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a. \(\dfrac{-2}{3}+\dfrac{-1}{5}+\dfrac{3}{4}-\dfrac{5}{6}-\dfrac{7}{10}\)
= \(\dfrac{-4}{6}+\dfrac{-2}{10}+\dfrac{3}{4}-\dfrac{5}{6}-\dfrac{7}{10}\)
= \(\dfrac{-3}{2}+\dfrac{1}{2}+\dfrac{3}{4}\)
= (-1) + \(\dfrac{3}{4}\)
= \(\dfrac{-4}{4}+\dfrac{3}{4}\)
= \(\dfrac{-1}{4}\)
b; 0,5 + \(\dfrac{1}{3}\) + 0,4 + \(\dfrac{5}{7}\) + \(\dfrac{1}{6}\) - \(\dfrac{4}{35}\)
= (\(\dfrac{1}{3}\)+ \(\dfrac{1}{6}\) + \(\dfrac{1}{2}\)) + (\(\dfrac{5}{7}\)- \(\dfrac{4}{35}\)+ \(\dfrac{2}{5}\))
= ( \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\)) + (\(\dfrac{3}{5}\) + \(\dfrac{2}{5}\))
= 1 + 1
= 2
A = 6 -2/3 + 1 /2 - 5 -5/3 +3/2 -3 + 7/3 - 5/2
= (6 - 5 - 3) - ( 2/3 -5/3 + 7/3 ) + ( 1/2 +3/2 - 5/2)
= -2 + 0 -1/2 = -5/2
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 )
Số số hạng của dãy: (2012-2):1+1=2011(số)
Vì 2011 là số lẻ nên A là số âm.
a)1,5.(1/3-2/3)
=3/2.(-1/3)
=-1/2
b)2/5+3/5:(-3/2)=1/2
=2/3+2/5+1/2
=16/15+1/2
=47/30
c)1 và 2/5 - (-1/2)^2 + 7/10
=7/5 - 1/4 + 7/10
=23/20 + 7/10
=37/20