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1 tháng 2 2022
a: \(=2\cdot3+\sqrt{15}-2\sqrt{15}=6-\sqrt{15}\)
b: \(=5\sqrt{10}+2\cdot5-5\sqrt{10}=10\)
c: \(=2\sqrt{7}\cdot\sqrt{7}-\sqrt{12}\cdot\sqrt{7}-\sqrt{7}\cdot\sqrt{7}+2\sqrt{21}=2\cdot7-7=7\)
d: \(=\left(2\sqrt{11}-3\sqrt{2}\right)\cdot\sqrt{11}+3\sqrt{22}=2\cdot11=22\)
21 tháng 8 2016
\(2.3+\sqrt{15}-2\sqrt{15}=6-\sqrt{15}\)
\(5\sqrt{10}+2.5-5\sqrt{10}=10\)
\(14-2\sqrt{21}-7+2\sqrt{21}=7\)
\(33-3\sqrt{22}-11+3\sqrt{22}=22\)
NC
25 tháng 8 2020
1) \(\left(5\sqrt{2}+2\sqrt{5}\right)\sqrt{5}-\sqrt{250}\)
\(=5\sqrt{10}-10-5\sqrt{10}\)
\(=-10\)
2) \(\left(\sqrt{28}-\sqrt{12}-\sqrt{7}\right)\sqrt{7}+2\sqrt{21}\)
\(=14-2\sqrt{21}-7+2\sqrt{21}\)
\(=7\)
3) \(\left(\sqrt{99}-\sqrt{18}-\sqrt{11}\right)\sqrt{11}+3\sqrt{22}\) (hẳn đề là như thế này)
\(=33-3\sqrt{22}-11+3\sqrt{22}\)
\(=22\)
31 tháng 8 2022
a: Sửa đề: \(5\dfrac{1}{5}-\dfrac{1}{2}\sqrt{20}+\sqrt{5}\)
\(=5.2-\dfrac{1}{2}\cdot2\sqrt{5}+\sqrt{5}=5.2\)
b: \(=\dfrac{1}{2}\sqrt{2}+\dfrac{3}{2}\sqrt{2}+\dfrac{5}{2}\sqrt{2}=\dfrac{9}{2}\sqrt{2}\)
c: \(=2\sqrt{5}-3\sqrt{5}+9\sqrt{2}+\sqrt{77}=-\sqrt{5}+9\sqrt{2}+\sqrt{77}\)
d: \(=\dfrac{1}{10}\cdot10\sqrt{2}+\dfrac{2}{5}\sqrt{2}+0.4\cdot5\sqrt{2}\)
\(=\dfrac{17}{5}\sqrt{2}\)
S
5 tháng 7 2019
so sánh: \(4-\sqrt{2}\) và \(\sqrt{5}\)
\(\left(4-\sqrt{2}\right)^2=18-8\sqrt{2}>18-8\sqrt{2,25}=18-8.1,5=18-12=6>5=\sqrt{5}^2\Rightarrow4-\sqrt{2}>\sqrt{5}\left(vì:\left\{{}\begin{matrix}4-\sqrt{2}>0\\\sqrt{5}>0\end{matrix}\right.\right)\Rightarrow7+4-\sqrt{2}>7+\sqrt{5}\Rightarrow11-\sqrt{2}>7+\sqrt{5}\)
\(b,2006^2-2005.2007=2006^2-\left(2006-1\right)\left(2006+1\right)=2006^2-2006^2+1=1\Rightarrow2006^2>2005.2007\left(1\right)\)
\(\left(2\sqrt{2006}\right)^2=4.2006=8024;\left(\sqrt{2005}+\sqrt{2007}\right)^2=2005+2007+2\sqrt{2005.2007}=4012+2\sqrt{2005.2007}=4012+2\sqrt{2006.2006}\left(vì\left(1\right)\right)=8024=\left(2\sqrt{2006}\right)^2\)
\(\Rightarrow\sqrt{2005}+\sqrt{2007}< 2\sqrt{2006}\left(vì:\left\{{}\begin{matrix}\sqrt{2005}+\sqrt{2007}>0\\2\sqrt{2006}>0\end{matrix}\right.\right)\)
\(c,\left(\sqrt{10}+\sqrt{13}\right)^2=23+2\sqrt{130}>23+2\sqrt{121}\left(130>121\right)=23+2.11=45>4.11=\left(2\sqrt{11}\right)^2\Rightarrow\sqrt{10}+\sqrt{13}>2\sqrt{11}\left(vì\left\{{}\begin{matrix}\sqrt{10}+\sqrt{13}>0\\2\sqrt{11}>0\end{matrix}\right.\right)\)
\(d,\left(\sqrt{5}+\sqrt{7}\right)^2=12+2\sqrt{35}< 12+2\sqrt{36}=12+12=24< 15+6\sqrt{6}=\left(3+\sqrt{6}\right)^2\Rightarrow\sqrt{5}+\sqrt{7}< 3+\sqrt{6}\left(vì:\left\{{}\begin{matrix}\sqrt{5}+\sqrt{7}>0\\3+\sqrt{6}>0\end{matrix}\right.\right)\)
KT
4 tháng 7 2018
1) \(2\sqrt{2}=\sqrt{8}< \sqrt{9}=3\)
\(\Rightarrow\)\(6+2\sqrt{2}< 6+3=9\)
2) \(4\sqrt{5}=\sqrt{80}>\sqrt{49}=7\)
\(\Rightarrow\)\(9+4\sqrt{5}>9+7=16\)
3) \(2=\sqrt{4}>\sqrt{3}\)
\(\Rightarrow\)\(2-1>\sqrt{3}-1\)
hay \(1>\sqrt{3}-1\)
4) \(9-4\sqrt{5}< 16\)
5) \(\sqrt{2}>\sqrt{1}=1\)
\(\Rightarrow\)\(\sqrt{2}+1>2\)
21 tháng 6 2018
\(1)\) Ta có :
\(\left(\sqrt{3\sqrt{2}}\right)^4=\left[\left(\sqrt{3\sqrt{2}}\right)^2\right]^2=\left(3\sqrt{2}\right)^2=9.2=18\)
\(\left(\sqrt{2\sqrt{3}}\right)^4=\left[\left(\sqrt{2\sqrt{3}}\right)^2\right]^2=\left(2\sqrt{3}\right)^2=4.3=12\)
Vì \(18>12\) nên \(\left(\sqrt{3\sqrt{2}}\right)^4>\left(\sqrt{2\sqrt{3}}\right)^4\)
\(\Rightarrow\)\(\sqrt{3\sqrt{2}}>\sqrt{2\sqrt{3}}\)
Vậy \(\sqrt{3\sqrt{2}}>\sqrt{2\sqrt{3}}\)
Chúc bạn học tốt ~
AM
7 tháng 6 2019
\(\sqrt{11+6\sqrt{2}}-\sqrt{2}\text{=}\sqrt{9+6\sqrt{2}+4}-\sqrt{2}\)
=\(\sqrt{\left(3+\sqrt{2}\right)^2}-\sqrt{2}\text{=}3+\sqrt{2}-\sqrt{2}=3\)
HT
7 tháng 6 2019
a/ \(\sqrt{11+6\sqrt{2}}-\sqrt{2}=\sqrt{9+6\sqrt{2}+2}-\sqrt{2}=\sqrt{\left(3+\sqrt{2}\right)^2}-\sqrt{2}=3+\sqrt{2}-\sqrt{2}=3\)
b/ \(\sqrt{28-10\sqrt{3}}+5=\sqrt{25-10\sqrt{3}+3}+5\)
\(=\sqrt{\left(5-\sqrt{3}\right)^2}+5=5-\sqrt{3}+5=25-\sqrt{3}\)
c/ \(\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}=\sqrt{3+2\sqrt{3}+1}-\sqrt{3-2\sqrt{3}+1}\)
\(=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(=\sqrt{3}+1-\sqrt{3}+1=2\)
d/ \(3\sqrt{5}-\sqrt{6-2\sqrt{5}}=3\sqrt{5}-\sqrt{5-2\sqrt{5}+1}\)\
\(=3\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}=3\sqrt{5}-\sqrt{5}+1=2\sqrt{5}+1\)
\(\sqrt{28.9}\cdot\sqrt{40}\cdot\sqrt{11^2-5^2}\)
\(=\sqrt{289}\cdot\sqrt{4}\cdot4\sqrt{6}\)
\(=17\cdot2\cdot4\sqrt{6}\)
\(=136\sqrt{6}\)