1/ thực hiện phép tính:
a) (\(\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}\)) . (\(\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}\))
b) ( \(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\))2
c) \(\sqrt{5-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
giúp mjk nha m.n!! thks nhìu
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Những câu hỏi liên quan
24 tháng 8 2021
a: Ta có: \(\dfrac{4}{\sqrt{7}-\sqrt{3}}+\dfrac{6}{3+\sqrt{3}}+\dfrac{\sqrt{7}-7}{\sqrt{7}-1}\)
\(=\sqrt{7}+\sqrt{3}+3-\sqrt{3}-\sqrt{7}\)
=3
12 tháng 10 2021
\(a,=\dfrac{\sqrt{7}-5}{2}-\dfrac{3-\sqrt{7}}{2}+\dfrac{6\left(\sqrt{7}+2\right)}{3}-\dfrac{5\left(4-\sqrt{7}\right)}{9}\\ =\dfrac{\sqrt{7}-5-3+\sqrt{7}}{2}+2\sqrt{7}+4-\dfrac{20-5\sqrt{7}}{9}\\ =\dfrac{2\sqrt{7}-8}{2}+2\sqrt{7}+4-\dfrac{20-5\sqrt{7}}{9}\\ =\sqrt{7}-4+2\sqrt{7}+4-\dfrac{20-5\sqrt{7}}{9}\\ =\dfrac{27\sqrt{7}-20+5\sqrt{7}}{9}=\dfrac{32\sqrt{7}-20}{9}\)
\(b,=\dfrac{2\left(\sqrt{6}+2\right)}{2}+\dfrac{2\left(\sqrt{6}-2\right)}{2}+\dfrac{5\sqrt{6}}{6}\\ =\sqrt{6}+2+\sqrt{6}-2+\dfrac{5\sqrt{6}}{6}\\ =\dfrac{12\sqrt{6}+5\sqrt{6}}{6}=\dfrac{17\sqrt{6}}{6}\)
\(c,=\dfrac{\sqrt{3}+\sqrt{2}+\sqrt{5}-\sqrt{3}-\sqrt{2}+\sqrt{5}}{\left(\sqrt{3}+\sqrt{2}\right)^2-5}\\ =\dfrac{2\sqrt{5}}{5+2\sqrt{6}-5}=\dfrac{2\sqrt{5}}{2\sqrt{6}}=\dfrac{\sqrt{30}}{6}\)
27 tháng 7 2021
1: \(\sqrt{3+\sqrt{5}}\cdot\sqrt{2}=\sqrt{6+2\sqrt{5}}=\sqrt{5}+1\)
3) \(\left(\sqrt{\dfrac{3}{4}}-\sqrt{3}+5\cdot\sqrt{\dfrac{4}{3}}\right)\cdot\sqrt{12}\)
\(=\left(\dfrac{\sqrt{3}}{2}-\dfrac{2\sqrt{3}}{2}+5\cdot\dfrac{2}{\sqrt{3}}\right)\cdot\sqrt{12}\)
\(=\dfrac{17\sqrt{3}}{6}\cdot2\sqrt{3}\)
\(=\dfrac{34\cdot3}{6}=\dfrac{102}{6}=17\)
12 tháng 10 2021
c: Ta có: \(C=\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\)
\(=\dfrac{\sqrt{6-2\sqrt{5}}+\sqrt{6+2\sqrt{5}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{5}-1+\sqrt{5}+1}{\sqrt{2}}=\sqrt{10}\)
3 tháng 8 2023
a: =(2căn 3-8căn 3)(căn 3-1)
=-6căn 3*(căn 3-1)
=-18+6căn 3
b: \(=\dfrac{6-2\sqrt{5}}{\sqrt{5}-3}-\sqrt{5}+2\)
=-2-căn 5+2=-căn 5
c: \(=3\sqrt{2a}-3a\sqrt{2a}+2\sqrt{2a}-\dfrac{1}{4}\cdot8\sqrt{2a}\)
=\(3\sqrt{2a}-3a\cdot\sqrt{2a}\)
AH
Akai Haruma
Giáo viên
26 tháng 6 2021
\(A=\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}=\sqrt{3+1+2\sqrt{3.1}}-\sqrt{3+1-2\sqrt{3.1}}\)
\(=\sqrt{(\sqrt{3}+1)^2}-\sqrt{(\sqrt{3}-1)^2}=|\sqrt{3}+1|-|\sqrt{3}-1|=2\)
\(B=\sqrt{4+5-2\sqrt{4.5}}+\sqrt{4+5+2\sqrt{4.5}}=\sqrt{(\sqrt{4}-\sqrt{5})^2}+\sqrt{(\sqrt{4}+\sqrt{5})^2}\)
\(=|\sqrt{4}-\sqrt{5}|+|\sqrt{4}+\sqrt{5}|=2\sqrt{5}\)
AH
Akai Haruma
Giáo viên
26 tháng 6 2021
\(C\sqrt{2}=\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}=\sqrt{7+1-2\sqrt{7.1}}-\sqrt{7+1+2\sqrt{7.1}}\)
\(=\sqrt{(\sqrt{7}-1)^2}-\sqrt{(\sqrt{7}+1)^2}\)
\(=|\sqrt{7}-1|-|\sqrt{7}+1|=-2\Rightarrow C=-\sqrt{2}\)
----------------------------
\(7+4\sqrt{3}=(2+\sqrt{3})^2\Rightarrow 10\sqrt{7+4\sqrt{3}}=10(2+\sqrt{3})\)
\(\Rightarrow \sqrt{48-10\sqrt{7+4\sqrt{3}}}=\sqrt{28-10\sqrt{3}}=\sqrt{(5-\sqrt{3})^2}=5-\sqrt{3}\)
\(\Rightarrow 3+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}=3+5(5-\sqrt{3})=28-5\sqrt{3}\)
\(\Rightarrow D=\sqrt{5\sqrt{28-5\sqrt{3}}}\)
18 tháng 6 2021
`a)\sqrt{9-4sqrt5}-sqrt5`
`=sqrt{5-2.2sqrt5+4}-sqrt5`
`=sqrt{(sqrt5-2)^2}-sqrt5`
`=|\sqrt5-2|-sqrt5`
`=sqrt5-2-sqrt5=-2`
`b)\sqrt{7-4sqrt3}+sqrt{4-2sqrt3}`
`=\sqrt{4-2.2sqrt3+3}+\sqrt{3-2sqrt3+1}`
`=sqrt{(2-sqrt3)^2}+sqrt{(sqrt3-1)^2}`
`=|2-sqrt3|+|sqrt3-1|`
`=2-sqrt3+sqrt3-1=1`
`c)(x-49)/(sqrtx-7)(x>=0,x ne 49)`
`=((sqrtx-7)(sqrtx+7))/(sqrtx-7)`
`=sqrtx+7`
`d)\sqrt{4+2\sqrt3}-\sqrt{13+4sqrt3}`
`=\sqrt{3+2sqrt3+1}-\sqrt{12+2.2sqrt3+1}`
`=sqrt{(sqrt3+1)^2}-\sqrt{(2sqrt3+1)^2}`
`=sqrt3+1-2sqrt3-1=-sqrt3`
`e)2+sqrt{17-4sqrt{9+4sqrt{45}}}`(câu này hơi sai)
NH
21 tháng 5 2023
a)
\(M=\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(=\sqrt{4+4\sqrt{5}+5}-\sqrt{4-4\sqrt{5}+5}\)
\(=\sqrt{\left(2+\sqrt{5}\right)^2}-\sqrt{\left(2-\sqrt{5}\right)^2}\)
\(=\left|2+\sqrt{5}\right|-\left|2-\sqrt{5}\right|\)
\(=2+\sqrt{5}-\left(\sqrt{5}-2\right)\) (vì \(2+2\sqrt{5}>0;2-\sqrt{5}< 0\) )
\(=2+\sqrt{5}-\sqrt{5}+2\\ =4\)
b)
\(N=\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}\)
\(=\sqrt{7-2\sqrt{7}+1}-\sqrt{7+2\sqrt{7}+1}\)
\(=\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}\)
\(=\left|\sqrt{7}-1\right|-\left|\sqrt{7}+1\right|\)
\(=\sqrt{7}-1-\left(\sqrt{7}+1\right)\) (vì \(\sqrt{7}-1>0;\sqrt{7}+1>0\) )
\(=\sqrt{7}-1-\sqrt{7}-1\\ =-2\)
a.\(\left(\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}\right).\left(\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}\right)\)
\(=\left(\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-1\right)^2}\right).\left(\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{\left(\sqrt{3}+1\right)^2}\right)\)
\(=\left(\sqrt{3}+1-\sqrt{3}+1\right)\left(\sqrt{3}-1+\sqrt{3}+1\right)\)
\(=2.2\sqrt{3}=4\sqrt{3}\)
b.\(\left(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\right)^2=\left[\frac{\sqrt{8+2\sqrt{7}}}{\sqrt{2}}-\frac{\sqrt{8-2\sqrt{7}}}{\sqrt{2}}\right]^2\)
\(=\left(\frac{\sqrt{\left(\sqrt{7}+1\right)^2}}{\sqrt{2}}-\frac{\sqrt{\left(\sqrt{7}-1\right)^2}}{\sqrt{2}}\right)^2\)
\(=\left(\frac{\sqrt{7}+1-\sqrt{7}+1}{\sqrt{2}}\right)^2=\left(\sqrt{2}\right)^2=2\)
c.\(\sqrt{5-\sqrt{3-\sqrt{29-12\sqrt{5}}}}=\sqrt{5-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}\)
\(=\sqrt{5-\sqrt{3-\left(2\sqrt{5}-3\right)}}=\sqrt{5-\sqrt{6-2\sqrt{5}}}\)
\(=\sqrt{5-\sqrt{\left(\sqrt{5}-1\right)^2}}=\sqrt{5-\sqrt{5}+1}=\sqrt{6-\sqrt{5}}\)
!?
em ko biết làm!
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