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Ta có: \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2017}}\)
=>\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2016}}\)
=>\(A=2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2016}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2017}}\right)\)
\(A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2016}}-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-...-\frac{1}{2^{2017}}\)
\(A=1+\left(\frac{1}{2}-\frac{1}{2}\right)+\left(\frac{1}{2^2}-\frac{1}{2^2}\right)+\left(\frac{1}{2^3}-\frac{1}{2^3}\right)+...+\left(\frac{1}{2^{2016}}-\frac{1}{2^{2016}}\right)-\frac{1}{2^{2017}}\)
\(A=1-\frac{1}{2^{2017}}\)
Vậy: \(A=1-\frac{1}{2^{2017}}\)
2S = 2/1.3+1/3.5+2/5.7+...+2/99.100
2S = 1/1-1/3+1/3-1/5+....+1/99-1/100
2S = 1-1/100
2S = 99/100
S = 99/100:2
S = 99/200
ủng hộ mk nhé
= \(\frac{1}{2}x\left(\frac{1}{3}-\frac{1}{3}+\frac{1}{5}-\frac{1}{5}+\frac{1}{7}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{99}+\frac{1}{100}\right)\)
=\(\frac{1}{2}x\frac{1}{100}=\frac{1}{200}\)
vậy S = \(\frac{1}{200}\)
a) \(ĐKXĐ:\hept{\begin{cases}x\ne0\\x\ne-1\end{cases}}\)
\(M=\left(\frac{x+2}{3x}+\frac{2}{x+1}-3\right):\frac{2-4x}{x+1}-\frac{3x-x^2+1}{3x}\)
\(=\left[\frac{\left(x+2\right)\left(x+1\right)}{3x\left(x+1\right)}+\frac{6x}{3x\left(x+1\right)}-\frac{9x\left(x+1\right)}{3x\left(x+1\right)}\right].\frac{x+1}{2-4x}+\frac{x^2-3x-1}{3x}\)
\(=\left[\frac{x^2+3x+2}{3x\left(x+1\right)}+\frac{6x}{3x\left(x+1\right)}-\frac{9x^2+9x}{3x\left(x+1\right)}\right].\frac{x+1}{2-4x}+\frac{x^2-3x-1}{3x}\)
\(=\frac{x^2+3x+2+6x-9x^2-9x}{3x\left(x+1\right)}.\frac{x+1}{2-4x}+\frac{x^2-3x-1}{3x}\)
\(=\frac{2-8x^2}{3x}.\frac{1}{2\left(1-2x\right)}+\frac{x^2-3x-1}{3x}\)
\(=\frac{2\left(1-4x^2\right)}{3x}.\frac{1}{2\left(1-2x\right)}+\frac{x^2-3x-1}{3x}\)
\(=\frac{2\left(1-2x\right)\left(1+2x\right)}{3x}.\frac{1}{2\left(1-2x\right)}+\frac{x^2-3x-1}{3x}\)
\(=\frac{1+2x}{3x}+\frac{x^2-3x-1}{3x}\)
\(=\frac{1+2x+x^2-3x-1}{3x}=\frac{x^2-x}{3x}=\frac{x\left(x-1\right)}{3x}=\frac{x-1}{3}\)
b) Với \(x=6013\)( thỏa mãn ĐKXĐ )
Thay \(x=6013\)vào biểu thức ta được:
\(M=\frac{6013-1}{3}=\frac{6012}{3}=2004\)
\(S=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\)
=> 2S = \(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\)
=> 2S - S = ( \(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\) ) - ( \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\))
S = 1 - \(\frac{1}{2^{10}}\)
\(S=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{10}}\)
=> \(2S=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}\)
=> \(S=1-\frac{1}{2^{10}}\)
Study well ! >_<
Bài làm:
a) đkxđ: \(x\ne\pm1\)
Ta có:
\(M=\frac{x+1}{x^2-1}-\frac{x^2+2}{x^3-1}-\frac{x+1}{x^2+x+1}\)
\(M=\frac{1}{x-1}-\frac{x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{x+1}{x^2+x+1}\)
\(M=\frac{x^2+x+1-x^2-2-\left(x+1\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(M=\frac{x-1-x^2+1}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(M=\frac{x\left(1-x\right)}{\left(x-1\right)\left(x^2+x+1\right)}=-\frac{x}{x^2+x+1}\)
b) Mà x khác 1
=> x = -2, khi đó:
\(M=-\frac{-2}{4-2+1}=\frac{2}{3}\)
