\(\frac{1}{2}\sqrt{48}-2\sqrt{75}-\sqrt{54}+5\sqrt{1\frac{1}{3}}\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
27 tháng 9 2020
Bài 1: Thực hiện phép tính
a) Ta có: \(\frac{3+\sqrt{7}}{3-\sqrt{7}}-\frac{3-\sqrt{7}}{3+\sqrt{7}}\)
\(=\frac{\left(3+\sqrt{7}\right)^2}{\left(3-\sqrt{7}\right)\left(3+\sqrt{7}\right)}-\frac{\left(3-\sqrt{7}\right)^2}{\left(3+\sqrt{7}\right)\left(3-\sqrt{7}\right)}\)
\(=\frac{9+6\sqrt{7}+7-\left(9-6\sqrt{7}+7\right)}{9-7}\)
\(=\frac{16+6\sqrt{7}-16+6\sqrt{7}}{2}\)
\(=\frac{12\sqrt{7}}{2}=6\sqrt{7}\)
b)Sửa đề: \(\left(\frac{\sqrt{2}+5}{\sqrt{2}-5}-\frac{\sqrt{2}-5}{\sqrt{2}+5}\right):\frac{\sqrt{2}}{23}\)
Ta có: \(\left(\frac{\sqrt{2}+5}{\sqrt{2}-5}-\frac{\sqrt{2}-5}{\sqrt{2}+5}\right):\frac{\sqrt{2}}{23}\)
\(=\left(\frac{\left(\sqrt{2}+5\right)^2}{\left(\sqrt{2}-5\right)\left(\sqrt{2}+5\right)}-\frac{\left(\sqrt{2}-5\right)^2}{\left(\sqrt{2}+5\right)\left(\sqrt{2}-5\right)}\right)\cdot\frac{23}{\sqrt{2}}\)
\(=\left(\frac{27+10\sqrt{2}-\left(27-10\sqrt{2}\right)}{2-25}\right)\cdot\frac{23}{\sqrt{2}}\)
\(=\frac{27+10\sqrt{2}-27+10\sqrt{2}}{-23}\cdot\frac{23}{\sqrt{2}}\)
\(=\frac{20\sqrt{2}}{-\sqrt{2}}=-20\)
c) Ta có: \(5\sqrt{\frac{1}{5}}+\frac{1}{2}\sqrt{20}+\sqrt{5}\)
\(=\sqrt{25\cdot\frac{1}{5}}+\frac{1}{2}\cdot2\sqrt{5}+\sqrt{5}\)
\(=\sqrt{5}+\sqrt{5}+\sqrt{5}\)
\(=3\sqrt{5}\)
d) Ta có: \(\sqrt{\frac{1}{2}}+\sqrt{4.5}+12.5\)
\(=\frac{1}{\sqrt{2}}+\frac{3}{\sqrt{2}}+12.5\)
\(=2\sqrt{2}+12.5\)
e) Ta có: \(\frac{1}{2}\sqrt{48}-2\sqrt{75}-\sqrt{54}+5\sqrt{1\frac{1}{3}}\)
\(=\frac{1}{2}\cdot4\sqrt{3}-2\cdot5\sqrt{3}-3\sqrt{6}+5\cdot\sqrt{\frac{4}{3}}\)
\(=2\sqrt{3}-10\sqrt{3}-3\sqrt{6}+\frac{10}{\sqrt{3}}\)
\(=-8\sqrt{3}+\frac{10}{\sqrt{3}}-3\sqrt{6}\)
\(=\frac{-24+10}{\sqrt{3}}-\frac{9\sqrt{2}}{\sqrt{3}}\)
\(=\frac{-14-9\sqrt{2}}{\sqrt{3}}\)
QN
1
11 tháng 7 2017
\(=\frac{1}{2}\sqrt{48}-2\sqrt{75}-\frac{\sqrt{33}}{\sqrt{11}}+5\sqrt{1\frac{1}{3}}\)
\(=\frac{1}{2}4\sqrt{3}-2.5\sqrt{3}-\sqrt{3}+\frac{10}{\sqrt{3}}=-9\sqrt{3}+\frac{10}{\sqrt{3}}=\frac{-17\sqrt{3}}{3}\)
11 tháng 12 2019
a) = \(5\sqrt{2}-3\sqrt{6}+3\sqrt{2}+5\sqrt{6}\)
= \(8\sqrt{2}+2\sqrt{6}\)
b) = \(2\sqrt{3}-4\sqrt{2}-5\sqrt{3}-\sqrt{2}\)
= \(-3\sqrt{3}-5\sqrt{2}\)
c) = \(\frac{\left(\sqrt{2}-1\right)\left(2+\sqrt{2}\right)}{\left(2-\sqrt{2}\right)\left(2+\sqrt{2}\right)}\)
=\(\frac{2\sqrt{2}+2-2-\sqrt{2}}{2^2-\sqrt{2^2}}\)
=\(\frac{\sqrt{2}}{4-2}\) = \(\frac{\sqrt{2}}{2}\)
d) = \(2\sqrt{6}-5\sqrt{6}+2\sqrt{2}\)
=\(-3\sqrt{6}+2\sqrt{2}\)
e) = \(8\sqrt{6}+3\sqrt{6}-6\sqrt{6}=5\sqrt{6}\)
f) = \(4\sqrt{3}+9\sqrt{3}-4\sqrt{3}=9\sqrt{3}\)
g) = \(10+5\sqrt{10}-5\sqrt{10}=10\)
h) = \(\frac{\left(3+\sqrt{3}\right)\left(3+\sqrt{3}\right)}{\left(3-\sqrt{3}\right)\left(3+\sqrt{3}\right)}+\frac{\left(3-\sqrt{3}\right)\left(3-\sqrt{3}\right)}{\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)}\)
= \(\frac{9+3\sqrt{3}+3\sqrt{3}+3}{3^2-\sqrt{3^2}}+\frac{9-3\sqrt{3}-3\sqrt{3}+3}{3^2-\sqrt{3^2}}\)
= \(\frac{12+6\sqrt{3}}{9-3}+\frac{12-6\sqrt{3}}{9-3}\)
= \(\frac{12+6\sqrt{3}+12-6\sqrt{3}}{6}\)
= \(\frac{24}{6}=4\)
k) = \(\left(2\sqrt{7}-2\sqrt{3}+\sqrt{7}\right).\sqrt{7}+2\sqrt{21}\)
= \(\left(3\sqrt{7}-2\sqrt{3}\right).\sqrt{7}+2\sqrt{21}\)
= \(21-2\sqrt{21}+2\sqrt{21}=21\)
l) = \(\frac{\left(2\sqrt{3}-\sqrt{6}\right)\left(\sqrt{8}+2\right)}{\left(\sqrt{8}-2\right)\left(\sqrt{8}+2\right)}\)
= \(\frac{4\sqrt{6}+4\sqrt{3}-4\sqrt{3}-2\sqrt{6}}{\sqrt{8^2}-2^2}\)
= \(\frac{2\sqrt{6}}{8-4}=\frac{2\sqrt{6}}{4}=\frac{\sqrt{6}}{2}\)
13 tháng 9 2018
\(=\frac{1}{2}\sqrt{16.3}-2\sqrt{25.3}-\frac{\sqrt{3.11}}{\sqrt{11}}+5\sqrt{\frac{1.3+1}{3}}\)
\(=\frac{1}{2}\sqrt{4^2.3}-2\sqrt{5^2.3}-\frac{\sqrt{3}.\sqrt{11}}{\sqrt{11}}+5\sqrt{\frac{4}{3}}\)
\(=\frac{1}{2}.4\sqrt{3}-2.5\sqrt{3}-\sqrt{3}+5\frac{\sqrt{4}}{\sqrt{3}}\)
\(=\frac{4}{2}\sqrt{3}-10\sqrt{3}-\sqrt{3}+5\frac{\sqrt{4}.\sqrt{4}}{\sqrt{3.}\sqrt{3}}\)
\(=2\sqrt{3}-10\sqrt{3}-\sqrt{3}+5\frac{2\sqrt{3}}{3}\)
\(=2\sqrt{3}-10\sqrt{3}-\sqrt{3}+10\frac{\sqrt{3}}{3}\)
\(=\left(2-10-1+\frac{10}{3}\right)\sqrt{3}\)
\(=-\frac{17}{3}\)
27 tháng 8 2019
a)\(\sqrt{75}-\sqrt{5\frac{1}{3}}+\frac{9}{2}\sqrt{2\frac{2}{3}}+2\sqrt{27}=5\sqrt{3}-\frac{\sqrt{15}}{3}+3\sqrt{3}+6\sqrt{3}=14\sqrt{3}-\frac{\sqrt{15}}{3}\)
b) \(\sqrt{48}+\sqrt{5\frac{1}{3}}+2\sqrt{75}-5\sqrt{1\frac{1}{3}}=4\sqrt{3}+\frac{\sqrt{15}}{3}+10\sqrt{3}-\frac{5\sqrt{3}}{3}=\frac{12\sqrt{3}+30\sqrt{3}-5\sqrt{3}}{3}+\frac{\sqrt{15}}{3}=\frac{37\sqrt{3}+\sqrt{15}}{3}\)
c) \(\left(\sqrt{15}+2\sqrt{3}\right)^2+12\sqrt{5}=\left[\left(\sqrt{15}\right)^2+4\sqrt{45}+\left(2\sqrt{3}\right)^2\right]+12\sqrt{5}=15+12\sqrt{5}+12+12\sqrt{5}=27+24\sqrt{5}\)
d) \(\left(\sqrt{6}+2\right)\left(\sqrt{3}-\sqrt{2}\right)=\sqrt{18}-\sqrt{12}+\sqrt{6}-2\sqrt{2}=3\sqrt{2}-2\sqrt{3}+\sqrt{6}-2\sqrt{2}=\sqrt{2}-2\sqrt{3}+\sqrt{6}\)
e) \(\left(\sqrt{3}+1\right)^2-2\sqrt{3}+4=\left(\sqrt{3}\right)^2+2\sqrt{3}+1-2\sqrt{3}+4=3+2\sqrt{3}+1-2\sqrt{3}+4=8\)
f) \(\frac{1}{7+4\sqrt{3}}+\frac{1}{7-4\sqrt{3}}=\frac{7-4\sqrt{3}+7+4\sqrt{3}}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}=\frac{14}{1}=14\)
g) \(\left(\frac{1}{\sqrt{5}-\sqrt{2}}-\frac{1}{\sqrt{5}+\sqrt{2}}+1\right)\frac{1}{\left(\sqrt{2}+1\right)^2}=\left(\frac{\sqrt{5}+2-\sqrt{5}+2+5-2}{\left(\sqrt{5}-\sqrt{2}\right)\left(\sqrt{5}+\sqrt{2}\right)}\right)\frac{1}{3+2\sqrt{2}}=\frac{7}{3}.\frac{1}{3+2\sqrt{2}}=\frac{7}{9+6\sqrt{2}}\)
bạn thử tải app này xem có đáp án không nhé <3 https://giaingay.com.vn/downapp.html
\(\frac{1}{2}\sqrt{48}-2\sqrt{75}-\sqrt{54}+5\sqrt{1\frac{1}{3}}\) =\(2\sqrt{3}-10\sqrt{3}-3\sqrt{6}+\frac{10\sqrt{3}}{3}\)
=\(\frac{-24\sqrt{3}-9\sqrt{6}+10\sqrt{3}}{3}=\frac{-14\sqrt{3}-9\sqrt{6}}{3}\)