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RK
11 tháng 7 2021
d) \(\sqrt{17+12\sqrt{2}}=\sqrt{8+12\sqrt{2}+9}=\sqrt{\left(2\sqrt{2}+3\right)^2}=2\sqrt{2}+3\)
19 tháng 8 2023
1:
\(A=\sqrt{x^2+\dfrac{2x^2}{3}}=\sqrt{\dfrac{5x^2}{3}}=\left|\sqrt{\dfrac{5}{3}}x\right|=-x\sqrt{\dfrac{5}{3}}\)
2: \(=\left(\dfrac{\sqrt{100}+\sqrt{40}}{\sqrt{5}+\sqrt{2}}+\sqrt{6}\right)\cdot\dfrac{2\sqrt{5}-\sqrt{6}}{2}\)
\(=\dfrac{\left(2\sqrt{5}+\sqrt{6}\right)\left(2\sqrt{5}-\sqrt{6}\right)}{2}\)
\(=\dfrac{20-6}{2}=7\)
5 tháng 12 2021
Đk: x>0, x≠1
P=(√x/(√x -1) +√x/(√x +1)):√(4x)/(x-1)
P=((x+√x)/(x-1)+(x-√x)/(x-1)).(x-1)/√(4x)
P=(x+√x + x-√x)/(x-1).(x-1)/√(4x)
P=(2x)/(x-1).(x-1)/√(4x)
P=(2x)/√(4x)
P=√x
Vậy P=√x
23 tháng 10 2023
\(\dfrac{6}{\sqrt{2}-\sqrt{3}+3}\)
\(=\dfrac{6\left(\sqrt{2}-\sqrt{3}-3\right)}{\left(\sqrt{2}-\sqrt{3}\right)^2-9}\)
\(=\dfrac{6\left(\sqrt{2}-\sqrt{3}-3\right)}{-4-2\sqrt{6}}\)
\(=\dfrac{-3\left(\sqrt{2}-\sqrt{3}-3\right)}{2+\sqrt{6}}=\dfrac{-3\left(\sqrt{6}-2\right)\left(\sqrt{2}-\sqrt{3}-3\right)}{2}\)
TL
1
16 tháng 9 2021
\(A=\sqrt{5-2\sqrt{6}}-\sqrt{\left(\sqrt{2}-\sqrt{3}\right)^2}\)
\(=\sqrt{3}-\sqrt{2}-\sqrt{3}+\sqrt{2}\)
=0
TS
2
6 tháng 5 2022
gánh còng não :v
\(\left(\dfrac{\sqrt{y}}{x+\sqrt{y}}+\dfrac{\sqrt{y}}{x-\sqrt{xy}}\right):\dfrac{2\sqrt{xy}}{xy}=\left(\dfrac{\sqrt{y}}{x+\sqrt{y}}+\dfrac{\sqrt{y}}{\sqrt{x}\left(\sqrt{x}-\sqrt{y}\right)}\right):\dfrac{2}{\sqrt{xy}}=\dfrac{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)+\sqrt{x}\left(x+\sqrt{y}\right)}{\sqrt{x}\left(x+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}:\dfrac{2}{\sqrt{xy}}=\dfrac{x\sqrt{y}-y\sqrt{x}+x\sqrt{x}+\sqrt{xy}}{\sqrt{x}\left(x+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}:\dfrac{2}{\sqrt{xy}}=\dfrac{\sqrt{x}\left(\sqrt{xy}-y+x+\sqrt{y}\right)}{\sqrt{x}\left(x+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}:\dfrac{2}{\sqrt{xy}}=\dfrac{\sqrt{y}\left(\sqrt{x}-\sqrt{y}\right)+\left(x+\sqrt{y}\right)}{\left(x +\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}:\dfrac{2}{\sqrt{xy}}\) mình làm đc đó thôi ( mỏi tay :v )
H
11 tháng 7 2021
a) \(\sqrt{6+2\sqrt{5}}=\sqrt{5+2\sqrt{5}+1}=\sqrt{\left(\sqrt{5}+1\right)}=\sqrt{5}+1\)
b) \(\sqrt{8-2\sqrt{7}}=\sqrt{7-2\sqrt{7}+1}=\sqrt{\left(\sqrt{7}-1\right)^2}=\sqrt{7}-1\left(do\sqrt{7}>1\right)\)
c) \(\sqrt{9+4\sqrt{5}}=\sqrt{5+2\cdot\sqrt{5}\cdot2+4}=\sqrt{\left(\sqrt{5}+2\right)^2}=2+\sqrt{5}\)
a, \(\sqrt{23-8\sqrt{7}}\)
\(=\sqrt{16-8\sqrt{7}+7}\)
\(=\sqrt{4^2-2.4.\sqrt{7}+\left(\sqrt{7}\right)^2}\)
\(=\sqrt{\left(4-\sqrt{7}\right)^2}\)
\(=\left|4-\sqrt{7}\right|\)
\(=4-\sqrt{7}\)
b, \(\sqrt{18-8\sqrt{2}}\)
\(=\sqrt{16-8\sqrt{2}+2}\)
\(=\sqrt{4^2-2.4.\sqrt{2}+\left(\sqrt{2}\right)^2}\)
\(=\sqrt{\left(4-\sqrt{2}\right)^2}\)
\(=\left|4-\sqrt{2}\right|\)
\(=4-\sqrt{2}\)
c, \(\sqrt{12+6\sqrt{3}}\)
\(=\sqrt{9+6\sqrt{3}+3}\)
\(=\sqrt{3^2+2.3.\sqrt{3}+\left(\sqrt{3}\right)^2}\)
\(=\sqrt{\left(3+\sqrt{3}\right)^2}\)
\(=\left|3+\sqrt{3}\right|\)
\(=3+\sqrt{3}\)
d, \(\sqrt{17+12\sqrt{2}}\)
\(=\sqrt{9+12\sqrt{2}+8}\)
\(=\sqrt{3^2+3.2.2\sqrt{2}+\left(2\sqrt{2}\right)^2}\)
\(=\sqrt{\left(3+2\sqrt{2}\right)^2}\)
\(=\left|3+2\sqrt{2}\right|\)
\(=3+2\sqrt{2}\)
a) \(\sqrt{23-8\sqrt{7}}=4-\sqrt{7}\)
b) \(\sqrt{18-8\sqrt{2}}=4-\sqrt{2}\)
c) \(\sqrt{12+6\sqrt{3}}=3+\sqrt{3}\)
d) \(\sqrt{17+12\sqrt{2}}=3+2\sqrt{2}\)