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Lời giải:
a. \(=|\sqrt{7}-5|+|2-\sqrt{7}|=5-\sqrt{7}+(\sqrt{7}-2)=3\)
b. \(=\sqrt{(3+\sqrt{2})^2}-\sqrt{(3-\sqrt{2})^2}=|3+\sqrt{2}|-|3-\sqrt{2}|\)
\(=(3+\sqrt{2})-(3-\sqrt{2})=2\sqrt{2}\)
c.
\(=\sqrt{(3+2\sqrt{2})^2}+\sqrt{(3-2\sqrt{2})^2}=|3+2\sqrt{2}|+|3-2\sqrt{2}|\)
$=(3+2\sqrt{2})+(3-2\sqrt{2})=6$
d.
$=\sqrt{(\sqrt{5}+1)^2}-\sqrt{(\sqrt{5}-1)^2}$
$=|\sqrt{5}+1|-|\sqrt{5}-1|=\sqrt{5}+1-(\sqrt{5}-1)=2$
\(a,=\sqrt{17}-4-\sqrt{17}-2=-6\\ b,=7\left(\sqrt{3}+\sqrt{2}\right)-7\sqrt{3}-6\sqrt{2}\\ =7\sqrt{3}+7\sqrt{2}-7\sqrt{3}-6\sqrt{2}=\sqrt{2}\\ c,=\dfrac{6\sqrt{5}+12-6\sqrt{5}+12}{3}+2\sqrt{2}-\dfrac{4\sqrt{7}}{7}\\ =8+2\sqrt{2}-\dfrac{4\sqrt{7}}{7}=\dfrac{56+14\sqrt{2}-4\sqrt{7}}{7}\\ d,=\left(\dfrac{\sqrt{2}}{4}-\dfrac{6\sqrt{2}}{4}+8\sqrt{2}\right):\dfrac{1}{8}=\dfrac{-5\sqrt{2}+32\sqrt{2}}{4}\cdot8=54\sqrt{2}\)
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: \(17-12\sqrt{2}=8-12\sqrt{2}+9=\left(2\sqrt{2}\right)^2-12\sqrt{2}+3^2=\left(2\sqrt{2}-3\right)^2\)
\(VT=\left|2\sqrt{2}-3\right|+2\sqrt{2}=3-2\sqrt{2}+2\sqrt{2}=3=VP\)
`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)
\(A=\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}\)
\(=\sqrt{\left(2+\sqrt{3}\right)^2}-\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(=|2+\sqrt{3}|-|2-\sqrt{3}|\)
\(=2+\sqrt{3}-2+\sqrt{3}\)
\(=2\sqrt{3}\)
\(B=\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}\)
\(=\sqrt{\left(3+\sqrt{2}\right)^2}-\sqrt{\left(3-\sqrt{2}\right)^2}\)
\(=|3+\sqrt{2}|-|3-\sqrt{2}|\)
\(=3+\sqrt{2}-3+\sqrt{2}\)
\(=2\sqrt{2}\)
\(C=\sqrt{17+12\sqrt{2}}+\sqrt{17-12\sqrt{2}}\)
\(=\sqrt{\left(3+2\sqrt{2}\right)^2}+\sqrt{\left(3-2\sqrt{2}\right)^2}\)
\(=|3+2\sqrt{2}|+|3-2\sqrt{2}|\)
\(=3+2\sqrt{2}+3-2\sqrt{2}\)
\(=6\)
\(D=\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(=\sqrt{\left(2+\sqrt{5}\right)^2}-\sqrt{\left(2-\sqrt{5}\right)^2}\)
\(=|2+\sqrt{5}|-|2-\sqrt{5}|\)
\(=2+\sqrt{5}-\sqrt{5}+2\)
\(=4\)
\(E=\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}\)
\(=\sqrt{\left(1+\sqrt{5}\right)^2}-\sqrt{\left(1-\sqrt{5}\right)^2}\)
\(=|1+\sqrt{5}|-|1-\sqrt{5}|\)
\(=1+\sqrt{5}-\sqrt{5}+1\)
\(=2\)
\(A=\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}\)
\(A=\sqrt{3}+2+2-\sqrt{3}\)
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