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`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)
\(A=\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}=\sqrt{3+1+2\sqrt{3.1}}-\sqrt{3+1-2\sqrt{3.1}}\)
\(=\sqrt{(\sqrt{3}+1)^2}-\sqrt{(\sqrt{3}-1)^2}=|\sqrt{3}+1|-|\sqrt{3}-1|=2\)
\(B=\sqrt{4+5-2\sqrt{4.5}}+\sqrt{4+5+2\sqrt{4.5}}=\sqrt{(\sqrt{4}-\sqrt{5})^2}+\sqrt{(\sqrt{4}+\sqrt{5})^2}\)
\(=|\sqrt{4}-\sqrt{5}|+|\sqrt{4}+\sqrt{5}|=2\sqrt{5}\)
\(C\sqrt{2}=\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}=\sqrt{7+1-2\sqrt{7.1}}-\sqrt{7+1+2\sqrt{7.1}}\)
\(=\sqrt{(\sqrt{7}-1)^2}-\sqrt{(\sqrt{7}+1)^2}\)
\(=|\sqrt{7}-1|-|\sqrt{7}+1|=-2\Rightarrow C=-\sqrt{2}\)
----------------------------
\(7+4\sqrt{3}=(2+\sqrt{3})^2\Rightarrow 10\sqrt{7+4\sqrt{3}}=10(2+\sqrt{3})\)
\(\Rightarrow \sqrt{48-10\sqrt{7+4\sqrt{3}}}=\sqrt{28-10\sqrt{3}}=\sqrt{(5-\sqrt{3})^2}=5-\sqrt{3}\)
\(\Rightarrow 3+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}=3+5(5-\sqrt{3})=28-5\sqrt{3}\)
\(\Rightarrow D=\sqrt{5\sqrt{28-5\sqrt{3}}}\)
a) Ta có: \(9+4\sqrt{5}\)
\(=5+2\cdot\sqrt{5}\cdot2+4\)
\(=\left(\sqrt{5}+2\right)^2\)(đpcm)
b) Ta có: \(\sqrt{9-4\sqrt{5}}-\sqrt{5}\)
\(=\sqrt{\left(\sqrt{5}-2\right)^2}-\sqrt{5}\)
\(=\sqrt{5}-2-\sqrt{5}\)
=-2(ddpcm)
c) Ta có: \(\left(4-\sqrt{7}\right)^2\)
\(=16-2\cdot4\cdot\sqrt{7}+7\)
\(=23-8\sqrt{7}\)(đpcm)
d) Ta có: \(\sqrt{17-12\sqrt{2}}+2\sqrt{2}\)
\(=\sqrt{9-2\cdot3\cdot2\sqrt{2}+8}+2\sqrt{2}\)
\(=\sqrt{\left(3-2\sqrt{2}\right)^2}+2\sqrt{2}\)
\(=3-2\sqrt{2}+2\sqrt{2}=3\)(đpcm)
\(a.VT=4+4\sqrt{5}+5=2^2+4\sqrt{5}+\sqrt{5}^2=\left(2+\sqrt{5}\right)^2=VP\)
\(b.\) Dựa vào câu a ta có: \(9-4\sqrt{5}=\left(\sqrt{5}-2\right)^2\)
\(VT=\left|\sqrt{5}-2\right|-\sqrt{5}=\sqrt{5}-2-\sqrt{5}=-2=VP\)
\(c.VT=16-8\sqrt{7}+7=4^2-8\sqrt{7}+\sqrt{7}^2=\left(4-\sqrt{7}\right)^2=VP\)
\(d.\)
Ta có: \(17-12\sqrt{2}=8-12\sqrt{2}+9=\left(2\sqrt{2}\right)^2-12\sqrt{2}+3^2=\left(2\sqrt{2}-3\right)^2\)
\(VT=\left|2\sqrt{2}-3\right|+2\sqrt{2}=3-2\sqrt{2}+2\sqrt{2}=3=VP\)
Lời giải:
a. \(\sqrt{6-2\sqrt{5}}=\sqrt{5-2\sqrt{5}.\sqrt{1}+1}=\sqrt{(\sqrt{5}-1)^2}=\sqrt{5}-1\)
b. \(\sqrt{7-4\sqrt{3}}=\sqrt{4-2\sqrt{4}.\sqrt{3}+3}=\sqrt{(\sqrt{4}-\sqrt{3})^2}=\sqrt{4}-\sqrt{3}=2-\sqrt{3}\)
c.
\(\sqrt{3-2\sqrt{2}}-\sqrt{6-4\sqrt{2}}=\sqrt{2-2\sqrt{2}+1}-\sqrt{4-4\sqrt{2}+2}\)
\(=\sqrt{(\sqrt{2}-1)^2}-\sqrt{(\sqrt{4}-\sqrt{2})^2}\)
\(=|\sqrt{2}-1|-|\sqrt{4}-\sqrt{2}|=\sqrt{2}-1-(2-\sqrt{2})=2\sqrt{2}-3\)
d.
\(=\sqrt{13+30\sqrt{2+\sqrt{(\sqrt{8}+1)^2}}}=\sqrt{13+30\sqrt{2+\sqrt{8}+1}}\)
\(=\sqrt{13+30\sqrt{3+2\sqrt{2}}}=\sqrt{13+30\sqrt{(\sqrt{2}+1)^2}}\)
\(=\sqrt{13+30(\sqrt{2}+1)}=\sqrt{43+30\sqrt{2}}=\sqrt{18+2\sqrt{18.25}+25}\)
\(=\sqrt{(\sqrt{18}+\sqrt{25})^2}=\sqrt{18}+\sqrt{25}=5+3\sqrt{2}\)
a) \(\sqrt{6-2\sqrt{5}}=\sqrt{5}-1\)
b) \(\sqrt{7-4\sqrt{3}}=2-\sqrt{3}\)
c) \(\sqrt{3-2\sqrt{2}}-\sqrt{6-4\sqrt{2}}=\sqrt{2}-1-2+\sqrt{2}=-3+2\sqrt{2}\)
d) Ta có: \(\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}\)
\(=\sqrt{13+30\sqrt{2+1+2\sqrt{2}}}\)
\(=\sqrt{13+30\left(\sqrt{2}+1\right)}\)
\(=\sqrt{43+30\sqrt{2}}\)
\(=5+3\sqrt{2}\)
1. \(\sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}\)
\(=\sqrt{\left(\sqrt{2}+\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}\)
\(=\sqrt{2}+\sqrt{3}-\sqrt{3}+\sqrt{2}\)
\(=2\sqrt{2}\)
a) \(\dfrac{2\sqrt{3}+2}{4\sqrt{3}+4}=\dfrac{2\left(\sqrt{3}+1\right)}{4\left(\sqrt{3}+1\right)}=\dfrac{1}{2}\)
b) \(\dfrac{\sqrt{10}+\sqrt{15}}{\sqrt{8}+\sqrt{12}}=\dfrac{\sqrt{5}\left(\sqrt{2}+\sqrt{3}\right)}{\sqrt{4}\left(\sqrt{2}+\sqrt{3}\right)}=\dfrac{\sqrt{5}}{2}\)
c) \(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{4}+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\\ =\dfrac{\left(1+\sqrt{2}\right)\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}=1+\sqrt{2}\)
d) \(\sqrt{9+\sqrt{17}}.\sqrt{9-\sqrt{17}}=\sqrt{\left(9+\sqrt{17}\right)\left(9-\sqrt{17}\right)}\\ =\sqrt{81-17}=\sqrt{64}=8\)
\(a.\dfrac{2\sqrt{3}+2}{4\sqrt{3}+4}=\dfrac{2\left(\sqrt{3}+1\right)}{4\left(\sqrt{3}+1\right)}=\dfrac{2}{4}=\dfrac{1}{2}\)
\(b.\dfrac{\sqrt{10}+\sqrt{15}}{\sqrt{8}+\sqrt{12}}=\dfrac{\sqrt{5}\left(\sqrt{2}+\sqrt{3}\right)}{2\left(\sqrt{2}+\sqrt{3}\right)}=\dfrac{\sqrt{5}}{2}\)
\(c.\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\dfrac{\sqrt{2}+\sqrt{3}+2+2+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+2}=\dfrac{\sqrt{2}+\sqrt{3}+2}{\sqrt{2}+\sqrt{3}+2}+\dfrac{\sqrt{2}\left(\sqrt{2}+\sqrt{3}+2\right)}{\sqrt{2}+\sqrt{3}+2}=1+\sqrt{2}\)
\(d.\sqrt{9+\sqrt{17}}.\sqrt{9-\sqrt{17}}=\sqrt{\left(9+\sqrt{17}\right)\left(9-\sqrt{17}\right)}=\sqrt{81-17}=8\)
a.\(\sqrt{7+4\sqrt{3}}=\sqrt{\left(\sqrt{3}+2\right)^2}=\left|\sqrt{3}+2\right|=\sqrt{3}+2\)
b.\(\sqrt{9-4\sqrt{5}}=\sqrt{\left(\sqrt{5}-2\right)^2}=\left|\sqrt{5}-2\right|=\sqrt{5}-2\)
c.\(\sqrt{14+6\sqrt{5}}=\sqrt{\left(\sqrt{5}+3\right)^2}=\left|\sqrt{5}+3\right|=\sqrt{5}+3\)
d.\(\sqrt{17-12\sqrt{2}}=\sqrt{\left(2\sqrt{2}-3\right)^2}=\left|2\sqrt{2}-3\right|=3-2\sqrt{2}\)
\(A=\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}\)
\(=\sqrt{\left(2+\sqrt{3}\right)^2}-\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(=|2+\sqrt{3}|-|2-\sqrt{3}|\)
\(=2+\sqrt{3}-2+\sqrt{3}\)
\(=2\sqrt{3}\)
\(B=\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}\)
\(=\sqrt{\left(3+\sqrt{2}\right)^2}-\sqrt{\left(3-\sqrt{2}\right)^2}\)
\(=|3+\sqrt{2}|-|3-\sqrt{2}|\)
\(=3+\sqrt{2}-3+\sqrt{2}\)
\(=2\sqrt{2}\)
\(C=\sqrt{17+12\sqrt{2}}+\sqrt{17-12\sqrt{2}}\)
\(=\sqrt{\left(3+2\sqrt{2}\right)^2}+\sqrt{\left(3-2\sqrt{2}\right)^2}\)
\(=|3+2\sqrt{2}|+|3-2\sqrt{2}|\)
\(=3+2\sqrt{2}+3-2\sqrt{2}\)
\(=6\)
\(D=\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(=\sqrt{\left(2+\sqrt{5}\right)^2}-\sqrt{\left(2-\sqrt{5}\right)^2}\)
\(=|2+\sqrt{5}|-|2-\sqrt{5}|\)
\(=2+\sqrt{5}-\sqrt{5}+2\)
\(=4\)
\(E=\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}\)
\(=\sqrt{\left(1+\sqrt{5}\right)^2}-\sqrt{\left(1-\sqrt{5}\right)^2}\)
\(=|1+\sqrt{5}|-|1-\sqrt{5}|\)
\(=1+\sqrt{5}-\sqrt{5}+1\)
\(=2\)
\(A=\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}\)
\(A=\sqrt{3}+2+2-\sqrt{3}\)
A = 2 + 2
A = 4
\(B=\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}\)
\(B=\sqrt{2}+3+3-\sqrt{2}\)
B = 3 + 3
B = 6
\(C=\sqrt{17+12\sqrt{2}}+\sqrt{17-12\sqrt{2}}\)
\(C=3+2\sqrt{2}+3-2\sqrt{2}\)
C = 3 + 3
C = 6
\(D=\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(D=\sqrt{5}+2-\sqrt{5}+2\)
D = 2 + 2
D = 4
\(E=\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}\)
\(E=\sqrt{5}+1-\sqrt{5}+1\)
E = 1 + 1
E = 2
`a)sqrt{4+sqrt7}-sqrt{4-sqrt7}`
`=sqrt{(8+2sqrt7)/2}-sqrt{(8-2sqrt7)/2}`
`=sqrt{(7+2sqrt7+1)/2}-sqrt{(7-2sqrt7+1)/2}`
`=sqrt{(sqrt7+1)^2/2}-sqrt{(sqrt7-1)^2/2}`
`=(sqrt7+1)/sqrt2-(sqrt7-1)/sqrt2`
`=2/sqrt2=sqrt2`
`b)sqrt{4--sqrt15}-sqrt{4+sqrt15}`
`=sqrt{(8-2sqrt15)/2}-sqrt{(8+2sqrt15)/2}`
`=sqrt{(5-2sqrt{5.3}+3)/2}-sqrt{(5+2sqrt{5.3}+3)/2}`
`=sqrt{(sqrt5-sqrt3)^2/2}-sqrt{(sqrt5+sqrt3)^2/2}`
`=(sqrt5-sqrt3)/sqrt2-(sqrt5+sqrt3)/sqrt2`
`=(-2sqrt3)/sqrt2=-sqrt6`
`c)sqrt{2+sqrt3}+sqrt{2-sqrt3}`
`=sqrt{(4+2sqrt3)/2}+sqrt{(4-2sqrt3)/2}`
`=sqrt{(3+2sqrt3+1)/2}+sqrt{(3-2sqrt3+1)/2}`
`=sqrt{(sqrt3+1)^2/2}+sqrt{(sqrt3-1)^2/2}`
`=(sqrt3+1)/sqrt2+(sqrt3-1)/sqrt2`
`=(2sqrt3)/sqrt2=sqrt6`
`d)sqrt{9+sqrt17}-sqrt{9-sqrt17}`
`=sqrt{(18+2sqrt17)/2}-sqrt{(18-2sqrt17)/2}`
`=sqrt{(17+2sqrt17+1)/2}-sqrt{(17-2sqrt17+1)/2}`
`=sqrt{(sqrt17+1)^2/2}-sqrt{(sqrt17-1)^2/2}`
`=(sqrt17+1)/sqrt2-(sqrt17-1)/sqrt2`
`=2/sqrt2=sqrt2`
a: Ta có: \(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\)
\(=\dfrac{\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{7}+1-\sqrt{7}+1}{\sqrt{2}}=\sqrt{2}\)
b: Ta có: \(\sqrt{4-\sqrt{15}}-\sqrt{4+\sqrt{15}}\)
\(=\dfrac{\sqrt{8-2\sqrt{15}}-\sqrt{8+2\sqrt{15}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{5}-\sqrt{3}-\sqrt{5}-\sqrt{3}}{\sqrt{2}}=-\sqrt{6}\)