|-100|+|-99|-100-99

= 100 + 99 - 100 - 99

= [100 - (100)] + [99 - (-99)]

= 0 + 0

= 0

|-100|+|-99|-100-99

= 100+99-100-99

= 199 - 100 - 99

= 99-99

=0

88+88+99+99+100+100=574

  88 + 88 + 99 + 99 + 100 + 100

= 88 x 2 + 99 x 2 + 100 x 2

= 176  +  198  +  200

= 574

Ta có: \(S=\frac13-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+\cdots+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)

=>\(3S=1-\frac23+\frac{3}{3^2}-\frac{4}{3^3}+\ldots+\frac{99}{3^{98}}-\frac{100}{3^{99}}\)

=>3S+S=\(1-\frac23+\frac{3}{3^2}-\frac{4}{3^3}+\cdots+\frac{99}{3^{98}}-\frac{100}{3^{99}}+\frac13-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+\cdots+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)

=>4S=\(1-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\cdots-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)

Đặt \(A=-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\cdots-\frac{1}{3^{99}}\)

=>3A=\(-1+\frac13-\frac{1}{3^2}+\cdots-\frac{1}{3^{98}}\)

=>3A+A=\(-1+\frac13-\frac{1}{3^2}+\cdots-\frac{1}{3^{98}}-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\cdots-\frac{1}{3^{99}}\)

=>4A=\(-1-\frac{1}{3^{99}}=\frac{-3^{99}-1}{3^{99}}\)

=>\(A=\frac{-3^{99}-1}{4\cdot3^{99}}\)

Ta có: \(4S=1-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\cdots-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)

\(=1+\frac{-3^{99}-1}{4\cdot3^{99}}-\frac{100}{3^{100}}=1+\frac{-3^{100}-3-400}{4\cdot3^{100}}=1-\frac14-\frac{403}{4\cdot3^{100}}<\frac34\)

=>\(S<\frac{3}{16}\)

mà 3/16<3/15=1/5

nên S<1/5

camon truoc

1057 nhé bạn

\(P=\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}\cdot\cdot\cdot\frac{99}{100}\)

\(P=\frac{1\cdot3}{2\cdot2}\cdot\frac{2\cdot4}{3\cdot3}\cdot\frac{3\cdot5}{4\cdot4}\cdot\cdot\cdot\frac{9\cdot11}{10\cdot10}\)

\(P=\frac{1\cdot3\cdot2\cdot4\cdot3\cdot5\cdot\cdot\cdot9\cdot11}{2\cdot2\cdot3\cdot3\cdot4\cdot4\cdot\cdot\cdot10\cdot10}\)

\(P=\frac{\left(1\cdot2\cdot3\cdot\cdot\cdot9\right)\cdot\left(3\cdot4\cdot5\cdot\cdot\cdot11\right)}{\left(2\cdot3\cdot4\cdot\cdot\cdot10\right)\cdot\left(2\cdot3\cdot4\cdot\cdot\cdot10\right)}\)

\(P=\frac{1\cdot11}{10\cdot2}=\frac{11}{20}\)

A=-1++(-1)+..+-(1) có 50 số -1

=>A=-1x50=-50

B=(1-2-3+4)+(5-6-7+8)+...+(97-98-99+100)

B=0+0+0+..+0

B=0

C=2^100-(2^99+2^98+...+1)

C=2^100-(2^100-1)

C=1