Biến đổi các căn thức sau
a) \(\sqrt{3-\sqrt{5}}\)
b) \(\sqrt{4+\sqrt{7}}\)
c) \(\sqrt{5+\sqrt{21}}\)
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28 tháng 6 2019
a. \(=\sqrt{2}.\left(\sqrt{7}+\sqrt{8}\right)\sqrt{5-\sqrt{3}\sqrt{7}}\)
\(=\left(\sqrt{7}+\sqrt{8}\right)\sqrt{3-2\sqrt{3}.\sqrt{7}+7}\)
\(=\left(\sqrt{7}+\sqrt{8}\right)\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}\)
\(=\left(\sqrt{7}+\sqrt{8}\right)\left(\sqrt{7}-\sqrt{3}\right)\)
Rồi nhân ra. bạn làm tiếp nhé. Tuy nhiên minh nghĩ bạn bị nhầm đề. là \(\sqrt{6}\) chứ không phải căn 16
b. \(=\frac{5\left(\sqrt{21}+1\right)}{21-16}+\frac{\sqrt{3}.\sqrt{7}\left(\sqrt{3}-\sqrt{7}\right)}{-\left(\sqrt{3}-\sqrt{7}\right)}\)
\(=\sqrt{21}+4-\sqrt{21}=4\)
28 tháng 7 2017
a)\(\frac{21}{\sqrt{14}}\)=\(\frac{21.\sqrt{14}}{14}\)=\(\frac{3\sqrt{14}}{2}\)
b)\(\frac{3}{\sqrt{2}}+\frac{\sqrt{2}}{3}=\frac{3\sqrt{2}}{2}+\frac{\sqrt{2}}{3}=\frac{9\sqrt{2}}{6}+\frac{2\sqrt{2}}{6}=\frac{11\sqrt{2}}{6}\)
c)=\(-46\sqrt{5}\)
27 tháng 8 2023
a: \(=\left(\sqrt{3}-2\right)\cdot\sqrt{\left(2+\sqrt{3}\right)^2}\)
\(=\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)\)
=3-4=-1
b: \(=\sqrt{6+4\sqrt{2}}-\sqrt{11-2\sqrt{18}}\)
\(=\sqrt{\left(2+\sqrt{2}\right)^2}-\sqrt{\left(3-\sqrt{2}\right)^2}\)
\(=2+\sqrt{2}-3+\sqrt{2}=2\sqrt{2}-1\)
c: \(=\sqrt{\left(2\sqrt{5}-1\right)^2}+\sqrt{\left(2\sqrt{5}+1\right)^2}\)
\(=2\sqrt{5}-1+2\sqrt{5}+1\)
\(=4\sqrt{5}\)
17 tháng 5 2017
Ta có:\(\left(\sqrt{7-\sqrt{5}}+\sqrt{7+\sqrt{5}}\right)^2=7-\sqrt{5}+7+\sqrt{5}+2\sqrt{\left(7-\sqrt{5}\right)\left(7+\sqrt{5}\right)}=14+2\sqrt{44}=14+4\sqrt{11}\)
=>\(\sqrt{7-\sqrt{5}}+\sqrt{7+\sqrt{5}}=\sqrt{14+4\sqrt{11}}=\sqrt{2}.\sqrt{7+2\sqrt{11}}\)
=>B=\(\dfrac{\sqrt{2}.\sqrt{7+2\sqrt{11}}}{\sqrt{7+2\sqrt{11}}}\cdot\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
=\(\sqrt{2}\cdot\dfrac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\left(\sqrt{4}+\sqrt{6}+\sqrt{8}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)(mình làm tắt tách 4=2+2=\(\sqrt{4}+\sqrt{4}\))
=\(\sqrt{2}\)\(\cdot\dfrac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)\left(1+\sqrt{2}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\sqrt{2}\cdot\left(1+\sqrt{2}\right)=2+\sqrt{2}\)
17 tháng 5 2017
\(B=\dfrac{\sqrt{7-\sqrt{5}}+\sqrt{7+\sqrt{5}}}{\sqrt{7+2\sqrt{11}}}.\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(B=\dfrac{\sqrt{14-2\sqrt{5}}+\sqrt{14+2\sqrt{5}}}{\sqrt{2}.\sqrt{7+2\sqrt{11}}}.\dfrac{\sqrt{2}+\sqrt{3}+2+\sqrt{6}+\sqrt{8}+2}{\sqrt{2}+\sqrt{3}+2}\)
\(B=\dfrac{\sqrt{\left(\left(\sqrt{7+2\sqrt{11}}\right)-\left(\sqrt{7-2\sqrt{11}}\right)\right)^2}+\sqrt{\left(\left(\sqrt{7+2\sqrt{11}}\right)+\left(7-2\sqrt{11}\right)\right)^2}}{\sqrt{2}.\sqrt{7+2\sqrt{11}}}.\dfrac{\sqrt{2}+\sqrt{3}+2+\sqrt{2}\left(\sqrt{3}+2+\sqrt{2}\right)}{\sqrt{2}+\sqrt{3}+2}\)
\(B=\dfrac{\sqrt{7+2\sqrt{11}}-\sqrt{7-2\sqrt{11}}+\sqrt{7+2\sqrt{11}}+\sqrt{7-2\sqrt{11}}}{\sqrt{2}.\sqrt{7+2\sqrt{11}}}.\dfrac{\left(\sqrt{2}+\sqrt{3}+2\right)\left(1+\sqrt{2}\right)}{\sqrt{2}+\sqrt{3}+2}\)
\(B=\dfrac{2.\sqrt{7+2\sqrt{11}}}{\sqrt{2}.\sqrt{7+2\sqrt{11}}}.\left(1+\sqrt{2}\right)\)
\(B=\sqrt{2}.\left(1+\sqrt{2}\right)=\sqrt{2}+2\)
21 tháng 9 2021
a) \(\sqrt{28+10\sqrt{3}}=\sqrt{\left(5+\sqrt{3}\right)^2}=\left|5+\sqrt{3}\right|=5+\sqrt{3}\)
b) \(\sqrt{7+3\sqrt{5}}=\sqrt{\left(\dfrac{3\sqrt{2}}{2}+\dfrac{\sqrt{10}}{2}\right)^2}=\dfrac{3\sqrt{2}}{2}+\dfrac{\sqrt{10}}{2}\)
20 tháng 6 2021
`a)A=(3-sqrt5)sqrt{3+sqrt5}+(3+sqrt5)sqrt{3-sqrt5}`
`=sqrt{3-sqrt5}sqrt{3+sqrt5}(sqrt{3+sqrt5}+sqrt{3-sqrt5})`
`=sqrt{9-5}(sqrt{3+sqrt5}+sqrt{3-sqrt5})`
`=2(sqrt{3+sqrt5}+sqrt{3-sqrt5})`
`=sqrt2(sqrt{6+2sqrt5}+sqrt{6-2sqrt5})`
`=sqrt2(sqrt{(sqrt5+1)^2}+sqrt{(sqrt5+1)^2})`
`=sqrt2(sqrt5+1+sqrt5-1)`
`=sqrt{2}.2sqrt5`
`=2sqrt{10}`
20 tháng 6 2021
`b)B=(5+sqrt{21})(sqrt{14}-sqrt6)sqrt{5-sqrt{21}}`
`=sqrt{5+sqrt{21}}sqrt{5-sqrt{21}}sqrt{5+sqrt{21}}(sqrt{14}-sqrt6)`
`=sqrt{25-21}sqrt{5+sqrt{21}}(sqrt{14}-sqrt6)`
`=2sqrt{5+sqrt{21}}(sqrt{14}-sqrt6)`
`=2sqrt2sqrt{5+sqrt{21}}(sqrt{7}-sqrt3)`
`=2sqrt{10+2sqrt{21}}(sqrt{7}-sqrt3)`
`=2sqrt{(sqrt3+sqrt7)^2}(sqrt{7}-sqrt3)`
`=2(sqrt3+sqrt7)(sqrt{7}-sqrt3)`
`=2(7-3)`
`=8`
`c)C=sqrt{4+sqrt7}-sqrt{4-sqrt7}`
`=sqrt{(8+2sqrt7)/2}-sqrt{(8-2sqrt7)/2}`
`=sqrt{(sqrt7+1)^2/2}-sqrt{(sqrt7+1)^2/2}`
`=(sqrt7+1)/sqrt2-(sqrt7-1)/2`
`=2/sqrt2=sqrt2`
16 tháng 8 2023
1)
a) \(\sqrt{2x-4}\) có nghĩa khi:
\(2x-4\ge0\)
\(\Leftrightarrow2x\ge4\)
\(\Leftrightarrow x\ge\dfrac{4}{2}\)
\(\Leftrightarrow x\ge2\)
b) \(\sqrt{\dfrac{-7}{4-x}}\) có nghĩa khi
\(\dfrac{-7}{4-x}\ge0\) mà \(-7< 0\)
\(\Rightarrow4-x\le0\)
\(\Leftrightarrow x\ge4\)
\(a,\sqrt{3-\sqrt{5}}=\frac{\sqrt{6-2\sqrt{5}}}{\sqrt{2}}=\frac{\sqrt{\left(\sqrt{5}-1\right)^2}}{\sqrt{2}}=\frac{\sqrt{5}-1}{\sqrt{2}}=\frac{\sqrt{10}-\sqrt{2}}{2}\)
\(b,\sqrt{4+\sqrt{7}}=\frac{\sqrt{8+2\sqrt{7}}}{\sqrt{2}}=\frac{\sqrt{7}+1}{\sqrt{2}}=\frac{\sqrt{14}+\sqrt{2}}{2}\)
\(c,\sqrt{5+\sqrt{21}}=\frac{\sqrt{30+6\sqrt{21}}}{\sqrt{6}}=\frac{\sqrt{21}+3}{\sqrt{6}}\)