Tính \(\sqrt{2\sqrt{6}+6+2\sqrt{3}+2\sqrt{2}}\)
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8 tháng 7 2017
Ta có A > 0
Từ đó \(A^2=2+\sqrt{2+\sqrt{2+...}}\Leftrightarrow A^2=2+A\Leftrightarrow A^2-A-2=0\)
\(\Leftrightarrow\left(A+1\right)\left(A-2\right)=0\Leftrightarrow\orbr{\begin{cases}A+1=0\\A-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}A=-1\\A=2\end{cases}}\)
Do A > 0 nên A= 2
8 tháng 7 2017
b, tương tự
c,\(C>2\)
Xét \(C^2=5+\sqrt{13+\sqrt{5+\sqrt{13...}}}\)
\(\left(C^2-5\right)^2=13+C\Leftrightarrow C^4-10C^2-C+12=0\Leftrightarrow\left(C^4-9C^2\right)-\left(C^2-9\right)-\left(C-3\right)=0\)
\(\Leftrightarrow\left(C-3\right)\left[\left(C+3\right)\left(C-1\right)\left(C+1\right)-1\right]=0\)
VÌ C> 2 => C-3 = 0 => C=3
1 tháng 10 2021
a: Ta có: \(\sqrt{\left(5-\sqrt{19}\right)^2}-\sqrt{\left(4-\sqrt{19}\right)^2}\)
\(=5-\sqrt{19}-\sqrt{19}+4\)
\(=9-2\sqrt{19}\)
b: Ta có: \(\sqrt{\left(3-2\sqrt{2}\right)^2}-\sqrt{\left(2\sqrt{2}-3\right)^2}\)
\(=3-2\sqrt{2}-3+2\sqrt{2}\)
=0
AH
Akai Haruma
Giáo viên
2 tháng 10 2021
c.
Căn bậc 2 không xác định do $2-\sqrt{5}< 0$
d.
\(=\sqrt{(3+\sqrt{3})^2}(3+\sqrt{3})=|3+\sqrt{3}|(3+\sqrt{3})=(3+\sqrt{3})^2=12+6\sqrt{3}\)
e.
\(=(2-\sqrt{5})\sqrt{(2+\sqrt{5})^2}=(2-\sqrt{5})|2+\sqrt{5}|=(2-\sqrt{5})(2+\sqrt{5})=4-5=-1\)
13 tháng 7 2017
\(C=\sqrt{x}+\frac{\sqrt[3]{2-\sqrt{3}}.\sqrt[6]{7+4\sqrt{3}}-x}{\sqrt[4]{9-4\sqrt{5}}.\sqrt{2+\sqrt{5}}+\sqrt{x}}\)
\(=\sqrt{x}+\frac{\sqrt[6]{\left(7-4\sqrt{3}\right).\left(7+4\sqrt{3}\right)}-x}{\sqrt[4]{\left(9+4\sqrt{5}\right).\left(9-4\sqrt{5}\right)}+\sqrt{x}}\)
\(=\sqrt{x}+\frac{1-x}{1+\sqrt{x}}=\sqrt{x}+\frac{\left(1+\sqrt{x}\right).\left(1-\sqrt{x}\right)}{1+\sqrt{x}}\)
\(=\sqrt{x}+1-\sqrt{x}=1\)
NT
3
DH
14 tháng 9 2020
\(T=\left(2\sqrt{3}+1\right)\left(3\sqrt{2}-1\right)\sqrt{13-4\sqrt{3}}.\sqrt{19+6\sqrt{6}}\)
\(T=\left(2\sqrt{3}+1\right)\left(3\sqrt{2}-1\right)\sqrt{\left(2\sqrt{3}-1\right)^2}.\sqrt{\left(3\sqrt{2}+1\right)^2}\)
\(T=\left(2\sqrt{3}+1\right)\left(3\sqrt{2}-1\right)\left|2\sqrt{3}-1\right|.\left|3\sqrt{2}+1\right|\)
\(T=\left(2\sqrt{3}+1\right)\left(3\sqrt{2}-1\right)\left(2\sqrt{3}-1\right)\left(3\sqrt{2}+1\right)\)
\(T=\left(2\sqrt{3}+1\right)\left(2\sqrt{3}-1\right)\left(3\sqrt{2}-1\right)\left(3\sqrt{2}+1\right)\)
\(T=11\cdot17\)
\(T=187\)
LQ
1
DN
31 tháng 5 2018
\(A=\left(3\sqrt{2}+\sqrt{6}\right)\sqrt{\frac{9-2.3\sqrt{3}+3}{2}}=\frac{\sqrt{2}\left(3+\sqrt{3}\right)}{\sqrt{2}}.\sqrt{\left(3-\sqrt{3}\right)^2}=\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)=9-3=6\)
QH
25 tháng 9 2016
a, \(\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}-\sqrt{2}=\sqrt{\frac{6+2\sqrt{5}}{2}}-\sqrt{\frac{6-2\sqrt{5}}{2}}-\sqrt{2}\)
\(=\sqrt{\frac{\left(\sqrt{2}+\sqrt{3}\right)^2}{2}}-\sqrt{\frac{\left(\sqrt{3}-\sqrt{2}\right)}{2}}-\sqrt{2}=\frac{\sqrt{\left(\sqrt{3}+\sqrt{2}\right)}}{\sqrt{2}}-\frac{\sqrt{\left(\sqrt{3}-\sqrt{2}\right)}}{\sqrt{2}}-\sqrt{2}\)
\(=\frac{\left|\sqrt{3}+\sqrt{2}\right|}{\sqrt{2}}-\frac{\left|\sqrt{3}-\sqrt{2}\right|}{\sqrt{2}}-\sqrt{2}=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{2}}-\frac{\sqrt{3}-\sqrt{2}}{\sqrt{2}}-\frac{2}{\sqrt{2}}\)
= \(\frac{2\sqrt{2}-2}{\sqrt{2}}=\sqrt{2}-1\)
b, Tương tự
NT
1
BT
12 tháng 12 2017
\(\dfrac{10\sqrt{6}-12}{\sqrt{6}-5}-3\sqrt{\dfrac{2}{3}}+\dfrac{15}{\sqrt{6}-1}\)
\(=\dfrac{2\sqrt{6}\left(5-\sqrt{6}\right)}{\sqrt{6}-5}-3.\dfrac{\sqrt{2}}{\sqrt{3}}+\dfrac{15\left(\sqrt{6}+1\right)}{\left(\sqrt{6}-1\right)\left(\sqrt{6}+1\right)}\)
\(=-2\sqrt{6}-\sqrt{3}.\sqrt{2}+\dfrac{15\left(\sqrt{6}+1\right)}{6-1}\)
\(=-2\sqrt{6}-\sqrt{6}+3\left(\sqrt{6}+1\right)\)
\(=3\).
\(\sqrt{2\sqrt{6}+6+2\sqrt{3}+2\sqrt{2}}\)
=\(\sqrt{2+3+1+2\sqrt{6}+2\sqrt{3}+2\sqrt{2}}\)
=\(\sqrt{\left(\sqrt{2}\right)^2+\left(\sqrt{3}\right)^2+1^2+2\sqrt{2}.\sqrt{3}+2\sqrt{3}.1+2\sqrt{2}.1}\)
=\(\sqrt{\left(\sqrt{2}+\sqrt{3}+1\right)^2}\)
=\(\sqrt{2}+\sqrt{3}+1\)(Vì \(\sqrt{2}+\sqrt{3}+1>0\))
Chắc đúng á :=)