\(\left(5\sqrt{7}+7\sqrt{5}\right):\sqrt{35}\)
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
b: Ta có: \(\left(\sqrt{7-3\sqrt{5}}\right)\cdot\left(7+3\sqrt{5}\right)\cdot\left(3\sqrt{2}+\sqrt{10}\right)\)
\(=\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)\left(7+3\sqrt{5}\right)\)
\(=4\left(7+3\sqrt{5}\right)\)
\(=28+12\sqrt{5}\)
Lời giải:
a.
$A=\sqrt{8+\sqrt{55}}-\sqrt{8-\sqrt{55}}-\sqrt{125}$
$\sqrt{2}A=\sqrt{16+2\sqrt{55}}-\sqrt{16-2\sqrt{55}}-\sqrt{250}$
$=\sqrt{(\sqrt{11}+\sqrt{5})^2}-\sqrt{(\sqrt{11}-\sqrt{5})^2}-5\sqrt{10}$
$=|\sqrt{11}+\sqrt{5}|-|\sqrt{11}-\sqrt{5}|-5\sqrt{10}$
$=2\sqrt{5}-5\sqrt{10}$
$\Rightarrow A=\sqrt{10}-5\sqrt{5}$
b.
$B=\sqrt{7-3\sqrt{5}}.(7+3\sqrt{5})(3\sqrt{2}+\sqrt{10})$
$B\sqrt{2}=\sqrt{14-6\sqrt{5}}(7+3\sqrt{5})(3\sqrt{2}+\sqrt{10})$
$=\sqrt{(3-\sqrt{5})^2}(7+3\sqrt{5}).\sqrt{2}(3+\sqrt{5})$
$=(3-\sqrt{5})(7\sqrt{2}+3\sqrt{10})(3+\sqrt{5})$
$=(3^2-5)(7\sqrt{2}+3\sqrt{10})$
$=4(7\sqrt{2}+3\sqrt{10})=28\sqrt{2}+12\sqrt{10}$
$\Rightarrow B=28+12\sqrt{5}$
c.
$C=\sqrt{2}(\sqrt{7}-\sqrt{5})(6-\sqrt{35})\sqrt{6+\sqrt{35}}$
$=(\sqrt{7}-\sqrt{5})(6-\sqrt{35})\sqrt{12+2\sqrt{35}}$
$=(\sqrt{7}-\sqrt{5})(6-\sqrt{35})\sqrt{(\sqrt{7}+\sqrt{5})^2}
$=(\sqrt{7}-\sqrt{5})(6-\sqrt{35})(\sqrt{7}+\sqrt{5})$
$=(7-5)(6-\sqrt{35})$
$=2(6-\sqrt{35})=12-2\sqrt{35}$
![](https://rs.olm.vn/images/avt/0.png?1311)
`(5sqrt7+7sqrt5):sqrt{35}`
`=(sqrt{5}.sqrt{7}.sqrt{5}+sqrt{5}.sqrt{7}.sqrt{7}):sqrt{35}`
`=sqrt{5}.sqrt{7}(sqrt5+sqrt7):sqrt{35}`
`=sqrt{35}(sqrt5+sqrt7):sqrt{35}`
`=sqrt5+sqrt7`
Ta có : \(\left(\sqrt{5}\sqrt{5}\sqrt{7}+\sqrt{7}\sqrt{7}\sqrt{5}\right):\sqrt{35}\)
\(=\left(\sqrt{5}\sqrt{35}+\sqrt{7}\sqrt{35}\right):\sqrt{35}\)
\(=\sqrt{5}+\sqrt{7}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1. \(=\sqrt{\left(\sqrt{\frac{7}{2}}+\sqrt{\frac{3}{2}}\right)^2}+\sqrt{\left(\sqrt{\frac{7}{2}}-\sqrt{\frac{3}{2}}\right)^2}-2\sqrt{4\sqrt{7}}=\frac{7}{2}+\frac{3}{2}+\frac{7}{2}-\frac{3}{2}-2\sqrt{4\sqrt{7}}\)
\(=7-2\sqrt{4\sqrt{7}}\)
cho hỏi tại sao có số \(\frac{7}{2};\frac{3}{2}\)zậy chỉ với
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\left(\sqrt{28}-5\sqrt{35}+7\sqrt{112}\right)2\sqrt{7}=2\sqrt{196}-10\sqrt{245}+14\sqrt{784}\)
\(=28-10\sqrt{49.5}+392=420-70\sqrt{5}\)
b) \(\left(\sqrt{72}-3\sqrt{24}+5\sqrt{8}\right)\sqrt{2}+4\sqrt{27}=\sqrt{144}-3\sqrt{48}+5\sqrt{16}+4\sqrt{9.3}\)
\(=12-3\sqrt{16.3}+20+12\sqrt{3}=32-12\sqrt{3}+12\sqrt{3}=32\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: Ta có: \(\sqrt{4+\sqrt{15}}\)
\(=\dfrac{\sqrt{8+2\sqrt{15}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{2}}=\dfrac{\sqrt{10}+\sqrt{6}}{2}\)
b: Ta có: \(\left(3-\sqrt{2}\right)\cdot\sqrt{11+6\sqrt{2}}\)
\(=\left(3-\sqrt{2}\right)\left(3+\sqrt{2}\right)\)
=9-2
=7
c: Ta có: \(\left(\sqrt{7}+\sqrt{5}\right)\cdot\sqrt{12-2\sqrt{35}}\)
\(=\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)\)
=2
\(\left(5\sqrt{7}+7\sqrt{5}\right):\sqrt{35}=\left(\sqrt{5^2.7}+\sqrt{7^2.5}\right):\sqrt{35}\)
\(=\left(\sqrt{35.5}+\sqrt{35.7}\right):\sqrt{35}\)
\(=\sqrt{35}\left(\sqrt{5}+\sqrt{7}\right):\sqrt{35}\)
\(=\sqrt{5}+\sqrt{7}\)
Toán Học Team