\(=\frac{32+16+8+4+2+1}{64}=\frac{63}{64}\)

Ghi đề khó hiểu quá Nguyen Ngoc Vy Phuong

moi hok lop 6

a) Đặt \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)

\(2A=\left(\frac{1}{2}\times2\right)+\left(\frac{1}{4}\times2\right)+\left(\frac{1}{8}\times2\right)+\left(\frac{1}{16}\times2\right)+\left(\frac{1}{32}\times2\right)\)

\(2A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\)

Ta lấy : \(2A-1A=1A\)

\(A=\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\right)\)

\(A=1-\frac{1}{32}\)

\(A=\frac{31}{32}\)

Vậy \(A=\frac{31}{32}\)

b) Đặt \(B=\frac{2}{1\times2}+\frac{2}{2\times3}+\frac{2}{3\times4}+...+\frac{2}{18\times19}+\frac{2}{19\times20}\)

\(B=2\times(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20})\)

\(B=2\times\left(1-\frac{1}{20}\right)\)

\(B=2\times\frac{19}{20}\)

\(B=\frac{19}{10}\)

Vậy \(B=\frac{19}{10}\)

Học tốt # ^-<

dau gach cheo la dau gach ngang

S₁ = 1-2+3-4+....+2017-2018

     = (-1)+(-1)+....+(-1)

     = (-1) x 1009

     =   -1009

S3=2019 nha, mình ko kip viết cách giai

16 tấn hàng ứng vs phân số :

     1 - 1/2 - 1/3 = 1/6  ( vậy 16 tấn hàng ứng vs 1/6 )

Đội xe phải chở số tấn hàng là :

    16 : 1/6 = 96 ( tấn hàng )

                Đ/S : 96 tấn hàng

tui nha

 mình không hiểu đề bài

tim gia tri bieu thuc sao cho la so nho nhat

A=1.1+2.2+3.3+...+99.99+100.100

3A=1.2.3+2.3.(4-1)+...99.100.(101-98)

3A=1.2.3+2.3.4-1.2.3+...+99.100.101-98.99.100

3A=99.100.101=999900

A= 999900:3=333300

\(A=1^2+2^2+3^2+...+100^2\)

\(\Rightarrow A=1\left(2-1\right)+2\left(3-1\right)+3\left(4-1\right)+...+100\left(101-1\right)\)

\(\Rightarrow A=1.2-1+2.3-2+3.4-3+...+100.101-100\)

\(\Rightarrow A=1.2+2.3+...+100.101-\left(1+2+3+...+100\right)\)

\(\Rightarrow A=\frac{100.101.102}{3}-\frac{\left(100+1\right)\left[\left(100-1\right):1+1\right]}{2}=\frac{100.101.102.2}{6}-\frac{101.100.3}{6}\)

\(\Rightarrow A=\frac{100.101\left(102.2-3\right)}{6}\)

\(\left(1-x\right)\left(x^{31}+x^{30}+...+x+1\right)=\left(1-x\right)\left(1+x\right)\left(1+x^2\right).....\left(1+x^{16}\right)\)

VP = 1 - x³²

Đặt  \(A=x^{31}+x^{30}+..+x+1\Leftrightarrow xA=x^{32}+x^{31}+.......+x^2+x\)

VT = \(A-xA=\left(1-x\right)A=1-x^{32}\)= VP   (dpcm)