\(A=\frac{4}{10\cdot2}+\frac{6}{2\cdot20}+\frac{15}{5\cdot20}+\frac{5}{5\cdot40};B=\frac{3}{1\cdot5}+\frac{5}{13\cdot1}+\frac{11}{13\cdot3}+\frac{2}{3\cdot26}\)
So sánh A với B
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12 tháng 12 2019
a) \(\left(-\frac{5}{2}\right)^2:\left(-15\right)-\left(-0,45+\frac{3}{4}\right).\left(-1\frac{5}{9}\right)\)
= \(-\frac{25}{4}:\left(-15\right)-\left(\frac{9}{20}+\frac{15}{20}\right).\left(-\frac{14}{9}\right)\)
=\(-\frac{25}{4}.\frac{1}{-15}-\frac{6}{5}.\left(-\frac{14}{9}\right)\)
= \(\frac{-5}{12}-\frac{8}{5}\)
= \(\frac{\left(-25\right)-96}{60}\)
= \(\frac{\left(-25\right)+\left(-96\right)}{60}\)
=\(\frac{121}{60}\)
b) \(\left(\frac{-1}{3}\right)-\left(\frac{-3}{5}\right)^0+\left(1-\frac{1}{2}\right)^2:2\)
= \(\left(\frac{-1}{3}\right)-1+\left(\frac{1}{2}\right)^2.\frac{1}{2}\)
=\(\left(\frac{-1}{3}\right)-\frac{3}{3}+\frac{1}{4}.\frac{1}{2}\)
= \(\frac{-4}{3}+\frac{1}{8}\)=\(\frac{-32+3}{24}\)
=\(\frac{-29}{24}\)
c) E=\(\frac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}\)
=\(\frac{\left(2^2\right)^5.\left(3^2\right)^4-2.6^9}{2^{10}.3^8+6^8.20}\)
=\(\frac{2^{10}.3^8-2.6^9}{2^{10}.3^8+6^8.20}\)
=\(\frac{3}{5}\)
d)\(\frac{5^4.20^4}{25^5.4^5}\)
=\(\frac{\left(5.20\right)^4}{\left(25.4\right)^5}\)
=\(\frac{100^4}{100^5}\)
=\(\frac{1}{100}\)
5 tháng 9 2017
a)\(\frac{25.4-0,5\cdot40\cdot0,2\cdot20\cdot0,25}{1+2+8+...+129+156}\)
\(=\frac{100-100}{1+2+8+...+156}\)
\(=\frac{0}{1+2+8+...+156}\)
\(=0\)
b)\(\frac{0,5\cdot40-0,5\cdot20\cdot8\cdot0,1\cdot0,25\cdot10}{128:8.16.4\left(4+52:4\right)}=\frac{20-20}{128:8.16.4.\left(4+52:4\right)}=\frac{0}{128:8.16.4.\left(4+52:4\right)}=0\)
1 tháng 6 2018
\(A=\frac{1.2+2.4+3.6+4.8+5.10}{3.4+6.8+9.12+12.16+15.20}\)
\(A=\frac{1.2.\left(1+2^2+3^2+4^2+5^2\right)}{3.4.\left(1+2^2+3^2+4^2+5^2\right)}\)
\(A=\frac{1.2}{3.4}\)
\(A=\frac{1}{6}\)
Ta thấy : \(B=\frac{111111}{666665}>\frac{111111}{666666}=\frac{1}{6}\)
Vậy B > A
LH
11 tháng 7 2018
Theo đề bài, ta có:
\(A=\frac{1\times2+2\times4+3\times6+4\times8+5\times10}{3\times4+6\times8+9\times12+12\times16+15\times20}\)
\(A=\frac{1\times2\times\left(1+2^2+3^2+4^2+5^2\right)}{3\times4\times\left(1+2^2+3^2+4^2+5^2\right)}\)
\(A=\frac{1\times2}{3\times4}\)
\(A=\frac{1}{6}\)
Ta thấy rằng: \(B=\frac{111111}{666665}>\frac{111111}{666666}=\frac{1}{6}\)
Vậy \(B>A\)
16 tháng 3 2017
\(1+2+3+...+100=10x-\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\left(6.20-3.40\right)\)
\(\Rightarrow5050=10x-\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right).0\)
\(\Rightarrow5050=10x\)
\(\Rightarrow x=505\)
15 tháng 10 2016
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{19.20}\)
\(\Rightarrow A=\frac{1}{1}.\frac{1}{2}+\frac{1}{2}.\frac{1}{3}+...+\frac{1}{19}.\frac{1}{20}\)
\(\Rightarrow A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{19}-\frac{1}{20}\)
\(\Rightarrow A=\frac{1}{1}-\frac{1}{20}=\frac{19}{20}\)
