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A. 2/3 - (- 1/4)+3/5-(+7/45)-(-5/9)+1/12+1/39
B. 6.(-2/3)^2 - 3(-2/3)^2 - 2 : (-3/2)+4
b) 1-2-3+4+5-6-7+8+......+1989-1990-1991+1992+1993
=(1-2-3+4)+(5-6-7+8)+.....+(1989-1990-1991+1992)+1993
=0+0+...+0+1993=1993.
\(A=\left(3-\dfrac{1}{4}+\dfrac{3}{2}\right)-\left(5+\dfrac{1}{3}-\dfrac{5}{6}\right)-\left(6-\dfrac{7}{4}+\dfrac{2}{3}\right)\\ \Rightarrow A=3-\dfrac{1}{4}+\dfrac{3}{2}-5-\dfrac{1}{3}+\dfrac{5}{6}-6+\dfrac{7}{4}-\dfrac{2}{3}\\ \Rightarrow A=\left(3-5-6\right)-\left(\dfrac{1}{4}+\dfrac{7}{4}\right)+\left(\dfrac{3}{2}+\dfrac{5}{6}-\dfrac{2}{3}\right)\\ \Rightarrow A=-8-\dfrac{3}{2}+\dfrac{5}{3}\\ =-\dfrac{47}{6}.\\ B=0,5+\dfrac{1}{3}+0,4+\dfrac{5}{7}+\dfrac{1}{6}-\dfrac{4}{35}+\dfrac{1}{41}\)
\(\Rightarrow B=\left(0,5+0,4\right)+\left(\dfrac{1}{3}+\dfrac{1}{6}\right)+\left(\dfrac{5}{7}-\dfrac{4}{35}\right)+\dfrac{1}{41}\\ \Rightarrow B=\dfrac{9}{10}+\dfrac{1}{2}+\dfrac{3}{5}+\dfrac{1}{41}\\ \Rightarrow B=2+\dfrac{1}{41}\\ \Rightarrow B=\dfrac{83}{41}.\)
a) \(-1\frac{5}{7}.15+\frac{2}{7}.\left(-15\right)+\left(-105\right).\left(\frac{2}{3}-\frac{4}{5}+\frac{1}{7}\right)\)
\(=1\frac{5}{7}.\left(-15\right)+\frac{2}{7}.\left(-15\right)+\left(-105\right).\left(\frac{70}{105}-\frac{84}{105}+\frac{15}{105}\right)\)
\(=\left(-15\right)\left(1+\frac{5}{7}+\frac{2}{7}\right)+\left(-105\right).\frac{1}{105}\)
\(=-30-1=-31\)
b) \(\frac{2}{3}+\frac{3}{4}.\left(-\frac{4}{9}\right)\)
= \(=\frac{2}{3}+\frac{3.\left(-4\right)}{4.9}=\frac{2}{3}+\frac{-1}{3}=\frac{1}{3}\)
c) \(\left(\frac{3}{4}-0,2\right).\left(0,4-\frac{4}{5}\right)\)
\(=\left(\frac{3}{4}-\frac{1}{5}\right).\left(\frac{2}{5}-\frac{4}{5}\right)=\frac{11}{20}.\left(\frac{-2}{5}\right)=\frac{11.\left(-1\right).2}{2.10.5}=\frac{-11}{50}\)
\(A=\frac{4}{3.5}+\frac{4}{5.7}+\frac{4}{7.9}+...+\frac{4}{99.101}.\)
\(A=\frac{4}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(A=2.\left(\frac{1}{3}-\frac{1}{101}\right)=2\cdot\frac{98}{303}=\frac{196}{303}\)
\(A=\frac{4}{3.5}+\frac{4}{5.7}+\frac{4}{7.9}+....+\frac{4}{99.101}.\)
\(=2.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+.....+\frac{2}{99.101}\right)\)
\(=2.\left(\frac{5-3}{3.5}+\frac{7-5}{5.7}+\frac{9-7}{7.9}+.....+\frac{101-99}{99.101}\right)\)
\(=2.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+....+\frac{1}{99}-\frac{1}{101}\right)\)
\(=2.\left(\frac{1}{3}-\frac{1}{101}\right)\)
\(=2.\frac{98}{303}=\frac{196}{303}\)
1. \(A=\frac{1}{2}-\frac{2}{5}+\frac{1}{3}+\frac{5}{7}-\frac{-1}{6}+\frac{-4}{35}+\frac{1}{41}\)
\(=\frac{1}{2}-\frac{2}{5}+\frac{1}{3}+\frac{5}{7}+\frac{1}{6}-\frac{4}{35}+\frac{1}{41}\)
\(=\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{6}\right)-\left(\frac{2}{5}-\frac{5}{7}+\frac{4}{35}\right)+\frac{1}{41}\)
\(=\left(\frac{5}{6}+\frac{1}{6}\right)-\left(\frac{-11}{35}+\frac{4}{35}\right)+\frac{1}{41}\)\(=1-\frac{-7}{35}+\frac{1}{41}=1+\frac{1}{5}+\frac{1}{41}=\frac{251}{205}\)
2. a) \(1+4+4^2+4^3+......+4^{99}=\left(1+4\right)+\left(4^2+4^3\right)+.......+\left(4^{98}+4^{99}\right)\)
\(=\left(1+4\right)+4^2\left(1+4\right)+.........+4^{98}\left(1+4\right)\)
\(=5+4^2.5+........+4^{98}.5=5\left(1+4^2+.....+4^{98}\right)⋮5\)( đpcm )
b) \(3^{n+2}-2^{n+2}+3^n-2^n=\left(3^{n+2}+3^n\right)-\left(2^{n+2}+2^n\right)\)
\(=3^n\left(3^2+1\right)-2^n\left(2^2+1\right)=3^n\left(9+1\right)-2^n\left(4+1\right)\)
\(=3^n.10-2^n.5=3^n.10-2^{n-1+1}.5=3^n.10-2^{n-1}.2.5\)
\(=3^n.10-2^{n-1}.10=10\left(3^n-2^{n-1}\right)⋮10\)( đpcm )
a: \(=\dfrac{28-2-3}{4}:\dfrac{40-2-5}{8}=\dfrac{23}{4}\cdot\dfrac{8}{33}=\dfrac{46}{33}\)
b: =78(0,65+0,35)+2020(2,2-2,2)
=78*1=78
\(A=\dfrac{4}{1\cdot3}+\dfrac{4}{3\cdot5}+\dfrac{4}{5\cdot7}+...+\dfrac{4}{99\cdot101}\)
\(A=2\cdot\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{4}{99\cdot101}\right)\)
\(A=2\cdot\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
\(A=2\cdot\left(1-\dfrac{1}{101}\right)\)
\(A=2\cdot\dfrac{100}{101}\)
\(A=\dfrac{200}{101}\)