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`a, (2 sqrt 3 + sqrt 5)sqrt 3 - sqrt 60`
`= 2 sqrt 3 . sqrt 3 + sqrt 5 . sqrt 3 - sqrt(4 . 15)`
`= 2 . 3 + sqrt 15 - 2 sqrt 15`.
`= 6 - sqrt 15`.
`b, (5 sqrt 2 + 2 sqrt 5)sqrt 5 - sqrt250`
`= 5 sqrt 2 . sqrt 5 + 2 sqrt 5 . sqrt 5 - sqrt(25.10)`
`= 5 sqrt 10 + 10 - 5 sqrt 10`
`= 10`.
5) \(\left(2\sqrt{3}+\sqrt{5}\right)\sqrt{3}-\sqrt{60}\)
\(=2\sqrt{3}\cdot\sqrt{3}+\sqrt{5}\cdot\sqrt{3}-\sqrt{2^2\cdot15}\)
\(=2\cdot3+\sqrt{15}-2\sqrt{15}\)
\(=6+\left(1-2\right)\sqrt{15}\)
\(=6-\sqrt{15}\)
6) \(\left(5\sqrt{2}+2\sqrt{5}\right)\sqrt{5}-\sqrt{250}\)
\(=5\sqrt{2}\cdot\sqrt{5}+2\sqrt{5}\cdot\sqrt{5}-\sqrt{5^2\cdot10}\)
\(=5\sqrt{10}+2\cdot5-5\sqrt{10}\)
\(=\left(5-5\right)\sqrt{10}+10\)
\(=0+10\)
\(=10\)
\(=\frac{\sqrt{6+2\sqrt{5}}}{\sqrt{2}}.\left(\sqrt{10}+\sqrt{2}\right).\frac{6-2\sqrt{5}}{2}\)
\(=\frac{\sqrt{5}+1}{\sqrt{2}}.\sqrt{2}\left(\sqrt{5}+1\right).\frac{\left(\sqrt{5}-1\right)^2}{2}\)
\(=\frac{\left(\sqrt{5}+1\right)^2.\left(\sqrt{5}-1\right)^2}{2}\)
\(=\frac{\left[\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)\right]^2}{2}\)
\(=\frac{4^2}{2}=8\)
Đặt \(\hept{\begin{cases}\sqrt{3-\sqrt{5}}=A\\\sqrt{3+\sqrt{5}}=B\end{cases}}\)
Ta có A.B = 2
(A + B)2 = 6 + 4 = 10 => A + B = \(\sqrt{10}\)
Ta có cái ban đầu
= A2 B + AB2 = AB(A + B) = \(2\sqrt{10}\)
\(1,\sqrt{\left(2+\sqrt{7}\right)^2-\sqrt{\left(2-\sqrt{7}\right)^2}}\) ( 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 :>
\(\left(5\sqrt{3}+3\sqrt{5}\right):15\)
\(=\sqrt{5}.\sqrt{3}\left(\sqrt{5}+\sqrt{3}\right):15\)
\(=\sqrt{15}\left(\sqrt{5}+\sqrt{3}\right):15=\frac{\sqrt{5}+\sqrt{3}}{\sqrt{15}}\)