Tính: (2-\(\sqrt{5}\))2
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T
1
H
23 tháng 7 2021
Hướng giải :
Điều kiện xác định x>=5/2
Nhân căn 2 cả 2 vế của pt ta được
\(\sqrt{2x+4+6\sqrt{2x-5}} + \sqrt{2x-4-2\sqrt{2x-5}} = 4 \\ \sqrt{(\sqrt{2x-5} + 3)^2} +\sqrt{(\sqrt{2x-5} -1)^2} =4\\ |\sqrt{2x-5} +3| + | \sqrt{2x-5} -1| =4\)
Ta có : |A| + |B| >= |A+B|
Suy ra VT >= VP
Dấu = xảy ra <=> AB >= 0
Mà A >= 0 ; suy ra B >=0
=> căn (2x-5) >= 1
=> 2x-5 >=1
=> 2x>=6
=> x>=3
RK
22 tháng 6 2021
a) A= \(\sqrt{\left(2-\sqrt{5}\right)^2}+\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}\)
Vì \(\left\{{}\begin{matrix}2=\sqrt{4}< \sqrt{5}\\2\sqrt{2}=\sqrt{8}>\sqrt{5}\end{matrix}\right.\) nên A = \(\sqrt{\left(\sqrt{5}-2\right)^2}+\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}\)
= \(\sqrt{5}-2+2\sqrt{2}-\sqrt{5}\)
= \(2\left(\sqrt{2}-1\right)\)
RK
22 tháng 6 2021
b) B = \(\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}\) (B > 0)
Ta có:
B2 = \(6+2\sqrt{5}-2\sqrt{\left(6+2\sqrt{5}\right)\left(6-2\sqrt{5}\right)}+6-2\sqrt{5}\)
= \(12-2\sqrt{36-20}\)
= \(12-8\)
= \(4\)
\(\Rightarrow\) B =\(\pm2\) nhưng vì B > 0 nên B = 2
Vậy B = 2
25 tháng 9 2021
a)A=\(2\sqrt{3}-8\sqrt{3}+7\sqrt{3}=\sqrt{3}\)
b)B\(=\sqrt{\left(3-\sqrt{5}\right)^2}+\sqrt{\left(2-\sqrt{5}\right)^2}=3-\sqrt{5}+\sqrt{5}-2=1\)
d)\(=\dfrac{\left(5+\sqrt{5}\right)\left(\sqrt{5}-2\right)}{1}+1-\sqrt{5}-\dfrac{11\left(2\sqrt{5}-3\right)}{11}=5\sqrt{5}+5-10-2\sqrt{5}+1-\sqrt{5}-2\sqrt{5}+3=-1\)
T
23 tháng 9 2023
\(a,\dfrac{2}{\sqrt{3}-\sqrt{5}}+\dfrac{3-2\sqrt{3}}{\sqrt{3}-2}\)
\(=\dfrac{5-3}{\sqrt{3}-\sqrt{5}}+\dfrac{\sqrt{3}\left(\sqrt{3}-2\right)}{\sqrt{3}-2}\)
\(=\dfrac{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}{\sqrt{5}-\sqrt{3}}+\sqrt{3}\)
\(=\sqrt{5}+\sqrt{3}+\sqrt{3}\)
\(=\sqrt{5}+2\sqrt{3}\)
\(b,\dfrac{5-\sqrt{5}}{\sqrt{5}-1}+\dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+3}\)
\(=\dfrac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}+\dfrac{\left(\sqrt{5}-\sqrt{3}\right)^2}{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}\)
\(=\sqrt{5}+\dfrac{5-2\sqrt{15}+3}{5-3}\)
\(=\dfrac{2\sqrt{5}+8-2\sqrt{15}}{2}\)
\(=\dfrac{2\cdot\left(\sqrt{5}+4-\sqrt{15}\right)}{2}\)
\(=\sqrt{5}-\sqrt{15}+4\)
#\(Toru\)
25 tháng 6 2021
a) Ta có: \(A=\sqrt{\sqrt{3}+\sqrt{2}}\cdot\sqrt{\sqrt{3}-\sqrt{2}}\)
\(=\sqrt{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}\)
\(=\sqrt{3-2}=1\)
b) Ta có: \(B=\sqrt{5-2\sqrt{6}}+\sqrt{5+2\sqrt{6}}\)
\(=\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}\)
\(=\sqrt{3}-\sqrt{2}+\sqrt{3}+\sqrt{2}\)
\(=2\sqrt{3}\)
25 tháng 6 2021
`A=sqrt{sqrt3+sqrt2}.sqrt{sqrt3-sqrt2}`
`=sqrt{(sqrt3+sqrt2)(sqrt3-sqrt2)}`
`=sqrt{3-2}=1`
`b)B=sqrt{5-2sqrt6}+sqrt{5+2sqrt6}`
`=sqrt{3-2sqrt6+2}+sqrt{3+2sqrt6+2}`
`=sqrt{(sqrt3-sqrt2)^2}+sqrt{(sqrt3+sqrt2)^2}`
`=sqrt3-sqrt2+sqrt3+sqrt2=2sqrt3`
`c)C=3-sqrt{3-sqrt5}`
`=3-sqrt{(6-2sqrt5)/2}`
`=3-sqrt{(sqrt5-1)^2/2}`
`=3-(sqrt5-1)/sqrt2`
`=3-(sqrt{10}-sqrt2)/2`
`=(6-sqrt{10}+sqrt2)/2`
25 tháng 9 2021
\(1,=20-7=13\\ b,=12-50=-38\\ c,=\sqrt{7}-2+\sqrt{7}+2=2\sqrt{7}\\ d,=\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}=2\sqrt{3}\\ e,=11+2\sqrt{30}\\ f,=8-2\sqrt{15}\\ g,=11+2\sqrt{6}\)
25 tháng 9 2021
1) \(=\left(2\sqrt{5}\right)^2-\left(\sqrt{7}\right)^2=20-7=13\)
2) \(=\left(2\sqrt{3}\right)^2-\left(5\sqrt{2}\right)^2=12-50=-38\)
3) \(=\sqrt{7}-2+\sqrt{7}+2=2\sqrt[]{7}\)
4) \(=\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}=2\sqrt{3}\)
5) \(=5+6-2\sqrt{5.6}=11-2\sqrt{30}\)
6) \(=3+5-2\sqrt{3.5}=8-4\sqrt{2}\)
7) \(=\left(2\sqrt{2}\right)^2+\left(\sqrt{3}\right)^2+2\sqrt{2\sqrt{2}.3}=11+2\sqrt{6\sqrt{2}}\)
3 tháng 7 2018
\(\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}=\frac{\left(\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}\right).\sqrt{\sqrt{5}-1}}{\sqrt{\sqrt{5}+1}.\sqrt{\sqrt{5}-1}}=\frac{\sqrt{3+\sqrt{5}}+\sqrt{7-3\sqrt{5}}}{\sqrt{5-1}}\)\(=\frac{\sqrt{3+\sqrt{5}}+\sqrt{7-3\sqrt{5}}}{2}=\frac{\sqrt{2}.\left(\sqrt{3+\sqrt{5}}+\sqrt{7-3\sqrt{5}}\right)}{2\sqrt{2}}=\frac{\sqrt{6+2\sqrt{5}}+\sqrt{14-6\sqrt{5}}}{2\sqrt{2}}\)
\(=\frac{\sqrt{\left(\sqrt{5}+1\right)^2}+\sqrt{\left(3-\sqrt{5}\right)^2}}{2\sqrt{2}}=\frac{\left|\sqrt{5}+1\right|+\left|3-\sqrt{5}\right|}{2\sqrt{2}}=\frac{\sqrt{5}+1+3-\sqrt{5}}{2\sqrt{2}}\)
\(=\frac{4}{2\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}\)
Lại có: \(\sqrt{3-2\sqrt{2}}=\sqrt{\left(\sqrt{2}-1\right)^2}=\left|\sqrt{2}-1\right|=\sqrt{2}-1\)
\(\Rightarrow\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}-\sqrt{3-2\sqrt{2}}=\sqrt{2}-\left(\sqrt{2}-1\right)=1\)
4 tháng 3 2022
sửa đề \(\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{\sqrt{5}+1}+\dfrac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}}-\dfrac{11\left(2\sqrt{3}-3\right)}{3}\)
\(=\sqrt{5}+\sqrt{5}-1-\dfrac{11\left(2\sqrt{3}-3\right)}{3}=2\sqrt{5}-1-\dfrac{11\left(2\sqrt{3}-3\right)}{3}\)
\(=\dfrac{6\sqrt{5}-3-22\sqrt{3}+33}{3}=\dfrac{30+6\sqrt{5}-22\sqrt{3}}{3}\)
\(\left(2-\sqrt{5}\right)^2=2^2-2.2.\sqrt{5}+\left(\sqrt{5}\right)^2=4-4\sqrt{5}+5=9-4\sqrt{5}\)
\(29-4\sqrt{5}\)