Cm
\(\sqrt{6\sqrt{6+\sqrt{6}}}\)<9
K
Khách
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BK
1
PT
24 tháng 9 2017
\(B=\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6+....+\sqrt{6}}}}}\)
\(\Rightarrow B^2=6+\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6+....+\sqrt{6}}}}}\)
\(\Rightarrow B^2=6+B\)
\(\Rightarrow B^2-B-6=0\)
\(\Rightarrow B^2-2B+3A-6=0\)
\(\Rightarrow B\left(B-2\right)+3\left(B-2\right)=0\)
\(\Rightarrow\left(B+3\right)\left(B-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}B+3=0\\B-2=0\end{matrix}\right.\)
\(\Rightarrow B=2\left(B\in Z\right)\)
\(\Rightarrow\) Đpcm.
12 tháng 9 2021
\(\left(\dfrac{2}{\sqrt{6}-1}+\dfrac{3}{\sqrt{6}-2}-\dfrac{3}{3-\sqrt{6}}\right)\cdot\dfrac{5}{9\sqrt{6}+4}\)
\(=\left(\dfrac{2+2\sqrt{6}}{5}+\dfrac{6+3\sqrt{6}}{2}-3-\sqrt{6}\right)\cdot\dfrac{5}{9\sqrt{6}+4}\)
\(=\dfrac{4+4\sqrt{6}+30+15\sqrt{6}-30-10\sqrt{6}}{10}\cdot\dfrac{5}{9\sqrt{6}+4}\)
\(=\dfrac{1}{2}\)
17 tháng 7 2022
a: \(=\dfrac{2+\sqrt{3}}{2}:\left(1+\sqrt{\dfrac{2+\sqrt{3}}{2}}\right)+\dfrac{2-\sqrt{3}}{2}:\left(1-\sqrt{\dfrac{2-\sqrt{3}}{2}}\right)\)
\(=\dfrac{2+\sqrt{3}}{2}:\left(1+\sqrt{\dfrac{4+2\sqrt{3}}{4}}\right)+\dfrac{2-\sqrt{3}}{2}:\left(1-\sqrt{\dfrac{4-2\sqrt{3}}{4}}\right)\)
\(=\dfrac{2+\sqrt{3}}{2}:\left(1+\dfrac{\sqrt{3}+1}{2}\right)+\dfrac{2-\sqrt{3}}{2}:\left(1-\dfrac{\sqrt{3}-1}{2}\right)\)
\(=\dfrac{2+\sqrt{3}}{2}\cdot\dfrac{2}{2+\sqrt{3}+1}+\dfrac{2-\sqrt{3}}{2}\cdot\dfrac{2}{2-\sqrt{3}+1}\)
\(=\dfrac{2+\sqrt{3}}{3+\sqrt{3}}+\dfrac{2-\sqrt{3}}{3-\sqrt{3}}\)
\(=\dfrac{\left(2+\sqrt{3}\right)\left(3-\sqrt{3}\right)+\left(2-\sqrt{3}\right)\left(3+\sqrt{3}\right)}{9-3}\)
\(=\dfrac{6-2\sqrt{3}+3\sqrt{3}-3+6+2\sqrt{3}-3\sqrt{3}-3}{6}\)
\(=\dfrac{6}{6}=1\)
17 tháng 6 2021
Bài 1
a) Đặt VT = A
<=> \(2\sqrt{2}A=\left(8+2\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{8-2\sqrt{15}}\)
<=> \(2\sqrt{2}A=\left(\sqrt{5}+\sqrt{3}\right)^2.\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right).\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)
<=> \(2A=\left(\sqrt{5}+\sqrt{3}\right)^2.\left(\sqrt{5}-\sqrt{3}\right)^2\)
<=> 2A = \(\left(5-3\right)^2=4\)
<=> A = 2
b) Đặt VT = B
<=> \(2\sqrt{2}B=\left(10+2\sqrt{21}\right).\left(\sqrt{14}-\sqrt{6}\right)\sqrt{10-2\sqrt{21}}\)
<=> \(2\sqrt{2}B=\left(\sqrt{7}+\sqrt{3}\right)^2.\sqrt{2}\left(\sqrt{7}-\sqrt{3}\right).\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}\)
<=> \(2B=\left(\sqrt{7}+\sqrt{3}\right)^2.\left(\sqrt{7}-\sqrt{3}\right)^2=\left(7-3\right)^2=16\)
<=> B = 8
Bài 2
Đặt VT = A
<=> A2 = \(\dfrac{\sqrt{5}+2+\sqrt{5}-2+2\sqrt{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}}{2}\)
<=> A2 = \(\dfrac{2\sqrt{5}+2\sqrt{5-4}}{2}=\dfrac{2\sqrt{5}+2}{2}=\sqrt{5}+1\)
<=> \(A=\sqrt{\sqrt{5}+1}\)
14 tháng 10 2016
\(x^2=6+2\sqrt{2}+2\sqrt{\left[\left(3+\sqrt{2}\right)+\left(\sqrt{3}+\sqrt{6}\right)\right].\left[\left(3+\sqrt{2}\right)-\left(\sqrt{3}+\sqrt{6}\right)\right]}\)
\(=6+2\sqrt{2}+2\sqrt{11+6\sqrt{2}-\left(9+6\sqrt{2}\right)}=6+2\sqrt{2}+2\sqrt{2}=6+4\sqrt{2}=\left(\sqrt{2}+2\right)^2\)
\(\Rightarrow x=\sqrt{2}+2\)
...............................................................
7 tháng 10 2022
a: \(=\dfrac{3}{2}\sqrt{6}+\dfrac{2}{3}\sqrt{6}-2\sqrt{6}\)
\(=\dfrac{1}{6}\sqrt{6}\)
b: \(VT=\dfrac{\sqrt{y}}{\sqrt{x}\left(\sqrt{x}-\sqrt{y}\right)}+\dfrac{\sqrt{x}}{\sqrt{y}\left(\sqrt{y}-\sqrt{x}\right)}\)
\(=\dfrac{y-x}{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}=\dfrac{-\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}}\)
BT
5 tháng 11 2016
tách 10 + 6 căn 3 = 1 + 3 căn 3 +3 căn 3 + 9 = ( căn 3 -1)3
6 + 2 căn 5 = ( căn 5+1)2
sau đó thay vô là được
5 tháng 11 2016
Ta có
\(\frac{\sqrt[3]{10+6\sqrt{3}}.\left(\sqrt{3}-1\right)}{\sqrt{6+2\sqrt{5}}-\sqrt{5}}\)
\(=\frac{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}{\sqrt{5}+1-\sqrt{5}}=2\)
Thế vào ta được
P = (23 - 4×2 - 1)2012 = 1
\(\sqrt{6\sqrt{6+\sqrt{6}}}< \sqrt{6\sqrt{6+\sqrt{9}}}\)
\(=\sqrt{6\sqrt{6+3}}=\sqrt{6\sqrt{9}}\)
\(=\sqrt{6.3}< \sqrt{81}=9\)
Vậy \(\sqrt{6\sqrt{6+\sqrt{6}}}< 9\)