\(\sqrt{\left(3-\sqrt{10}\right)^2}+\sqrt{11-2\sqrt{10}}\)
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b: =căn 10-3+4-căn 10=1
a: \(=\sqrt{11-4\sqrt{6}+\sqrt{15}}\)
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\(a,=\sqrt{17}-5\sqrt{2}+3\\ b,=\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)\sqrt{6-2\sqrt{5}}\\ =\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)\left(\sqrt{5}-1\right)\\ =\left(3+\sqrt{5}\right)\left(6-2\sqrt{5}\right)=8\\ c,=\left(\sqrt{2}-3\right)\left(3+\sqrt{2}\right)=2-9=-7\\ d,4+\sqrt{7}-\sqrt{2}\)
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câu E dễ nhất nên mình làm trước , các câu còn lại làm tương tự ( biến đổi thành hằng đẳng thức rồi rút gọn ) :
\(E=\sqrt{9-2.3.\sqrt{6}+6}+\sqrt{24-2.2\sqrt{6}.3+9}\)
\(=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(2\sqrt{6}-3\right)^2}\)
\(=\left|3-\sqrt{6}\right|+\left|2\sqrt{6}-3\right|\)
\(=3-\sqrt{6}+2\sqrt{6}-3\) ( vì \(3-\sqrt{6}>0;2\sqrt{6}-3>0\) )
\(=\sqrt{6}\)
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`F=sqrt{(3-sqrt2)^2}+sqrt{(1-sqrt2)^2}``
`=3-sqrt2+sqrt2-1=2`
`G=sqrt{(5+sqrt7)^2}-sqrt{(2-sqrt7)^2}`
`=5+sqrt7-(sqrt7-2)`
`=5+sqrt7-sqrt7+2=2`
`H=sqrt{(3-sqrt{10})^2}+sqrt{(2-sqrt{10})^2}`
`=sqrt{10}-3+sqrt{10}-2`
`=2\sqrt{10}-5`
\(F=\left|3-\sqrt{2}\right|+\left|1-\sqrt{2}\right|=3-\sqrt{3}+\sqrt{2}-1=2\)
\(G=\left|5+\sqrt{7}\right|-\left|2-\sqrt{7}\right|=5+\sqrt{7}-\sqrt{7}+2=7\)
\(H=\left|3-\sqrt{10}\right|+\left|2-\sqrt{10}\right|=\sqrt{10}-3+\sqrt{10}-2=2\sqrt{10}-5\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a/
\(=\sqrt{6-2\sqrt{5}}\left(\sqrt{5}-1\right)\frac{\left(6+2\sqrt{5}\right)}{2}=\sqrt{\left(\sqrt{5}-1\right)^2}\left(\sqrt{5}-1\right)\frac{\left(\sqrt{5}+1\right)^2}{2}\)
\(=\left(\sqrt{5}-1\right)^2.\frac{\left(\sqrt{5}+1\right)^2}{2}=\frac{\left(5-1\right)^2}{2}=8\)
b/
\(\frac{\sqrt{3}+\sqrt{\left(3+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}}{\sqrt{2}+\sqrt{\left(\sqrt{5}+1\right)^2}-\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}}=\frac{\sqrt{3}+3+\sqrt{2}-\sqrt{3}-\sqrt{2}}{\sqrt{2}+\sqrt{5}+1-\sqrt{5}-\sqrt{2}}=\frac{3}{1}=3\)
https://hoc24.vn/hoi-dap/question/819940.html giúp mk bài này vs
\(=\left|3-\sqrt{10}\right|+\sqrt{\left(\sqrt{10}-1\right)^2}\\ =\sqrt{10}-3+\sqrt{10}-1=\sqrt{10}-4\)