\(\sqrt{11+2\sqrt{18}}\) giải giúp nhé
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NV
Nguyễn Việt Lâm
Giáo viên
16 tháng 8 2021
Nhìn nó có dạng \(\sqrt{\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2}=\sqrt{a+b+c+2\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)}\)
Thấy \(2\left(\sqrt{6}+\sqrt{12}+\sqrt{18}\right)=2\left(\sqrt{2.3}+\sqrt{2.6}+\sqrt{3.6}\right)\)
nên:
\(\sqrt{11+2\left(\sqrt{6}+\sqrt{12}+\sqrt{18}\right)}=\sqrt{\sqrt{2}^2+\sqrt{6}^2+\sqrt{3}^2+2\left(\sqrt{2.3}+\sqrt{2.6}+\sqrt{3.6}\right)}\)
\(=\sqrt{\left(\sqrt{2}+\sqrt{3}+\sqrt{6}\right)^2}=\sqrt{2}+\sqrt{3}+\sqrt{6}\)
25 tháng 9 2021
1) \(=\sqrt{\left(\sqrt{3}-1\right)^2}=\sqrt{3}-1\)
2) \(=\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}=\sqrt{3}+\sqrt{2}\)
3) \(=\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}=\sqrt{5}-\sqrt{2}\)
5) \(=\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}=\sqrt{5}+\sqrt{3}\)
6) \(=\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}=\sqrt{7}-\sqrt{3}\)
7) \(=\sqrt{\left(3+\sqrt{2}\right)^2}=3+\sqrt{2}\)
LT
22 tháng 10 2021
a)√x−2+12√4x−8=√9x−18−2
=>√x−2+12√4(x−2)=√9(x−2)−2
=>√x−2+12√22(x−2)=√32(x−2)−2
=>√x−2+12.2√(x−2)=3√(x−2)−2
=>√x−2+24√(x−2)=3√(x−2)−2
=>√x−2+24√(x−2)-3√(x−2)=-2
=>√x−2(1+24-3)=-2
=>22√x−2=-2
=>√x−2=-2/22
=>√x−2=-1/11
=>x−2=1/121
=>x=1/121+2=243/121
b)√(3x−1)2=5
=>|3x−1|=5
=>3x−1=5 hoặc 3x−1=-5
=>3x=6 hoặc 3x=-4
=>x=2 hoặc x=-4/3
21 tháng 7 2023
`a, (sqrt 28 - sqrt 12 - sqrt 7) sqrt 7 + 2 sqrt 21`.
`= sqrt(28.7) - sqrt(12.7) - sqrt(7.7) + 2 sqrt 21`.
`= sqrt(4. 7.7) - sqrt (12.7) - 7 + 2 sqrt 21`.
`= 14 - sqrt(4.3.7) - 7 + 2 sqrt 21`.
`= 7`.
`b, (sqrt99-sqrt18-sqrt11)sqrt11+3sqrt22`
`= sqrt(99.11)- sqrt(18.11)-sqrt(11.11) +3sqrt22`
`= sqrt(9.11.11)-sqrt(2.9.11)-11+3sqrt22`
`= 33 - 11 = 22`.
25 tháng 6 2017
1. \(\frac{1}{2}\sqrt{48}-2\sqrt{75}-\frac{\sqrt{33}}{\sqrt{11}}+\sqrt{84}\)= -6,423305878
2. \(\sqrt{150}+\sqrt{1,6}\sqrt{60}+4,5\sqrt{2\frac{2}{3}}-\sqrt{6}\)= 24,79207036
NHA 
Vũ Hoàng Thiên An ! ! !
K VÀ KB NHA !
NQ
5
TN
3 tháng 2 2016
\(\Leftrightarrow\sqrt{14x+18}+\sqrt{10x+11}+8x^2=11\)
\(\Rightarrow\sqrt{14x+18}+\sqrt{10x+11}+8x^2-11=0\)
\(\Rightarrow-\frac{\sqrt{-\sqrt{14x+18}-\sqrt{10x+11}+11}-\sqrt{2^3}x}{\sqrt{2^3}}=0\)
\(\Rightarrow\sqrt{-\sqrt{14x+18}-\sqrt{10x+11}+11}-\sqrt{2^3}x=0\)
\(\Rightarrow\frac{\sqrt{-\sqrt{14x+18}-\sqrt{10x+11}+11}+\sqrt{2^3}x}{\sqrt{2^3}}=0\)
\(\Rightarrow\sqrt{-\sqrt{14x+18}-\sqrt{10x+11}+11}+\sqrt{2^3}x=0\)
=>2x=1
=>x=-1
=>x=1:2=\(\frac{1}{2}\)
AT
4 tháng 7 2021
a) \(A=\sqrt{4-2\sqrt{3}}-\sqrt{4+2\sqrt{3}}\)
\(=\sqrt{\left(\sqrt{3}\right)^2-2.\sqrt{3}.1+1^2}-\sqrt{\left(\sqrt{3}\right)^2+2.\sqrt{3}.1+1^2}\)
\(=\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{\left(\sqrt{3}+1\right)^2}=\left|\sqrt{3}-1\right|-\left|\sqrt{3}+1\right|\)
\(=\sqrt{3}-1+-\sqrt{3}-1=-2\)
b) \(B=\sqrt{11-6\sqrt{2}}-\sqrt{3-2\sqrt{2}}\)
\(=\sqrt{3^2-2.3.\sqrt{2}+\left(\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{2}\right)^2-2.\sqrt{2}.1+1^2}\)
\(=\sqrt{\left(3-\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{2}-1\right)^2}=\left|3-\sqrt{2}\right|-\left|\sqrt{2}-1\right|\)
\(=3-\sqrt{2}-\sqrt{2}+1=4-2\sqrt{2}\)
c) \(C=\left(\sqrt{3}+\sqrt{5}\right)\sqrt{7-2\sqrt{10}}\)
\(=\left(\sqrt{5}+\sqrt{3}\right)\sqrt{\left(\sqrt{5}\right)^2-2.\sqrt{5}.\sqrt{2}+\left(\sqrt{2}\right)^2}\)
\(=\left(\sqrt{5}+\sqrt{3}\right)\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}=\left(\sqrt{5}+\sqrt{3}\right)\left|\sqrt{5}-\sqrt{2}\right|\)
\(=\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{2}\right)=5-\sqrt{10}+\sqrt{15}-\sqrt{6}\)
25 tháng 6 2021
`A=(sqrtx-1)/(sqrtx+1)-(sqrtx+3)/(sqrtx-2)-(x+5)/(x-sqrtx-2)`
`đk:x>=0,x ne 4`
`A=((sqrtx-1)(sqrtx-2)-(sqrtx+3)(sqrtx+1)-x-5)/(x-sqrtx-2)`
`=(x-3sqrtx+2-x-4sqrtx-3-x-5)/(x-sqrtx-2)`
`=(-x-7sqrtx-6)/(x-sqrtx-2)`
`=(-(sqrtx+1)(sqrtx+6))/((sqrtx+1)(sqrtx-2))`
`=(-(sqrtx+6))/(sqrtx-2)`
\(\sqrt{11+2\sqrt{18}}=\sqrt{11+2\sqrt{9.2}}\)
\(=\sqrt{\left(\sqrt{9}\right)^2+2\sqrt{9.2}+\left(\sqrt{2}\right)^2}\)
\(=\sqrt{\left(\sqrt{9}+\sqrt{2}\right)^2}=\left|\sqrt{9}+\sqrt{2}\right|=\sqrt{9}+\sqrt{2}\)