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a: \(D=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{8-2\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)
\(=32-8\sqrt{15}+8\sqrt{15}-30=2\)
b: \(E=\sqrt{6-2\sqrt{5}}\cdot\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)\)
\(=\left(6-2\sqrt{5}\right)\left(3+\sqrt{5}\right)\)
\(=18+6\sqrt{5}-6\sqrt{5}-10=8\)
\(1,\sqrt{\left(2+\sqrt{7}\right)^2-\sqrt{\left(2-\sqrt{7}\right)^2}}\) ( áp dụng hđt thứ 3 \(a^2-b^2=\left(a-b\right)\left(a+b\right)\))
\(=\sqrt{\left(2+\sqrt{7}+2-\sqrt{7}\right)\left(2+\sqrt{7}-2+\sqrt{7}\right)}\)
\(=\sqrt{4\cdot\sqrt{7}}\)
\(2,\sqrt{\left(3\sqrt{5}-5\sqrt{2}\right)^2}-\sqrt{\left(5\sqrt{2}+3\sqrt{5}\right)^2}\)
\(\Leftrightarrow\sqrt{\left(3\sqrt{5}-5\sqrt{2}\right)^2}=\sqrt{\left(5\sqrt{2}+3\sqrt{5}\right)^2}\)
\(\Leftrightarrow\left(3\sqrt{5}-5\sqrt{2}\right)^2=\left(5\sqrt{2}+3\sqrt{5}\right)^2\)
\(\Leftrightarrow\left(3\sqrt{5}-5\sqrt{2}\right)^2-\left(5\sqrt{2}+3\sqrt{5}\right)^2\)
\(=\left(3\sqrt{5}-5\sqrt{2}+5\sqrt{2}+3\sqrt{5}\right)\left(3\sqrt{5}-5\sqrt{2}-5\sqrt{2}-3\sqrt{5}\right)\)
\(=6\sqrt{5}\cdot\left(-10\sqrt{2}\right)\)
\(3,\sqrt{10+2\sqrt{21}}-\sqrt{10-2\sqrt{21}}\)
\(\Leftrightarrow\sqrt{10+2\sqrt{21}}=\sqrt{10-2\sqrt{21}}\)
\(\Leftrightarrow10+2\sqrt{21}=10-2\sqrt{21}\)
\(\Leftrightarrow4\sqrt{21}\)
cuối lười tính nên thôi nhá :>
a) \(\Leftrightarrow\left(\sqrt{3+\sqrt{5}}\right)^2\left(\sqrt{10}-\sqrt{2}\right)\sqrt{3-\sqrt{5}}=8\)
\(\Leftrightarrow\sqrt{3+\sqrt{5}}\cdot\left(\sqrt{10}-\sqrt{2}\right)\sqrt{\left(3-\sqrt{5}\right)\cdot\left(3+\sqrt{5}\right)}=8\)
\(\Leftrightarrow\sqrt{\frac{6+2\sqrt{5}}{2}}\cdot\left(\sqrt{5}\sqrt{2}-\sqrt{2}\right)\sqrt{3^2-5}=8\).
\(\Leftrightarrow\sqrt{\frac{5+2\sqrt{5}+1}{2}}\cdot\sqrt{2}\cdot\left(\sqrt{5}-1\right)\cdot\sqrt{4}=8\)
\(\Leftrightarrow\frac{\sqrt{\left(\sqrt{5}+1\right)^2}}{\sqrt{2}}\cdot\sqrt{2}\cdot\left(\sqrt{5}-1\right)\cdot2=8\)
\(\Leftrightarrow\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)=4\Leftrightarrow\left(\sqrt{5}\right)^2-1=4\Leftrightarrow5-1=4\)Đúng -ĐPCM.
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\(f,\sqrt{\dfrac{3-\sqrt{5}}{2-\sqrt{3}}}\\ =\sqrt{\dfrac{\left(3-\sqrt{5}\right)\left(2+\sqrt{3}\right)}{4-3}}\\ =\sqrt{\left(3-\sqrt{5}\right)\left(2+\sqrt{3}\right)}\\ =\sqrt{\dfrac{\left(6-2\sqrt{5}\right)\left(4+2\sqrt{3}\right)}{4}}\\ =\dfrac{\left(\sqrt{5}-1\right)\left(\sqrt{3}+1\right)}{2}\)
\(a,\sqrt{3+\sqrt{5}}\left(\sqrt{10}+\sqrt{2}\right)\left(3-\sqrt{5}\right)\\ =\sqrt{3+\sqrt{5}}.\sqrt{3-\sqrt{5}}.\sqrt{3-\sqrt{5}}.\sqrt{2}\left(\sqrt{5}+1\right)\\ =\sqrt{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}.\sqrt{6-2\sqrt{5}}.\left(\sqrt{5}+1\right)\\ =\sqrt{9-5}.\sqrt{\left(\sqrt{5}-1\right)^2}.\left(\sqrt{5}+1\right)\\ =2\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)\\ =2.4\\ =8\)
a) Ta có: \(\left(\sqrt{6}+\sqrt{2}\right)\cdot\left(\sqrt{3}-2\right)\cdot\left(\sqrt{2+\sqrt{3}}\right)\)
\(=\sqrt{2}\cdot\left(\sqrt{3}+1\right)\cdot\left(\sqrt{3}-2\right)\cdot\sqrt{2+\sqrt{3}}\)
\(=\sqrt{4+2\sqrt{3}}\cdot\left(\sqrt{3}+1\right)\cdot\left(\sqrt{3}-2\right)\)
\(=\sqrt{3+2\cdot\sqrt{3}\cdot1+1}\cdot\left(\sqrt{3}+1\right)\cdot\left(\sqrt{3}-2\right)\)
\(=\sqrt{\left(\sqrt{3}+1\right)^2}\cdot\left(\sqrt{3}+1\right)\cdot\left(\sqrt{3}-2\right)\)
\(=\left|\sqrt{3}+1\right|\cdot\left(\sqrt{3}+1\right)\cdot\left(\sqrt{3}-2\right)\)
\(=\left(\sqrt{3}+1\right)\cdot\left(\sqrt{3}+1\right)\cdot\left(\sqrt{3}-2\right)\)(Vì \(\sqrt{3}>1>0\))
\(=\left(4+2\sqrt{3}\right)\cdot\left(\sqrt{3}-2\right)\)
\(=2\cdot\left(\sqrt{3}+2\right)\left(\sqrt{3}-2\right)\)
\(=2\cdot\left(3-4\right)\)
\(=-2\)
b) Ta có: \(\sqrt{2}\cdot\left(\sqrt{2-\sqrt{3}}\right)\cdot\left(\sqrt{3}+1\right)\)
\(=\sqrt{4-2\sqrt{3}}\cdot\left(\sqrt{3}+1\right)\)
\(=\sqrt{3-2\cdot\sqrt{3}\cdot1+1}\cdot\left(\sqrt{3}+1\right)\)
\(=\sqrt{\left(\sqrt{3}-1\right)^2}\cdot\left(\sqrt{3}+1\right)\)
\(=\left|\sqrt{3}-1\right|\cdot\left(\sqrt{3}+1\right)\)
\(=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)\)(Vì \(\sqrt{3}>1\))
\(=3-1=2\)
c) Ta có: \(\left(\sqrt{10}-\sqrt{6}\right)\cdot\left(\sqrt{4-\sqrt{15}}\right)\)
\(=\sqrt{2}\cdot\sqrt{4-\sqrt{15}}\cdot\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\sqrt{8-2\sqrt{15}}\cdot\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\sqrt{5-2\cdot\sqrt{5}\cdot\sqrt{3}+3}\cdot\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\cdot\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\left|\sqrt{5}-\sqrt{3}\right|\cdot\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)\)(Vì \(\sqrt{5}>\sqrt{3}\))
\(=8-2\sqrt{15}\)
d) Ta có: \(\left(\sqrt{3}-\sqrt{12}\right)\cdot\left(\sqrt{5+2\sqrt{6}}\right)\)
\(=\sqrt{3}\cdot\left(1-2\right)\cdot\sqrt{3+2\cdot\sqrt{3}\cdot\sqrt{2}+2}\)
\(=-\sqrt{3}\cdot\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}\)
\(=-\sqrt{3}\cdot\left|\sqrt{3}+\sqrt{2}\right|\)
\(=-\sqrt{3}\cdot\left(\sqrt{3}+\sqrt{2}\right)\)(Vì \(\sqrt{3}>\sqrt{2}>0\))
\(=-3-\sqrt{6}\)
e) Ta có: \(\sqrt{2-\sqrt{3}}\cdot\left(\sqrt{6}-\sqrt{2}\right)\cdot\left(2+\sqrt{3}\right)\)
\(=\sqrt{2}\cdot\sqrt{2-\sqrt{3}}\cdot\left(\sqrt{3}-1\right)\cdot\left(2+\sqrt{3}\right)\)
\(=\sqrt{4-2\sqrt{3}}\cdot\left(\sqrt{3}-1\right)\cdot\left(\sqrt{3}+2\right)\)
\(=\sqrt{3-2\cdot\sqrt{3}\cdot1+1}\cdot\left(\sqrt{3}-1\right)\cdot\left(\sqrt{3}+2\right)\)
\(=\sqrt{\left(\sqrt{3}-1\right)^2}\cdot\left(\sqrt{3}-1\right)\cdot\left(\sqrt{3}+2\right)\)
\(=\left|\sqrt{3}-1\right|\cdot\left(\sqrt{3}-1\right)\cdot\left(\sqrt{3}+2\right)\)
\(=\left(\sqrt{3}-1\right)\left(\sqrt{3}-1\right)\left(\sqrt{3}+2\right)\)(Vì \(\sqrt{3}>1\))
\(=\frac{\left(4-2\sqrt{3}\right)\left(4+2\sqrt{3}\right)}{2}\)
\(=\frac{16-12}{2}=\frac{4}{2}=2\)
f) Ta có: \(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{4+2\cdot2\cdot\sqrt{3}+3}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\left|2+\sqrt{3}\right|}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\left(2+\sqrt{3}\right)}}}\)(Vì \(2>\sqrt{3}>0\))
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-20-10\sqrt{3}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{25-2\cdot5\cdot\sqrt{3}+3}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\left|5-\sqrt{3}\right|}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\left(5-\sqrt{3}\right)}}\)(Vì \(5>\sqrt{3}\))
\(=\sqrt{4+\sqrt{5\sqrt{3}+25-5\sqrt{3}}}\)
\(=\sqrt{4+\sqrt{25}}\)
\(=\sqrt{4+5}=\sqrt{9}=3\)
b,\(\sqrt{8\sqrt{3}}-2\sqrt{25\sqrt{12}}+4\sqrt{\sqrt{192}}\) \(=\sqrt{8\sqrt{3}}-2\sqrt{50\sqrt{3}}+4\sqrt{8\sqrt{3}}\)
\(=2\sqrt{2\sqrt{3}}-10\sqrt{2\sqrt{3}}+8\sqrt{2\sqrt{3}}\)
\(=0\)
d,\(A=\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\)
\(\sqrt{2}A=\sqrt{2}(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}})\)
\(\sqrt2A=\sqrt{6-2\sqrt{5}}+\sqrt{6+2\sqrt{5}}\)
\(\sqrt2A=\sqrt{(\sqrt5-1)^2}\) \(+\sqrt{(\sqrt5+1)^2}\) \(=\sqrt5-1 +\sqrt5+1=2\sqrt5\)
\(\Rightarrow A=\dfrac{2\sqrt5}{\sqrt2}\) \(=\sqrt{10}\)
a. \(\frac{\sqrt{3-\sqrt{5}}\left(3+\sqrt{5}\right)}{\sqrt{10}+\sqrt{2}}=\frac{\sqrt{3-\sqrt{5}}\left(3+\sqrt{5}\right)}{\sqrt{2}\left(\sqrt{5}+1\right)}\)
\(=\frac{\sqrt{6-2\sqrt{5}}\left(3+\sqrt{5}\right)}{2\left(\sqrt{5}+1\right)}=\frac{\sqrt{\left(\sqrt{5}-1\right)^2}\left(3+\sqrt{5}\right)}{2\left(\sqrt{5}+1\right)}\)
\(=\frac{\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)}{2\left(\sqrt{5}+1\right)}=\frac{3\sqrt{5}-3+5-\sqrt{5}}{2\left(\sqrt{5}+1\right)}\)
\(=\frac{2\sqrt{5}+2}{2\left(\sqrt{5}+1\right)}=\frac{2\left(\sqrt{5}+1\right)}{2\left(\sqrt{5}+1\right)}=1\)
a) \(\sqrt{3+\sqrt{5}}\left(\sqrt{10}+\sqrt{2}\right)\left(3-\sqrt{5}\right)\)
\(=\sqrt{3+\sqrt{5}}.\sqrt{3-\sqrt{5}}.\sqrt{3-\sqrt{5}}.\left(\sqrt{10}+\sqrt{2}\right)\)
\(=\left(9-5\right).\sqrt{3-\sqrt{5}}.\sqrt{2}\left(\sqrt{5}+1\right)\)
\(=4.\sqrt{6-2\sqrt{5}}.\left(\sqrt{5}+1\right)\)
\(=4.\sqrt{5-2\sqrt{5}+1}.\left(\sqrt{5}+1\right)\)
\(=4.\sqrt{\left(\sqrt{5}-1\right)^2}.\left(\sqrt{5}+1\right)\)
\(=4.\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)=4.\left(5-1\right)=16\)
b) \(2\sqrt{4+\sqrt{6-2\sqrt{5}}}.\left(\sqrt{10}-\sqrt{2}\right)\)
\(=2\sqrt{4+\sqrt{5-2\sqrt{5}+1}}.\left(\sqrt{10}-\sqrt{2}\right)\)
\(=2\sqrt{4+\sqrt{\left(\sqrt{5}-1\right)^2}}.\left(\sqrt{10}-\sqrt{2}\right)\)
\(=2\sqrt{3+\sqrt{5}}.\sqrt{2}.\left(\sqrt{5}-\sqrt{1}\right)\)
\(=2\sqrt{6+2\sqrt{5}}.\left(\sqrt{5}-1\right)\)
\(=2\sqrt{5+2\sqrt{5}+1}.\left(\sqrt{5}-1\right)\)
\(=2\sqrt{\left(\sqrt{5}+1\right)^2}.\left(\sqrt{5}-1\right)=2.\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)\)
\(=2.\left(5-1\right)=2.4=8\)