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KV
29 tháng 10 2020
a) \(\left(\frac{2^2}{5}\right)+5\frac{1}{2}.\left(4,5-2,5\right)+\frac{2^3}{-4}\)
\(=\frac{4}{5}+\frac{11}{2}.2+\frac{-8}{4}\)
\(=\frac{4}{5}+11-2\)
\(=\frac{4}{5}+9\)
\(=\frac{49}{9}\)
b) \(\left(-2^3\right)+\frac{1}{2}:\frac{1}{8}-\sqrt{25}+\left|-64\right|\)
\(=-8+4-5+64\)
= 55
c) \(\frac{\sqrt{3^2+\sqrt{39}^2}}{\sqrt{91^2}-\sqrt{\left(-7\right)^2}}\)
\(=\frac{\sqrt{9+39}}{91-\sqrt{49}}\)
\(=\frac{\sqrt{48}}{91-7}\)
\(=\frac{4\sqrt{3}}{84}\)
\(=\frac{\sqrt{3}}{41}\)
d) Xem lại đề nhé em!
e) \(\sqrt{25}-3\sqrt{\frac{4}{9}}\)
\(=5-3.\frac{2}{3}\)
= 5 - 2
= 3
h) \(\left(-3^2\right).\frac{1}{3}-\sqrt{49}+\left(5^3\right):\sqrt{25}\)
\(=-9.\frac{1}{3}-7+125:5\)
\(=-3-7+25\)
= 15
VM
4 tháng 8 2019
a) \(\sqrt{\frac{4}{81}}:\sqrt{\frac{25}{81}}-1\frac{2}{5}\)
\(=\frac{2}{9}:\frac{5}{9}-\frac{7}{5}\)
\(=\frac{2}{5}-\frac{7}{5}\)
\(=-1.\)
b) \(\sqrt{36}.\sqrt{\frac{25}{16}}+\frac{1}{4}\)
\(=6.\frac{5}{4}+\frac{1}{4}\)
\(=\frac{15}{2}+\frac{1}{4}\)
\(=\frac{31}{4}.\)
c) \(1\frac{1}{2}+\frac{4}{7}:\left(-\frac{8}{9}\right)\)
\(=\frac{3}{2}+\frac{4}{7}:\left(-\frac{8}{9}\right)\)
\(=\frac{3}{2}+\left(-\frac{9}{14}\right)\)
\(=\frac{6}{7}.\)
d) \(1,17-0,4.\left(\frac{1}{2}\right)^2-\frac{1}{-5}\)
\(=\frac{117}{100}-\frac{2}{5}.\frac{1}{4}-\left(-\frac{1}{5}\right)\)
\(=\frac{117}{100}-\frac{1}{10}+\frac{1}{5}\)
\(=\frac{107}{100}+\frac{1}{5}\)
\(=\frac{127}{100}.\)
Chúc bạn học tốt!
4 tháng 8 2019
a, \(\frac{4}{81}:\sqrt{\frac{25}{81}-1\frac{2}{5}}\)
\(\Rightarrow\frac{4}{81}:\frac{5}{9}-\frac{7}{5}\)
\(\Rightarrow\frac{4}{81}.\frac{9}{5}-\frac{7}{5}\)
\(\Rightarrow\frac{4}{9}.\frac{1}{5}-\frac{7}{5}\)
\(\Rightarrow\frac{-59}{45}\)
b,\(\sqrt{36}.\sqrt{\frac{25}{16}+\frac{1}{4}}\)
\(\Rightarrow6.\frac{5}{4}+\frac{1}{4}\)
\(\Rightarrow\frac{15}{2}+\frac{1}{4}\)
\(\Rightarrow\frac{31}{4}\)
c,\(1\frac{1}{2}+\frac{4}{7}:\frac{-8}{9}\)
\(\Rightarrow\frac{3}{2}-\frac{4}{7}.\frac{-8}{9}\)
\(\Rightarrow\frac{3}{2}-\frac{9}{14}\)
\(\Rightarrow\frac{6}{7}\)
d, \(1,17-\left(0,4.\frac{1}{2}\right)^2-\frac{1}{5}\)
\(\Rightarrow\frac{117}{100}-\left(\frac{1}{5}\right)^2-\frac{1}{5}\)
\(\Rightarrow\frac{117}{100}-\frac{1}{25}-\frac{1}{5}\)
\(\Rightarrow\frac{93}{100}\)
8 tháng 11 2016
Ta có:
\(A=\frac{1}{\sqrt{2}+\sqrt{1}}+\frac{1}{\sqrt{3}+\sqrt{2}}+...+\frac{1}{\sqrt{25}+\sqrt{24}}\)
\(=\frac{\sqrt{2}-\sqrt{1}}{2-1}+\frac{\sqrt{3}-\sqrt{2}}{3-2}+...+\frac{\sqrt{25}-\sqrt{24}}{25-24}\)
\(=\frac{\sqrt{2}-\sqrt{1}}{1}+\frac{\sqrt{3}-\sqrt{2}}{1}+...+\frac{\sqrt{25}-\sqrt{24}}{1}\)
\(=5-1=4\)
2 tháng 7 2020
Bài 1:
a) Ta có: \(\left(0.125\right)\cdot\left(-3\cdot7\right)\cdot\left(-2\right)^3\)
\(=\frac{1}{8}\cdot\left(-21\right)\cdot\left(-8\right)\)
\(=\frac{1}{8}\cdot168\)
\(=21\)
b) Ta có: \(\sqrt{36}\cdot\sqrt{\frac{25}{16}}+\frac{1}{4}\)
\(=\sqrt{36\cdot\frac{25}{16}}+\frac{1}{4}\)
\(=\sqrt{\frac{225}{4}}+\frac{1}{4}\)
\(=\frac{15}{2}+\frac{1}{4}\)
\(=\frac{31}{4}\)
c) Ta có: \(\sqrt{\frac{4}{81}}:\sqrt{\frac{25}{81}}-1\frac{2}{5}\)
\(=\frac{2}{9}:\frac{5}{9}-\frac{7}{5}\)
\(=\frac{2}{5}-\frac{7}{5}=-1\)
d) Ta có: \(0,1\cdot\sqrt{225}\cdot\sqrt{\frac{1}{4}}\)
\(=0,1\cdot15\cdot\frac{1}{2}=\frac{3}{4}\)
cảm ơn bạn nhé 
NQ
1
LM
27 tháng 8 2016
\(\sqrt[2]{\frac{1}{4}}-\sqrt{25}+\sqrt{\frac{4}{9}}=\frac{1}{2}-5+\frac{2}{3}=\)\(\frac{-23}{6}\)
5 tháng 4 2017
a, \(-\frac{187}{70}\)
b,\(\frac{27}{70}\)
c,\(\frac{53}{14}\)
d,\(\frac{27}{4}\)
e,1
f,\(\frac{23}{4}\)
g,-1
i,6
k,315
l,\(\frac{9}{2}\)
2 tháng 6 2015
a) \(\sqrt{0,09}-\sqrt{0,64}=\frac{-1}{2}=-0,5\)
b) \(0,1\cdot\sqrt{225}-\sqrt{\frac{1}{4}}=0,1\cdot15-\frac{1}{2}=1\)
c) \(\sqrt{0,36}\cdot\sqrt{\frac{25}{16}+\frac{1}{4}}=\frac{3\sqrt{29}}{20}\)
d) đề baì có sai ko ban?
1 tháng 12 2019
\(\sqrt{\frac{25}{81}}-1\frac{5}{9}+\left(-1\right)^{2015}+4\sqrt{\frac{1}{4}}\)
\(=\frac{5}{9}-\frac{6}{9}-1+2\)
\(=\frac{-1}{9}+1\)
\(=\frac{8}{9}\)
=1/5
Chúc bn học tốt.
.........................
\(= 1/5\)