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6 tháng 4 2017
\(2\sqrt{3}\). (\(3\sqrt{3}+8\sqrt{3}-5\))+ \(10\sqrt{3}\)
= \(2\sqrt{3}.3\sqrt{3}+8\sqrt{3}+10\sqrt{3}-5\)
=\(18+2\sqrt{3}-5\)
=\(13+21\sqrt{3}\)
8 tháng 10 2018
Ta có công thức tổng quát là \(\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\dfrac{1}{\sqrt{n}.\sqrt{n+1}\left(\sqrt{n+1}+\sqrt{n}\right)}=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}.\sqrt{n+1}\left(n+1-n\right)}=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}.\sqrt{n+1}}=\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)
Vậy \(\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+...+\dfrac{1}{400\sqrt{399}+399\sqrt{400}}=\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{399}}-\dfrac{1}{\sqrt{400}}=\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{400}}=1-\dfrac{1}{20}=\dfrac{19}{20}\)
MV
13 tháng 7 2017
a,
\(\sqrt{4-2\sqrt{3}}-\sqrt{3}\\ =\sqrt{3-2\cdot1\cdot\sqrt{3}+1}-\sqrt{3}\\ =\sqrt{\left(\sqrt{3}\right)^2-2\cdot1\cdot\sqrt{3}+1^2}-\sqrt{3}\\ =\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{3}\\ =\sqrt{3}-1-\sqrt{3}\\ =-1\)
b,
\(\sqrt{11+6\sqrt{2}}-3+\sqrt{2}\\ =\sqrt{9+2\cdot3\cdot\sqrt{2}+2}-3+\sqrt{2}\\ =\sqrt{3^2+2\cdot3\cdot\sqrt{2}+\left(\sqrt{2}\right)^2}-3+\sqrt{2}\\ =\sqrt{\left(3+\sqrt{2}\right)^2}-3+\sqrt{2}\\ =3+\sqrt{2}-3+\sqrt{2}\\ =2\sqrt{2}\)
c,
\(\sqrt{7+2\sqrt{10}}-\sqrt{7-2\sqrt{10}}\\ =\sqrt{5+2\cdot\sqrt{2\cdot5}+2}-\sqrt{5-2\cdot\sqrt{2\cdot5}+2}\\ =\sqrt{\left(\sqrt{5}\right)^2+2\cdot\sqrt{2}\cdot\sqrt{5}+\left(\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{5}\right)^2-2\cdot\sqrt{2}\cdot\sqrt{5}+\left(\sqrt{2}\right)^2}\\ =\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}\\ =\sqrt{5}+\sqrt{2}-\sqrt{5}+\sqrt{2}\\ =2\sqrt{2}\)
d,
\(\left(20\sqrt{300}-15\sqrt{675}+5\sqrt{75}\right):\sqrt{15}\\ =\left(20\cdot\sqrt{20}\cdot\sqrt{15}-15\cdot\sqrt{45}\cdot\sqrt{15}+5\cdot\sqrt{5}\cdot\sqrt{15}\right):\sqrt{15}\\ =\left(20\cdot2\cdot\sqrt{5}\cdot\sqrt{15}-15\cdot3\cdot\sqrt{5}\cdot\sqrt{15}+5\cdot\sqrt{5}\cdot\sqrt{15}\right):\sqrt{15}\\ =\sqrt{15}\cdot\left(20\cdot2\cdot\sqrt{5}-15\cdot3\cdot\sqrt{5}+5\cdot\sqrt{5}\right):\sqrt{15}\\ =20\cdot2\cdot\sqrt{5}-15\cdot3\cdot\sqrt{5}+5\cdot\sqrt{5}\\ =40\sqrt{5}-45\sqrt{5}+5\sqrt{5}\\ =0\)
25 tháng 7 2019
\(\sqrt{300}-\sqrt{27}+4\sqrt{3}\)
=\(10\sqrt{3}-3\sqrt{3}+4\sqrt{3}\)
=\(11\sqrt{3}\)
B
25 tháng 7 2019
\(\sqrt{300}-\sqrt{27}+4\sqrt{3}\)
\(=\sqrt{10^2.3}-\sqrt{3^2.3}+4\sqrt{3}\)
\(=10\sqrt{3}-3\sqrt{3}+4\sqrt{3}\)
\(=11\sqrt{3}\)
\(\frac{2+2\sqrt{5}}{3-\sqrt{5}}=\frac{\left(2+2\sqrt{5}\right)\left(3+\sqrt{5}\right)}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}=\frac{6+2\sqrt{5}+6\sqrt{5}+10}{3^2-\sqrt{5}^2}=\frac{16+8\sqrt{5}}{4}=\frac{4\left(4+2\sqrt{5}\right)}{4}=4+2\sqrt{5}\)
30 tháng 8 2022
\(=2\sqrt{4\sqrt{3}}-5\sqrt{20\sqrt{3}}-3\sqrt{500\sqrt{3}}\)
\(=\sqrt{\sqrt{3}}\left(2\sqrt{4}-5\sqrt{20}-3\sqrt{500}\right)\)
\(=\sqrt{\sqrt{3}}\cdot\left(4-40\sqrt{5}\right)\)
VN
14 tháng 12 2016
a) = \(\frac{\sqrt{2}.\sqrt{4+\sqrt{15}}}{\sqrt{2}}+\frac{\sqrt{2}.\sqrt{4-\sqrt{15}}}{\sqrt{2}}\)
= \(\frac{\sqrt{8+2\sqrt{15}}}{\sqrt{2}}+\frac{\sqrt{8-2\sqrt{15}}}{\sqrt{2}}\)
= \(\frac{\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}}{\sqrt{2}}+\frac{\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}}{\sqrt{2}}\)
= \(\frac{\sqrt{5}+\sqrt{3}}{\sqrt{2}}+\frac{\sqrt{5}-\sqrt{3}}{\sqrt{2}}\)
= \(\frac{2\sqrt{5}}{\sqrt{2}}\)
= \(\sqrt{10}\)
b) = \(5\sqrt{3}+4\sqrt{3}-10\sqrt{3}\)
= \(-\sqrt{3}\)
c) = \(\sqrt{\left(\sqrt{2}-\sqrt{1}\right)^2}+\frac{1\left(\sqrt{2}+1\right)}{2-1}\)
= \(\sqrt{2}-\sqrt{1}+\frac{1\left(\sqrt{2}+1\right)}{1}\)
= \(\sqrt{2}-\sqrt{1}+\sqrt{2}+\sqrt{1}\)
= \(2\sqrt{2}\)
Chúc bạn học tốt ^^
TT
1
23 tháng 12 2016
\(\sqrt{75}+\sqrt{48}+\frac{1}{2}.\sqrt{300}\)
\(=5\sqrt{3}+4\sqrt{3}+5\sqrt{3}\)
\(=\left(5+4+5\right)\sqrt{3}\)
\(=14\sqrt{3}\)
S
12 tháng 12 2021
Ta có:
\(\frac{1}{n\sqrt{n+1}+\left(n+1\right)\sqrt{n}}=\frac{1}{\sqrt{n\left(n+1\right)}\left(\sqrt{n+1}+\sqrt{n}\right)}=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left(n+1\right)}}\)
\(=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}.\text{ Vì thế, }A=1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-...-\frac{1}{\sqrt{401}}< 1.\)
\(\sqrt{1600^2-400^2}=1549,193338\)
\(\sqrt{1600^2-400^2}=400\sqrt{15}\)
\(\sqrt{1200^2-300^2}=2^2\cdot3\cdot5^2\sqrt{15}\)