\(\frac{\sqrt{4+\sqrt{15}}+\sqrt{5-\sqrt{21}}}{\sqrt{6+\sqrt{35}}}\)+1
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ML
22 tháng 5 2016
\(\frac{\sqrt{\sqrt{4+\sqrt{15}}+\sqrt{5-\sqrt{21}}}}{\sqrt{6+\sqrt{35}}}\)+\(\sqrt{\frac{1}{4-2\sqrt{3}}}\)-\(\sqrt{\frac{1}{4+2\sqrt{3}}}\)
=\(\frac{\sqrt{\sqrt{\frac{1}{2}\left(8+2\sqrt{15}\right)}+\sqrt{\frac{1}{2}\left(10-2\sqrt{21}\right)}}}{\sqrt{\frac{1}{2}\left(12+2\sqrt{35}\right)}}\)+\(\sqrt{\frac{1}{3-2\sqrt{3}.1+1}}\)-\(\sqrt{\frac{1}{3+2\sqrt{3}.1+1}}\)
=\(\frac{\sqrt{\sqrt{\frac{1}{2}\left(5+2\sqrt{5}.\sqrt{3}+3\right)}+\sqrt{\frac{1}{2}\left(7-2\sqrt{7}.\sqrt{3}+3\right)}}}{\sqrt{\frac{1}{2}\left(7+2\sqrt{7}.\sqrt{5}+5\right)}}\)+\(\sqrt{\frac{1}{\left(\sqrt{3}-1\right)^2}}\)-\(\sqrt{\frac{1}{\left(\sqrt{3}+1\right)^2}}\)
=\(\frac{\sqrt{\sqrt{\frac{1}{2}\left(\sqrt{5}+\sqrt{3}\right)^2}+\sqrt{\frac{1}{2}\left(\sqrt{7}-\sqrt{3}\right)^2}}}{\sqrt{\frac{1}{2}\left(\sqrt{7}+\sqrt{5}\right)^2}}\)+\(\frac{1}{\sqrt{3}-1}\)-\(\frac{1}{\sqrt{3}+1}\)
=\(\frac{\sqrt{\sqrt{\frac{1}{2}}.\left(\sqrt{5}+\sqrt{3}\right)+\sqrt{\frac{1}{2}}.\left(\sqrt{7}-\sqrt{3}\right)}}{\sqrt{\frac{1}{2}}.\left(\sqrt{7}+\sqrt{5}\right)}\)+\(\frac{\sqrt{3}+1-\sqrt{3}+1}{3-1}\)
=\(\frac{\sqrt{\sqrt{\frac{1}{2}}.\left(\sqrt{7}+\sqrt{5}\right)}}{\sqrt{\frac{1}{2}}.\left(\sqrt{7}+\sqrt{5}\right)}\)+1
=\(\frac{1}{\sqrt{\sqrt{\frac{1}{2}}.\left(\sqrt{7}+\sqrt{5}\right)}}\)+1
21 tháng 8 2020
a) Ta có: \(A=\frac{8+2\sqrt{15}+\sqrt{21}+\sqrt{35}}{\sqrt{3}+\sqrt{5}+\sqrt{7}}\)
\(=\frac{\left(\sqrt{3}+\sqrt{5}\right)^2+\sqrt{7}\cdot\left(\sqrt{3}+\sqrt{5}\right)}{\sqrt{3}+\sqrt{5}+\sqrt{7}}\)
\(=\frac{\left(\sqrt{3}+\sqrt{5}\right)\left(\sqrt{3}+\sqrt{5}+\sqrt{7}\right)}{\sqrt{3}+\sqrt{5}+\sqrt{7}}\)
\(=\sqrt{3}+\sqrt{5}\)
b) Ta có: \(B=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{4}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{6}}\)
\(=\frac{\sqrt{2}-1}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}+\frac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}+\frac{\sqrt{4}-\sqrt{3}}{\left(\sqrt{4}+\sqrt{3}\right)\left(\sqrt{4}-\sqrt{3}\right)}+\frac{\sqrt{5}-\sqrt{4}}{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}+\frac{\sqrt{6}-\sqrt{5}}{\left(\sqrt{6}+\sqrt{5}\right)\left(\sqrt{6}-\sqrt{5}\right)}\)
\(=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+2-\sqrt{3}+\sqrt{5}-2+\sqrt{6}-\sqrt{5}\)
\(=-1+\sqrt{6}\)
VN
1 tháng 5 2017
Làm tới dòng thứ 3 máy đơ, 2 lần rồi T,T
Mình chia làm 2 phần tính nhé
\(A=\frac{4\sqrt{2}}{\sqrt{10-2\sqrt{21}}}+\frac{3}{\sqrt{15+6\sqrt{6}}}-\frac{1}{\sqrt{19-6\sqrt{10}}}\)
\(A=\frac{4\sqrt{2}}{\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}}+\frac{3}{\sqrt{\left(\sqrt{9}+\sqrt{6}\right)^2}}-\frac{1}{\sqrt{\left(\sqrt{10}-\sqrt{9}\right)^2}}\)
\(A=\frac{4\sqrt{2}}{\sqrt{7}-\sqrt{3}}+\frac{3}{3+\sqrt{6}}-\frac{1}{\sqrt{10}-3}\)
\(A=\frac{4\sqrt{2}\left(\sqrt{7}+\sqrt{3}\right)}{7-3}+\frac{3\left(3-\sqrt{6}\right)}{9-6}-\frac{1\left(\sqrt{10}+3\right)}{10-9}\)
\(A=\frac{4\sqrt{14}+4\sqrt{6}}{4}+\frac{9-3\sqrt{6}}{3}-\sqrt{10}-3\)
\(A=\sqrt{14}+\sqrt{6}+3-\sqrt{6}-\sqrt{10}-3\)
\(A=\sqrt{14}-\sqrt{10}\)
\(B=\sqrt{6+\sqrt{35}}\)
\(B=\frac{\sqrt{2}\left(\sqrt{6+\sqrt{35}}\right)}{\sqrt{2}}\)
\(B=\frac{\sqrt{12+2\sqrt{35}}}{\sqrt{2}}\)
\(B=\frac{\sqrt{\left(\sqrt{7}+\sqrt{5}\right)^2}}{\sqrt{2}}\)
\(B=\frac{\sqrt{7}+\sqrt{5}}{\sqrt{2}}\)
\(\Rightarrow M=A.B=\left(\sqrt{14}-\sqrt{10}\right).\frac{\sqrt{7}+\sqrt{5}}{\sqrt{2}}\)
\(M=\sqrt{2}\left(\sqrt{7}-\sqrt{5}\right).\frac{\sqrt{7}+\sqrt{5}}{\sqrt{2}}\)
\(M=\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)\)
\(M=\left(\sqrt{7}\right)^2-\left(\sqrt{5}\right)^2\)
\(M=7-5=2\)
26 tháng 6 2021
`1)A=sqrt{4+sqrt{10+2sqrt5}}+sqrt{4-sqrt{10+2sqrt5}}`
`<=>A^2=4+sqrt{10+2sqrt5}+4-sqrt{10+2sqrt5}+2sqrt{16-10-2sqrt5}`
`<=>A^2=8+2sqrt{6-2sqrt5}`
`<=>A^2=8+2sqrt{(sqrt5-1)^2}`
`<=>A^2=8+2(sqrt5-1)`
`<=>A^2=6+2sqrt5=(sqrt5+1)^2`
`<=>A=sqrt5+1(do \ A>0)`
`b)B=sqrt{35+12sqrt6}-sqrt{35-12sqrt6}`
Vì `35+12sqrt6>35-12sqrt6`
`=>B>0`
`B^2=35+12sqrt6+35-12sqrt6-2sqrt{35^2-(12sqrt6)^2}`
`<=>B^2=70-2sqrt{361}`
`<=>B^2=70-2sqrt{19^2}=70-38=32`
`<=>B=sqrt{32}=4sqrt2(do \ B>0)`
`3)(4+sqrt{15})(sqrt{10}-sqrt6)sqrt{4-sqrt{15}}`
`=sqrt{4+sqrt{15}}.sqrt{4-sqrt{15}}.sqrt{4+sqrt{15}}(sqrt{10}-sqrt6)`
`=sqrt{16-15}.sqrt2(sqrt5-sqrt3).sqrt{4+sqrt{15}}`
`=sqrt{8+2sqrt{15}}(sqrt5-sqrt3)`
`=sqrt{5+2sqrt{5.3}+3}(sqrt5-sqrt3)`
`=sqrt{(sqrt5+sqrt3)^2}(sqrt5-sqrt3)`
`=(sqrt5+sqrt3)(sqrt5-sqrt3)`
`=5-3=2`
25 tháng 8 2019
a,\(\left(5+4\sqrt{2}\right)\left(3+2\sqrt{1+\sqrt{2}}\right)\left(3-2\sqrt{1+\sqrt{2}}\right)\)
=\(\left(5+4\sqrt{2}\right)\left(9-4\left(1+\sqrt{2}\right)\right)\)
=\(\left(5+4\sqrt{2}\right)\left(9-4-4\sqrt{2}\right)\)
=\(\left(5+4\sqrt{2}\right)\left(5-4\sqrt{2}\right)=25-\left(4\sqrt{2}\right)^2\)
=-7
b, \(\sqrt{\frac{9}{4}-\sqrt{2}}=\sqrt{\frac{9-4\sqrt{2}}{4}}=\frac{\sqrt{9-4\sqrt{2}}}{2}=\frac{\sqrt{9-2\sqrt{8}}}{2}=\frac{\sqrt{\left(\sqrt{8}-1\right)^2}}{2}=\frac{\left|\sqrt{8}-1\right|}{2}=\frac{\sqrt{8}-1}{2}\)
26 tháng 8 2019
So sánh:
1) \(2\sqrt{27}\) và \(\sqrt{147}\)
+ \(2\sqrt{27}\) = \(6\sqrt{3}\)
+ \(\sqrt{147}\) = \(7\sqrt{3}\)
⇒ \(6\sqrt{3}\) < \(7\sqrt{3}\)
Vậy: \(2\sqrt{27}\)< \(\sqrt{147}\)
2) \(2\sqrt{15}\) và \(\sqrt{59}\)
+ \(2\sqrt{15}\) = \(\sqrt{60}\)
⇒ \(\sqrt{60}\) > \(\sqrt{59}\)
Vậy: \(2\sqrt{15}\) > \(\sqrt{59}\)
3) \(2\sqrt{2}-1\) và 2
\(giống\left(-1\right)\left\{{}\begin{matrix}3-1\\2\sqrt{2}-1\end{matrix}\right.\)
So sánh: 3 và \(2\sqrt{2}\)
+ 3 = \(\sqrt{9}\)
+ \(2\sqrt{2}=\sqrt{8}\)
⇒ \(\sqrt{8}\) < \(\sqrt{9}\)
⇒ \(\sqrt{8}\) -1 < \(\sqrt{9}\) -1
⇒ \(2\sqrt{2}\) - 1 < 3 - 1
Vậy: \(2\sqrt{2}-1< 2\)
4) \(\frac{\sqrt{3}}{2}\) và 1
+ 1 = \(\frac{2}{2}\)
⇒ \(\frac{\sqrt{3}}{2}\) < \(\frac{2}{2}\)
Vậy: \(\frac{\sqrt{3}}{2}\) < 1
5) \(\frac{-\sqrt{10}}{2}\) và \(-2\sqrt{5}\)
+ \(-2\sqrt{5}\) = \(\frac{-4\sqrt{5}}{2}\) = \(\frac{-\sqrt{80}}{2}\)
⇒ \(\frac{-\sqrt{10}}{2}\) > \(\frac{-\sqrt{80}}{2}\)
Vậy: \(\frac{-\sqrt{10}}{2}\) > \(-2\sqrt{5}\)
20 tháng 10 2018
a) \(\frac{\sqrt{2}\left(\sqrt{3}+\sqrt{5}\right)}{\sqrt{7\left(\sqrt{3}+\sqrt{5}\right)}}=\) \(\frac{\sqrt{2}}{\sqrt{7}}\)
b ) \(\frac{15\sqrt{2}+9\sqrt{3}}{3\sqrt{3}+3\sqrt{5}}=\frac{3\left(5\sqrt{2}+3\sqrt{3}\right)}{3\left(\sqrt{3}+\sqrt{5}\right)}\)\(=\frac{5\sqrt{2}+3\sqrt{3}}{\sqrt{3}+\sqrt{5}}\)
c)\(\frac{\sqrt{2}-\sqrt{6}+\sqrt{3}-\sqrt{9}+\sqrt{4}-\sqrt{12}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\) = \(\frac{\sqrt{2}\left(1-\sqrt{3}\right)+\sqrt{3}\left(1-\sqrt{3}\right)+\sqrt{4}\left(1-\sqrt{3}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)\(=\frac{\left(1-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=1-\sqrt{3}\)
d) \(\frac{\sqrt{\left(\sqrt{5}-1\right)^2}}{\sqrt{5}-1}=\frac{\sqrt{5}-1}{\sqrt{5}-1}=1\)
1 tháng 7 2017
a) \(\sqrt{5+\sqrt{21}}-\sqrt{6-\sqrt{35}}\) = \(\dfrac{\sqrt{10+2\sqrt{21}}}{\sqrt{2}}-\dfrac{\sqrt{12-2\sqrt{35}}}{\sqrt{2}}\)
= \(\dfrac{\sqrt{\left(\sqrt{7}+\sqrt{3}\right)^2}}{\sqrt{2}}-\dfrac{\sqrt{\left(\sqrt{7}-\sqrt{5}\right)^2}}{\sqrt{2}}\)
= \(\dfrac{\sqrt{7}+\sqrt{3}}{\sqrt{2}}-\dfrac{\sqrt{7}-\sqrt{5}}{\sqrt{2}}\) = \(\dfrac{\sqrt{7}+\sqrt{3}-\left(\sqrt{7}-\sqrt{5}\right)}{\sqrt{2}}\)
= \(\dfrac{\sqrt{7}+\sqrt{3}-\sqrt{7}+\sqrt{5}}{\sqrt{2}}=\dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{2}}\)
câu b) hình như đề sai
k mình đê :) Rồi mình giải cho :v