+\sqrt{15}}\right).\left(\sqrt{4+\sqrt{15}}.\sqrt{4-\sqrt{15}}\right).\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\sqrt{8+2\sqrt{15}}.\left(16-15\right).\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\sqrt{\sqrt{5}+2\sqrt{5}.\sqrt{3}+\sqrt{3}}.\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}.\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\left(\sqrt{5}+\sqrt{3}\right).\left(\sqrt{5}-\sqrt{3}\right)\)
\(=5-3=2\)
Học tốt nha ^_^
Bài 2:
Ta có: \(B=\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}-\sqrt{3-2\sqrt{2}}\)
\(=\frac{\sqrt{\sqrt{5}-1}\left(\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}\right)}{2}-\sqrt{2-2\cdot\sqrt{2}\cdot1+1}\)
\(=\frac{\sqrt{3+\sqrt{5}}+\sqrt{7-3\sqrt{5}}}{2}-\sqrt{\left(\sqrt{2}-1\right)^2}\)
\(=\frac{\sqrt{6+2\sqrt{5}}+\sqrt{14-6\sqrt{5}}}{2\sqrt{2}}-\left(\sqrt{2}-1\right)\)
\(=\frac{\sqrt{5}+1+3-\sqrt{5}}{2\sqrt{2}}-\sqrt{2}+1\)
\(=\frac{4}{2\sqrt{2}}-\sqrt{2}+1\)
\(=\sqrt{2}-\sqrt{2}+1\)
=1
câu 3: C = \(\frac{\sqrt{3-\sqrt{5}}\left(\sqrt{10}-\sqrt{2}\right)\left(3+\sqrt{5}\right)}{\left(\text{4+\sqrt{15}}\right)\left(\sqrt{10-\sqrt{6}}\right)\sqrt{4-\sqrt{15}}}\)
\(=\frac{\sqrt{3-\sqrt{5}}\left(\sqrt{10}-\sqrt{2}\right)\sqrt{3+\sqrt{5}}.\sqrt{3+\sqrt{5}}}{\sqrt{4+\sqrt{15}}.\sqrt{4+\sqrt{15}}\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}}\)
=\(\frac{\sqrt{9-\left(\sqrt{5}\right)^2}\left(\sqrt{10}-\sqrt{2}\right)\sqrt{3+\sqrt{5}}}{\sqrt{16-\left(\sqrt{15}\right)^2}.\left(\sqrt{10}-\sqrt{6}\right).\sqrt{4+\sqrt{15}}}\)
\(=\frac{2\left(\sqrt{30+10\sqrt{5}}-\sqrt{6+2\sqrt{5}}\right)}{\sqrt{40+10\sqrt{15}}-\sqrt{24-6\sqrt{15}}}\)
\(=2.\frac{\left(\sqrt{5}+5\right)-\left(\sqrt{5}+1\right)}{\left(\sqrt{15}+5\right)-\left(\sqrt{15}+3\right)}\)
= 4
3: \(=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{8-2\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)
\(=32-8\sqrt{15}+8\sqrt{15}-30=2\)
4: \(=\dfrac{\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{7}-1-\sqrt{7}-1}{\sqrt{2}}=-\sqrt{2}\)
5: \(=\dfrac{\sqrt{23-8\sqrt{7}}}{3}+\dfrac{\sqrt{23+8\sqrt{7}}}{3}\)
\(=\dfrac{4-\sqrt{7}+4+\sqrt{7}}{3}=\dfrac{8}{3}\)
\(A=\sqrt{2}\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right).\sqrt{4-\sqrt{15}}\)
\(=\sqrt{2}\sqrt{4+\sqrt{15}}\left(\sqrt{5}-\sqrt{3}\right).\left(16-15\right)\)
\(=\sqrt{8+2\sqrt{15}}\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\sqrt{\left(\sqrt{3}+\sqrt{5}\right)^2}\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\left(\sqrt{3}+\sqrt{5}\right).\left(\sqrt{5}-\sqrt{3}\right)=5-3=2\)
\(B^2=8+2\sqrt{16-\left(10+2\sqrt{5}\right)}\)
\(=8+2\sqrt{6-2\sqrt{5}}\)
\(=8+2\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=8+2\sqrt{5}-2=6+2\sqrt{5}\)
Mà \(B>0\) \(\Rightarrow B=\sqrt{6+2\sqrt{5}}=\sqrt{\left(\sqrt{5}-1\right)^2}=\sqrt{5}-1\)
\(\sqrt{31+8\sqrt{15}}=\sqrt{31+2\sqrt{240}}=\sqrt{16+2\sqrt{15.16}+15}=4+\sqrt{15}\)
Thay zô ta đc
\(\frac{4+\sqrt{15}}{\sqrt{4+\sqrt{15}}}.\sqrt{4-\sqrt{15}}=\sqrt{4+\sqrt{15}}.\sqrt{4-\sqrt{15}}=1\)