Rút gọn các biểu thức sau:
A =\(\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}\)
B = \(\sqrt{3-\sqrt{5}}.\left(\sqrt{10}-\sqrt{2}\right)\left(3+\sqrt{5}\right)\)
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a) Ta có: \(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{20-2\cdot\sqrt{20}\cdot3+9}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{3-\left(2\sqrt{5}-3\right)}}\)
\(=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{5-2\cdot\sqrt{5}\cdot1+1}}\)
\(=\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}\)
\(=\sqrt{\sqrt{5}-\left(\sqrt{5}-1\right)}\)
\(=\sqrt{\sqrt{5}-\sqrt{5}+1}\)
\(=\sqrt{1}=1\)
b) Ta có: \(\sqrt{6+2\sqrt{5}-\sqrt{29-12\sqrt{5}}}\)
\(=\sqrt{6+2\sqrt{5}-\sqrt{20-2\cdot2\sqrt{5}\cdot3+9}}\)
\(=\sqrt{6+2\sqrt{5}-\sqrt{\left(2\sqrt{5}-3\right)^2}}\)
\(=\sqrt{6+2\sqrt{5}-\left(2\sqrt{5}-3\right)}\)
\(=\sqrt{6+3}=3\)
c) Sửa đề: \(\sqrt{2+\sqrt{5+\sqrt{13-\sqrt{48}}}}\)
Ta có: \(\sqrt{2+\sqrt{5+\sqrt{13-\sqrt{48}}}}\)
\(=\sqrt{2+\sqrt{5+\sqrt{12-2\cdot2\sqrt{3}\cdot1+1}}}\)
\(=\sqrt{2+\sqrt{5+\sqrt{\left(2\sqrt{3}-1\right)^2}}}\)
\(=\sqrt{2+\sqrt{5+2\sqrt{3}-1}}\)
\(=\sqrt{2+\sqrt{3+2\sqrt{3}\cdot1+1}}\)
\(=\sqrt{2+\sqrt{\left(\sqrt{3}+1\right)^2}}\)
\(=\sqrt{3+\sqrt{3}}\)
d) Ta có: \(\left(3-\sqrt{5}\right)\sqrt{3+\sqrt{5}}+\left(3+\sqrt{5}\right)\sqrt{3-\sqrt{5}}\)
\(=\dfrac{\left(6-2\sqrt{5}\right)\sqrt{6+2\sqrt{5}}+\left(6+2\sqrt{5}\right)\sqrt{6-2\sqrt{5}}}{2\sqrt{2}}\)
\(=\dfrac{\left(\sqrt{5}-1\right)^2\cdot\left(\sqrt{5}+1\right)+\left(\sqrt{5}+1\right)^2\cdot\left(\sqrt{5}-1\right)}{2\sqrt{2}}\)
\(=\dfrac{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)\left(\sqrt{5}-1+\sqrt{5}+1\right)}{2\sqrt{2}}\)
\(=\dfrac{4\cdot2\sqrt{5}}{2\sqrt{2}}\)
\(=\dfrac{8\sqrt{5}}{2\sqrt{2}}=\dfrac{4\sqrt{5}}{\sqrt{2}}=2\sqrt{10}\)
a: \(\sqrt{\left(4-\sqrt{15}\right)^2}+\sqrt{15}\)
\(=4-\sqrt{15}+\sqrt{15}=4\)
b: \(\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}\)
\(=2+\sqrt{3}-2+\sqrt{3}\)
\(=2\sqrt{3}\)
c: \(\sqrt{29+12\sqrt{5}}-\sqrt{29-12\sqrt{5}}\)
\(=\sqrt{\left(2\sqrt{5}+3\right)^2}-\sqrt{\left(2\sqrt{5}-3\right)^2}\)
\(=2\sqrt{5}+3-2\sqrt{5}+3=6\)
a)
\(\left(3-\sqrt{15}\right)\sqrt{4+\sqrt{15}}\\ =\left(3-\sqrt{15}\right)\cdot\dfrac{\sqrt{8+2\sqrt{15}}}{\sqrt{2}}\\ =\left(3-\sqrt{15}\right)\cdot\dfrac{\sqrt{5+2\sqrt{15}+3}}{\sqrt{2}}\\ =\left(3-\sqrt{15}\right)\cdot\dfrac{\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}}{\sqrt{2}}\\ =\left(\sqrt{9}-\sqrt{15}\right)\cdot\dfrac{\left|\sqrt{5}+\sqrt{3}\right|}{\sqrt{2}}\)
\(=\sqrt{3}\left(\sqrt{3}-\sqrt{5}\right)\cdot\dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{2}}\) (vì \(\sqrt{5}+\sqrt{3}>0\))
\(=\sqrt{3}\cdot\dfrac{3-5}{\sqrt{2}}\\ =\sqrt{3}\cdot\dfrac{-2}{\sqrt{2}}\\ =\sqrt{3}\cdot\dfrac{-\sqrt{4}}{\sqrt{2}}\\ =-\sqrt{6}\)
b)
\(\sqrt{29-12\sqrt{5}}-\sqrt{24-8\sqrt{5}}\\ =\sqrt{20-2\cdot3\cdot2\sqrt{5}+9}-\sqrt{20-2\cdot2\cdot2\sqrt{5}+4}\\ =\sqrt{\left(2\sqrt{5}-3\right)^2}-\sqrt{\left(2\sqrt{5}-2\right)^2}\\ =\left|2\sqrt{5}-3\right|-\left|2\sqrt{5}-2\right|\)
\(=2\sqrt{5}-3-\left(2\sqrt{5}-2\right)\) (vì \(2\sqrt{5}-3>0;2\sqrt{5}-2>0\))
\(=2\sqrt{5}-3-2\sqrt{5}+2\\ =-1\)
9: \(A=\dfrac{\sqrt{8+2\sqrt{15}}-\sqrt{14-6\sqrt{5}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{5}+\sqrt{3}-3+\sqrt{5}}{\sqrt{2}}=\dfrac{2\sqrt{10}+\sqrt{6}-3\sqrt{2}}{2}\)
10: \(A=\dfrac{\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{3}+1+\sqrt{3}-1}{\sqrt{2}}=\dfrac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}\)
11: \(A=\dfrac{\sqrt{24-6\sqrt{7}}-\sqrt{24+6\sqrt{7}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{21}-\sqrt{3}-\sqrt{21}-\sqrt{3}}{\sqrt{2}}=-\dfrac{2\sqrt{3}}{\sqrt{2}}=-\sqrt{6}\)
12: \(B=\left(3+\sqrt{3}\right)\sqrt{12-6\sqrt{3}}\)
\(=\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)\)
=9-3=6
13: \(A=\sqrt{5}-2-\left(3-\sqrt{5}\right)\)
\(=\sqrt{5}-2-3+\sqrt{5}=2\sqrt{5}-5\)
Phần a sai đề sửa đề
\(A=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-{12\sqrt{5}}}}}\)
=\(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{(2\sqrt{5}-3)^2 } } } \)
=\(\sqrt{5-\sqrt{3-2\sqrt{5}+3 }}\)
=\(\sqrt{\sqrt{5}-\sqrt{(\sqrt{5}-1)^2 } } \)
=\(\sqrt{\sqrt{5}-\sqrt{5}+1 } \)
=1
B=\((\sqrt{4+\sqrt{15} }) \sqrt{2}(\sqrt{5}-\sqrt{3})(\sqrt{4-\sqrt{15} })({\sqrt{4+\sqrt{15} }) } \)
=(\((\sqrt{4+\sqrt{15} })\sqrt{2}(\sqrt{5}-\sqrt{3}) \)
=\((\sqrt{8+2\sqrt{15} })(\sqrt{5}-\sqrt{3}) \)
=\((\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3}) \)
=2
\(\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}\)
\(=\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}\)
\(=\sqrt{5}-\sqrt{6-2\sqrt{5}}\)
\(=\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}\)
= 1
\(\sqrt{3-\sqrt{5}}\left(\sqrt{10}-\sqrt{2}\right)\left(3+\sqrt{5}\right)\)
\(=\sqrt{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}\times\sqrt{3+\sqrt{5}}\times\sqrt{2}\left(\sqrt{5}-1\right)\)
\(=\sqrt{9-5}\times\sqrt{6+2\sqrt{5}}\left(\sqrt{5}-1\right)\)
\(=2\sqrt{\left(\sqrt{5}+1\right)^2}\left(\sqrt{5}-1\right)\)
\(=2\left(5-1\right)\)
= 8
a) \(A=\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}\)
\(=\sqrt{5}-\sqrt{3-\sqrt{\left(3-2\sqrt{5}\right)^2}}\)
\(=\sqrt{5}-\sqrt{3-\left(2\sqrt{5}-3\right)}\)
\(=\sqrt{5}-\sqrt{3-2\sqrt{5}+3}\)
\(=\sqrt{5}-\sqrt{6-2\sqrt{5}}\)
\(=\sqrt{5}-\sqrt{\left(1-\sqrt{5}\right)^2}\)
\(=\sqrt{5}-\left(\sqrt{5}-1\right)\)
\(=\sqrt{5}-\sqrt{5}+1\)
\(=1\)
b) \(B=\sqrt{3-\sqrt{5}}\cdot\left(\sqrt{10}-\sqrt{2}\right)\left(3+\sqrt{5}\right)\)
\(=\sqrt{3-\sqrt{5}}\left(3\sqrt{10}+\sqrt{50}-3\sqrt{2}-\sqrt{10}\right)\)
\(=\sqrt{3-\sqrt{5}}\left(3\sqrt{10}+5\sqrt{2}-3\sqrt{2}-\sqrt{10}\right)\)
\(=\sqrt{3-\sqrt{5}}\left(2\sqrt{10}+2\sqrt{2}\right)\)
\(=\sqrt{3-\sqrt{5}}\sqrt{\left(2\sqrt{10}+2\sqrt{2}\right)^2}\)
\(=\sqrt{\left(3-\sqrt{5}\right)\left(2\sqrt{10}+2\sqrt{2}\right)^2}\)
\(=\sqrt{\left(3-\sqrt{5}\right)\left(4\cdot10+8\sqrt{20}+4\cdot2\right)}\)
\(=\sqrt{\left(3-\sqrt{5}\right)\left(40+16\sqrt{5}+8\right)}\)
\(=\sqrt{\left(3-\sqrt{5}\right)\left(48+16\sqrt{5}\right)}\)
\(=\sqrt{\left(3-\sqrt{5}\right)\cdot16\left(3+\sqrt{5}\right)}\)
\(=\sqrt{\left(9-5\right)\cdot16}\)
\(=\sqrt{4\cdot16}\)
\(=\sqrt{64}\)
\(=8\)