VV
0
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
2 tháng 11 2017
\(\frac{1}{2}\sqrt{48}-2\sqrt{75}-\frac{\sqrt{33}}{\sqrt{11}}+5\sqrt{1\frac{1}{3}}\)
=\(\frac{1}{2}\sqrt{3.4^2}-2\sqrt{3.5^2}-\sqrt{\frac{33}{11}}+5\sqrt{\frac{4}{3}}\)
\(=2\sqrt{3}-10\sqrt{3}-\sqrt{3}+10\sqrt{\frac{1}{3}}\)
\(=2\sqrt{3}-10\sqrt{3}-\sqrt{3}+\frac{10}{3}\sqrt{3}\)
\(=\left(2-10-1+\frac{10}{3}\right)\sqrt{3}\)
\(=\frac{-17}{3}\sqrt{3}\)
\(\sqrt{150}+\sqrt{1,6}.\sqrt{60}+4,5\sqrt{2\frac{2}{3}}-\sqrt{6}\)
\(=\sqrt{6.5^2}+\sqrt{96}+4,5\sqrt{\frac{8}{3}}-\sqrt{6}\)
\(=5\sqrt{6}+\sqrt{6.4^2}+4,5\frac{\sqrt{24}}{3}-\sqrt{6}\)
\(=5\sqrt{6}+4\sqrt{6}+\frac{4,5.2\sqrt{6}}{3}-\sqrt{6}\)
\(=8\sqrt{6}+3\sqrt{6}\)
\(=11\sqrt{6}\)
L
25 tháng 11 2020
Tự hòi tự trl :D ?
\(\frac{1}{2}\sqrt{48}-2\sqrt{75}-\frac{\sqrt{33}}{\sqrt{11}}+5\sqrt{1\frac{1}{3}}\)
\(=\frac{1}{2}\sqrt{16.3}-2.5\sqrt{3}-\sqrt{3}-\frac{10}{3}\sqrt{3}\)
\(=2\sqrt{3}-10\sqrt{3}-\sqrt{3}-\frac{10}{3}\sqrt{3}\)
\(=-9\sqrt{3}+\frac{10}{3}\sqrt{3}=\left(-9+\frac{10}{3}\right)\sqrt{3}\)
\(=-\frac{17}{3}\sqrt{3}\)
\(\sqrt{150}+\sqrt{1,6}.\sqrt{60}+4,5.\sqrt{2\frac{2}{3}}-\sqrt{6}\)
\(=\sqrt{25.6}+\sqrt{1,6.60}+4,8\sqrt{\frac{8}{3}}-\sqrt{6}\)
\(=5\sqrt{6}+\sqrt{16.6}+4,5.\frac{1}{3}\sqrt{3^2.\frac{4.2}{3}}-\sqrt{6}\)
\(=9\sqrt{6}+3\sqrt{6}-\sqrt{6}=11\sqrt{6}\)
14 tháng 5 2018
1/ \(7-2\sqrt{6}=\left(\sqrt{6}\right)^2-2\sqrt{6}+1\)
\(=\left(\sqrt{6}-1\right)^2\)
2/ \(10+2\sqrt{21}=\left(\sqrt{7}\right)^2+2.\sqrt{7}.\sqrt{3}+\left(\sqrt{3}\right)^2\)
\(=\left(\sqrt{7}+\sqrt{3}\right)^2\)
4/ \(10+4\sqrt{6}=2^2+2.2.\sqrt{6}+\left(\sqrt{6}\right)^2\)
\(=\left(2+\sqrt{6}\right)^2\)
5/ \(11-2\sqrt{30}=\left(\sqrt{6}\right)^2-2.\sqrt{6}.\sqrt{5}+\left(\sqrt{5}\right)^2\)
= \(\left(\sqrt{6}-\sqrt{5}\right)^2\)
8/ \(11+4\sqrt{7}=2^2+2.2.\sqrt{7}+\left(\sqrt{7}\right)^2\)
= \(\left(2+\sqrt{7}\right)^2\)
10/ \(12+6\sqrt{3}=3^2+2.3.\sqrt{3}+\left(\sqrt{3}\right)^2\)
= \(\left(3+\sqrt{3}\right)^2\)
11 tháng 7 2017
\(=\frac{1}{2}\sqrt{48}-2\sqrt{75}-\frac{\sqrt{33}}{\sqrt{11}}+5\sqrt{1\frac{1}{3}}\)
\(=\frac{1}{2}4\sqrt{3}-2.5\sqrt{3}-\sqrt{3}+\frac{10}{\sqrt{3}}=-9\sqrt{3}+\frac{10}{\sqrt{3}}=\frac{-17\sqrt{3}}{3}\)
13 tháng 9 2018
\(=\frac{1}{2}\sqrt{16.3}-2\sqrt{25.3}-\frac{\sqrt{3.11}}{\sqrt{11}}+5\sqrt{\frac{1.3+1}{3}}\)
\(=\frac{1}{2}\sqrt{4^2.3}-2\sqrt{5^2.3}-\frac{\sqrt{3}.\sqrt{11}}{\sqrt{11}}+5\sqrt{\frac{4}{3}}\)
\(=\frac{1}{2}.4\sqrt{3}-2.5\sqrt{3}-\sqrt{3}+5\frac{\sqrt{4}}{\sqrt{3}}\)
\(=\frac{4}{2}\sqrt{3}-10\sqrt{3}-\sqrt{3}+5\frac{\sqrt{4}.\sqrt{4}}{\sqrt{3.}\sqrt{3}}\)
\(=2\sqrt{3}-10\sqrt{3}-\sqrt{3}+5\frac{2\sqrt{3}}{3}\)
\(=2\sqrt{3}-10\sqrt{3}-\sqrt{3}+10\frac{\sqrt{3}}{3}\)
\(=\left(2-10-1+\frac{10}{3}\right)\sqrt{3}\)
\(=-\frac{17}{3}\)
KT
1
HS
11 tháng 9 2018
\(\dfrac{1}{2}\sqrt{48}-2\sqrt{75}-\dfrac{\sqrt{33}}{\sqrt{11}}+5\sqrt{1\dfrac{1}{3}}\\ =\dfrac{1}{2}\sqrt{16.3}-2\sqrt{25.3}-\sqrt{3}+5\sqrt{\dfrac{4}{3}}\\ =4.\dfrac{1}{2}-5.2\sqrt{3}-\sqrt{3}+5\sqrt{\dfrac{4}{9}}.\sqrt{3}\\ =2\sqrt{3}-10\sqrt{3}-\sqrt{3}+\dfrac{10}{3}.\sqrt{3}=-\dfrac{17}{3}\sqrt{3}\)
23 tháng 8 2020
1) Ta có: \(\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\sqrt{2}+\sqrt{3}+2+2+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+2}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}+2\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+2\right)}{\left(\sqrt{2}+\sqrt{3}+2\right)}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}+2\right)\left(1+\sqrt{2}\right)}{\left(\sqrt{2}+\sqrt{3}+2\right)}\)
\(=1+\sqrt{2}\)
2) Ta có: \(2\sqrt{27}-6\sqrt{\frac{4}{3}}+\frac{3}{5}\sqrt{75}\)
\(=\sqrt{108}-\sqrt{36\cdot\frac{4}{3}}+\sqrt{75\cdot\frac{9}{25}}\)
\(=\sqrt{108}-\sqrt{48}+\sqrt{27}\)
\(=\sqrt{3}\left(6-4+3\right)\)
\(=5\sqrt{3}\)
3) Sửa đề: \(\sqrt{8\sqrt{3}}-2\sqrt{25\sqrt{12}}+4\sqrt{192}\)
Ta có: \(\sqrt{8\sqrt{3}}-2\sqrt{25\sqrt{12}}+4\sqrt{192}\)
\(=\sqrt{2}\cdot\sqrt{4}\cdot\sqrt{3}-10\sqrt{4}\cdot\sqrt{3}+16\cdot\sqrt{4}\cdot\sqrt{3}\)
\(=\sqrt{2}\cdot\sqrt{12}-10\sqrt{12}+16\sqrt{12}\)
\(=\sqrt{12}\left(\sqrt{2}-10+16\right)\)
\(=2\sqrt{3}\left(\sqrt{2}-6\right)\)
\(=2\sqrt{6}-12\sqrt{3}\)
4) Ta có: \(\frac{1}{2+\sqrt{3}}+\frac{\sqrt{2}}{\sqrt{6}}-\frac{2}{3+\sqrt{3}}\)
\(=\frac{2-\sqrt{3}}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}+\frac{\sqrt{12}}{6}-\frac{2\left(3-\sqrt{3}\right)}{\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)}\)
\(=\frac{6\left(2-\sqrt{3}\right)+2\sqrt{3}-6+2\sqrt{3}}{6}\)
\(=\frac{12-6\sqrt{3}+2\sqrt{3}-6+2\sqrt{3}}{6}\)
\(=\frac{6-2\sqrt{3}}{6}\)
\(=\frac{2\sqrt{3}\left(\sqrt{3}-1\right)}{2\sqrt{3}\cdot\sqrt{3}}\)
\(=\frac{\sqrt{3}-1}{\sqrt{3}}\)
5) Ta có: \(\left(\sqrt{12}+\sqrt{75}+\sqrt{27}\right):\sqrt{15}\)
\(=\frac{\sqrt{3}\left(2+5+3\right)}{\sqrt{15}}=\frac{10}{\sqrt{5}}=2\sqrt{5}\)
6) Ta có: \(\frac{1}{2}\sqrt{48}-2\sqrt{75}-\frac{\sqrt{33}}{\sqrt{11}}+5\sqrt{1\frac{1}{3}}\)
\(=\sqrt{48\cdot\frac{1}{4}}-\sqrt{75\cdot4}-\sqrt{3}+5\sqrt{\frac{4}{3}}\)
\(=\sqrt{12}-\sqrt{300}-\sqrt{3}+\sqrt{25\cdot\frac{4}{3}}\)
\(=\sqrt{12}-\sqrt{300}-\sqrt{3}+\sqrt{\frac{100}{3}}\)
\(=\sqrt{3}\left(2-10-1+\frac{10}{3}\right)\)
\(=-\frac{17\sqrt{3}}{3}=-\frac{17}{\sqrt{3}}\)
4 tháng 7 2018
\(1.A=\dfrac{1}{2}\sqrt{48}-2\sqrt{75}-\dfrac{\sqrt{33}}{\sqrt{11}}+5\sqrt{1\dfrac{1}{3}}=\dfrac{1}{2}\sqrt{\dfrac{1}{3}.144}-2\sqrt{\dfrac{1}{3}.225}-\sqrt{\dfrac{1}{3}.9}+5\sqrt{\dfrac{4}{3}}=6\sqrt{\dfrac{1}{3}}-30\sqrt{\dfrac{1}{3}}-3\sqrt{\dfrac{1}{3}}+10\sqrt{\dfrac{1}{3}}=-17\sqrt{\dfrac{1}{3}}\) \(2.B=\left(2\sqrt{27}-3\sqrt{48}+3\sqrt{75}-\sqrt{192}\right)\left(1-\sqrt{3}\right)=\left(6\sqrt{3}-12\sqrt{3}+15\sqrt{3}-8\sqrt{3}\right)\left(1-\sqrt{3}\right)=\sqrt{3}\left(1-\sqrt{3}\right)=\sqrt{3}-3\) \(3.C=\left(2\sqrt{7}-2\sqrt{6}\right).\sqrt{6}-\sqrt{168}=2\sqrt{42}-12-2\sqrt{42}=-12\) \(4.D=\left(\sqrt{28}-2\sqrt{8}+\sqrt{7}\right).\sqrt{7}+4\sqrt{14}=\left(3\sqrt{7}-4\sqrt{2}\right)\sqrt{7}=21-4\sqrt{14}+4\sqrt{14}=21\)
Q
24 tháng 6 2017
\(\dfrac{1}{2}\sqrt{48}-2\sqrt{75}-\dfrac{\sqrt{33}}{\sqrt{11}}+5\sqrt{1\dfrac{1}{3}}\)
\(=2\sqrt{3}-10\sqrt{3}-\sqrt{3}+5\sqrt{\dfrac{4}{3}}\)
\(=2\sqrt{3}-10\sqrt{3}-\sqrt{3}+5\cdot\dfrac{2}{\sqrt{3}}\)
\(=2\sqrt{3}-10\sqrt{3}-\sqrt{3}+\dfrac{10}{\sqrt{3}}\)
\(=2\sqrt{3}-10\sqrt{3}-\sqrt{3}+\dfrac{10\sqrt{3}}{3}\)
\(=-\dfrac{17\sqrt{3}}{3}\)
E
3 tháng 7 2017
Ta có: \(B-\dfrac{1}{3}=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}-\dfrac{1}{3}=\dfrac{3\sqrt{x}+3-\sqrt{x}-3}{3\left(\sqrt{x}+3\right)}=\dfrac{2\sqrt{x}}{3\left(\sqrt{x}+3\right)}\)Vì x > 0 \(\Rightarrow2\sqrt{x}>0;3\left(\sqrt{x}+3\right)>0\Rightarrow\dfrac{2\sqrt{x}}{3\left(\sqrt{x}+3\right)}>0\)
\(\Rightarrow B-\dfrac{1}{3}>0\Rightarrow B>\dfrac{1}{3}\)