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4 tháng 8 2020
\(\Rightarrow A=4.\left[\frac{6}{2.\left(2.4\right)}+\frac{5}{\left(2.4\right).13}+\frac{3}{13.\left(4.4\right)}+\frac{2}{\left(4.4\right).18}+\frac{10}{18.\left(7.4\right)}\right]\)
\(=4.\left(\frac{6}{2.8}+\frac{5}{8.13}+\frac{3}{13.16}+\frac{2}{16.18}+\frac{10}{18.28}\right)=4.\left(\frac{1}{2}-\frac{1}{8}+\frac{1}{8}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}+\frac{1}{16}-\frac{1}{18}+\frac{1}{18}-\frac{1}{28}\right)\)
\(=4.\left(\frac{1}{2}-\frac{1}{28}\right)=4.\frac{13}{28}=\frac{13}{7}\)
1 tháng 4 2019
A=1+(2-3-3+5)+(6-7-8+9)+....+(98-99-100+101)+102
=1+0+0+....+102=103
b) |1-2x|>7
=> 1-2x>7 hoặc 1-2x<-7
=> 2x<-6 hoặc 2x>8
=> x<-3 hoặc x>4
XO
1 tháng 9 2019
\(A=\left(3+\frac{1}{2}-\frac{2}{3}\right)-\left(2-\frac{2}{3}+\frac{5}{2}\right)+\left(-5+\frac{5}{2}-\frac{4}{3}\right)\)
\(=3+\frac{1}{2}-\frac{2}{3}-2+\frac{2}{3}-\frac{5}{2}-5+\frac{5}{2}-\frac{4}{3}\)
\(=\left(3-2-5\right)+\left(\frac{1}{2}-\frac{5}{2}+\frac{5}{2}\right)-\left(\frac{2}{3}-\frac{2}{3}+\frac{4}{3}\right)\)
\(=-4-\frac{1}{2}\)
\(=-\frac{9}{2}\)
\(B=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{72}+\frac{1}{90}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}\)
\(=\frac{9}{10}\)
KK
1 tháng 9 2019
\(A=\left(3+\frac{1}{2}-\frac{2}{3}\right)-\left(2-\frac{2}{3}+\frac{5}{2}\right)+\left(-5+\frac{5}{2}-\frac{4}{3}\right)\)
\(A=3+\frac{1}{2}-\frac{2}{3}-2+\frac{2}{3}-\frac{5}{2}-5+\frac{5}{2}-\frac{4}{3}\)
\(A=\left(3-2-5\right)+\left(\frac{1}{2}-\frac{5}{2}+\frac{5}{2}\right)-\left(\frac{2}{3}-\frac{2}{3}+\frac{4}{3}\right)\)
\(A=-4+\frac{1}{2}-\frac{4}{3}\)
\(A=-\frac{29}{6}\)
11 tháng 2 2019
a, \(A=\frac{2^{12}\cdot3^5-4^6\cdot9^2}{(2^2\cdot3)^6+8^4\cdot3^5}-\frac{5^{10}\cdot7^3-25^5\cdot49^2}{(125\cdot7)^3+5^9\cdot14^3}\)
\(A=\frac{2^{12}\cdot3^5-2^{12}\cdot3^4}{2^{12}\cdot3^6+2^{12}\cdot3^5}-\frac{5^{10}\cdot7^3-5^{10}\cdot7^4}{5^9\cdot7^3+5^9\cdot2^3\cdot7^3}\)
\(A=\frac{2^{12}\cdot3^4(3-1)}{2^{12}\cdot3^5(3+1)}-\frac{5^{10}\cdot7^3(1-7)}{5^9\cdot7^3(1+2^3)}\)
\(A=\frac{2^{12}\cdot3^4\cdot2}{2^{12}\cdot3^5\cdot4}-\frac{5^{10}\cdot7^3\cdot(-6)}{5^9\cdot7^3\cdot9}=\frac{1}{6}-\frac{-10}{3}=\frac{7}{2}\)
11 tháng 2 2019
b,\(3^{n+2}-2^{n+2}+3^n-2^n\)
\(=(3^{n+2}+3^n)-(2^{n+2}-2^n)\)
\(=(3^n\cdot3^2+3^n)-(2^n\cdot2^2-2^n)\)
\(=3^n\cdot(3^2+1)-2^n\cdot(2^2+1)\)
\(=3^n\cdot9+1-2^n\cdot4+1\)
\(=3^n\cdot10-2^n\cdot5\)
Vì \(2\cdot5⋮10\Rightarrow2^n\cdot5⋮10\)
\(3^n\cdot10⋮10\)
Vậy : ....
29 tháng 3 2019
\(\frac{2^5.7+2^5}{2^5.5^2-2^5.3}=\frac{2^5.\left(7+1\right)}{2^5.\left(5^2-3\right)}=\frac{8}{22}\)
29 tháng 3 2019
\(\frac{2^5.7+2^5}{2^5.5^2-2^5.3}\)
\(\frac{2^5.\left(7+1\right)}{2^5.\left(5^2-3\right)}\)=\(\frac{8}{22}=\frac{4}{11}\)
Hok tốt
13 tháng 9 2019
\(B=\frac{\left[\frac{2}{3}\right]^3\cdot\left[-\frac{3}{4}\right]^2\cdot\left[-1\right]^5}{\left[\frac{2}{5}\right]^2\cdot\left[-\frac{5}{12}\right]^3}\)
\(=\frac{\frac{2^3}{3^3}\cdot\frac{\left[-3\right]^2}{4^2}\cdot\left[-1\right]}{\frac{2^2}{5^2}\cdot\frac{\left[-5\right]^3}{12^3}}\)
\(=\frac{\frac{8}{27}\cdot\frac{9}{16}\cdot\left[-1\right]}{\frac{4}{25}\cdot\frac{-125}{\left[2^2\cdot3\right]^3}}\)
\(=\frac{\frac{1}{3}\cdot\frac{1}{2}\cdot\left[-1\right]}{\frac{4}{25}\cdot\frac{-125}{\left[2^2\right]^3\cdot3^3}}\)
\(=\frac{\frac{1\cdot1\cdot\left[-1\right]}{3\cdot2\cdot1}}{\frac{4}{25}\cdot\frac{-125}{4^3\cdot3^3}}\)
\(=\frac{\frac{-1}{6}}{\frac{4}{25}\cdot\frac{-125}{64\cdot27}}=\frac{\frac{-1}{6}}{\frac{4}{1}\cdot\frac{-5}{64\cdot27}}\)
\(=\frac{\frac{-1}{6}}{4\cdot\frac{-5}{64\cdot27}}=\frac{\frac{-1}{6}}{-\frac{20}{64\cdot27}}=\frac{72}{5}\)
2 tháng 10 2021
a: \(\left(0.5\right)^3\cdot2^3=1\)
b: \(\left(0.25\right)^2\cdot16=1\)
c: \(\left(\dfrac{3}{5}\right)^3:\left(-\dfrac{27}{1000}\right)=\dfrac{3^3}{5^3}\cdot\dfrac{-1000}{27}=\dfrac{-1000}{125}=-8\)
\(=\dfrac{2^4\cdot5^4+2^5\cdot5^3}{2^{10}\cdot16}=\dfrac{2^4\cdot5^3\left(5+2\right)}{2^{10}\cdot2^4}=\dfrac{2^4\cdot5^3\cdot7}{2^{14}}=\dfrac{5^3\cdot7}{2^{10}}=\dfrac{875}{1024}\)