\(\sqrt{\sqrt{17+12\sqrt{2}}}\)và \(\sqrt{2}\)\(+\)\(1\)
\(\sqrt{\sqrt{28-16\sqrt{3}}}\)và \(\sqrt{3}\)\(-\)\(2\)
So sánh cho e vs ạ . e hứa sẽ click cho ạ . E cảm ơn ạ !!
K
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KB
1 tháng 5 2019
a ) Để ý thấy \(16\sqrt{3}=2.2\sqrt{3}.4=2.\sqrt{12}.4\) , như vậy , ta sẽ tách :
\(28=12+16\) \(\Rightarrow\sqrt{\sqrt{28+16\sqrt{3}}=\sqrt{\sqrt{12+16+16\sqrt{3}}}}=\sqrt{\sqrt{\left(\sqrt{12}+4\right)^2}}=\sqrt{\sqrt{12}+4}\)
\(=\sqrt{3+2.\sqrt{3}+1}=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)
b ) \(4\sqrt{3}=2.2\sqrt{3}\), tách \(7=4+3\)
c ) \(24\sqrt{5}=2.\sqrt{5}.12=2.\sqrt{5}.2.6=2.\sqrt{20}.6\) , tách : \(56=20+36\)
d ) \(2\sqrt{11}=2.11.1\) , tách : \(12=11+1\)
e ) \(4\sqrt{2}=2.\sqrt{2}.2.1=2.\sqrt{8}.1\) , tách : \(9=8+1\)
Y
1 tháng 5 2019
a) \(\sqrt{\sqrt{28+16\sqrt{3}}}\)
\(=\sqrt{\sqrt{\left(2\sqrt{3}\right)^2+2\cdot2\sqrt{3}\cdot4+16}}\)
\(=\sqrt{\sqrt{\left(2\sqrt{3}+4\right)^2}}\) \(=\sqrt{2\sqrt{3}+4}\)
\(=\sqrt{3+2\sqrt{3}+1}=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)
b)\(\sqrt{7+4\sqrt{3}}=\sqrt{4+4\sqrt{3}+3}\)
\(=\sqrt{\left(2+\sqrt{3}\right)^2}=2+\sqrt{3}\)
c) \(\sqrt{\sqrt{56-24\sqrt{5}}}=\sqrt{\sqrt{\left(2\sqrt{5}\right)^2-2\cdot2\sqrt{5}\cdot6}+36}\)
\(=\sqrt{\sqrt{\left(2\sqrt{5}-6\right)^2}}=\sqrt{6-2\sqrt{5}}\)
\(=\sqrt{5-2\sqrt{5}+1}=\sqrt{\left(\sqrt{5}-1\right)^2}=\sqrt{5}-1\)
d) \(\sqrt{12-2\sqrt{11}}=\sqrt{11-2\sqrt{11}+1}\)
\(=\sqrt{\left(\sqrt{11}-1\right)^2}=\sqrt{11}-1\)
e) \(\sqrt{9+4\sqrt{2}}=\sqrt{\left(2\sqrt{2}\right)^2+2\cdot2\sqrt{2}+1}\)
\(=\sqrt{\left(2\sqrt{2}+1\right)^2}=2\sqrt{2}+1\)
19 tháng 9 2021
a) \(1=\sqrt{1}< \sqrt{2}\)
b) \(2=\sqrt{4}>\sqrt{3}\)
c) \(6=\sqrt{36}< \sqrt{41}\)
d) \(7=\sqrt{49}>\sqrt{47}\)
e) \(2=1+1=\sqrt{1}+1< \sqrt{2}+1\)
f) \(1=2-1=\sqrt{4}-1>\sqrt{3}-1\)
g) \(2\sqrt{31}=\sqrt{4.31}=\sqrt{124}>\sqrt{100}=10\)
h) \(\sqrt{3}>0>-\sqrt{12}\)
i) \(5=\sqrt{25}< \sqrt{29}\)
\(\Rightarrow-5>-\sqrt{29}\)
30 tháng 4 2019
\(\sqrt{\sqrt{\left(3\right)^8}}\)=\(\sqrt{\sqrt{6561}}=\sqrt{81}=9\)
\(\sqrt[2]{\left(-5^4\right)}=\sqrt[2]{625}=25\)
NT
19 tháng 6 2019
1.
\(\sqrt{\frac{2+\sqrt{3}}{2}}\\ =\frac{\sqrt{2+\sqrt{3}}}{\sqrt{2}}\\ =\frac{\sqrt{4+2\sqrt{3}}}{2}\\ =\frac{\sqrt{\left(1+\sqrt{3}\right)^2}}{2}\\ =\frac{1+\sqrt{3}}{2}\)
2.
\(\sqrt{\frac{14+5\sqrt{3}}{2}}\\ =\frac{\sqrt{14+5\sqrt{3}}}{\sqrt{2}}\\ =\frac{\sqrt{28+10\sqrt{3}}}{2}\\ =\frac{\sqrt{\left(5+\sqrt{3}\right)^2}}{2}\\ =\frac{5+\sqrt{3}}{2}\)
11 tháng 8 2021
a: \(4\sqrt{7}=\sqrt{4^2\cdot7}=\sqrt{112}\)
\(3\sqrt{13}=\sqrt{3^2\cdot13}=\sqrt{117}\)
mà 112<117
nên \(4\sqrt{7}< 3\sqrt{13}\)
b: \(3\sqrt{12}=\sqrt{3^2\cdot12}=\sqrt{108}\)
\(2\sqrt{16}=\sqrt{16\cdot2^2}=\sqrt{64}\)
mà 108>64
nên \(3\sqrt{12}>2\sqrt{16}\)
c: \(\dfrac{1}{4}\sqrt{84}=\sqrt{\dfrac{1}{16}\cdot84}=\sqrt{\dfrac{21}{4}}\)
\(6\sqrt{\dfrac{1}{7}}=\sqrt{36\cdot\dfrac{1}{7}}=\sqrt{\dfrac{36}{7}}\)
mà \(\dfrac{21}{4}>\dfrac{36}{7}\)
nên \(\dfrac{1}{4}\sqrt{84}>6\sqrt{\dfrac{1}{7}}\)
d: \(3\sqrt{12}=\sqrt{3^2\cdot12}=\sqrt{108}\)
\(2\sqrt{16}=\sqrt{16\cdot2^2}=\sqrt{64}\)
mà 108>64
nên \(3\sqrt{12}>2\sqrt{16}\)
HP
25 tháng 8 2021
\(\left|\vec{AD}+\vec{AB}\right|=\left|\vec{AC}\right|=AC=a\sqrt{2}\)
19 tháng 6 2019
1/\(\sqrt{\frac{4}{5}}\)+\(\sqrt{\frac{1}{2}}\)
=\(\sqrt{\frac{4.5}{5.5}}\)+\(\sqrt{\frac{1.2}{2.2}}\)
= \(5.2.\sqrt{5}\)+\(2\sqrt{2}\)
=\(10\sqrt{5}+2\sqrt{2}\)
19 tháng 6 2019
2.
\(\sqrt{\frac{1}{12}}\)+\(\sqrt{\frac{1}{3}}\)
=\(\sqrt{\frac{1.12}{12.12}}\)+\(\sqrt{\frac{1.3}{3.3}}\)
=\(12.2\sqrt{3}\)+\(3\sqrt{3}\)
=\(\sqrt{3}\left(24+3\right)\)
=\(27\sqrt{3}\)
6 tháng 7 2018
\(Q=\dfrac{x-3\sqrt{x}-4}{x-\sqrt{x}-12}\left(ĐK:x\ge0;x\ne16\right)\\ =\dfrac{x-4\sqrt{x}+\sqrt{x}-4}{x-4\sqrt{x}+3\sqrt{x}-12}\\ =\dfrac{\sqrt{x}\left(\sqrt{x}-4\right)+\left(\sqrt{x}-4\right)}{\sqrt{x}\left(\sqrt{x}-4\right)+3\left(\sqrt{x}-4\right)}\\ =\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-4\right)}\\ =\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\)
\(P=\dfrac{x\sqrt{y}-y\sqrt{x}}{\sqrt{x}-\sqrt{y}}=\dfrac{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}=\sqrt{xy}\)