CMR: A=1x2-1/2! + 2x3-1/3! +..+ 99x100-1/100! <2
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chứng minh rằng:
1x2-1/2! + 2x3-1/3!+.....+99x100-1/100! <2
Chứng minh B/A thuộc Z
A= 1/1x2+1/2x3+...+1/99x100
B=2017/51+2017/52+...+2017/100
A=1 - 1/2 + 1/2 - 1/3 +...+ 1/99 - 1/100
A=1 - 1/100
A=100/100 - 1/100
A=99/100
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A=1x2+2x3+3x4+...+99x100
B=1^2+2^2+...+99^2+100^2
C=5+55+555+...+555...5(có 100 chữ số 5)
1. chứng minh rằng
1/2!+2/3!+3/4!+....+99/100! <1
1x2-1/2!+2x3-1/3!+3x4-1/4!+...+99x100-1/100!
2.có tồn tại hay k 2 số dương a,b khác nhau sao cho 1/a-1/b=1/a-b
1/2!+1/3!+...+1/100!<1/1*2+1/2*3+1/3*4+...+1/99*100
1-1/100<1
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A= 1/1x2+ 1/2x3 + 1/3x4 +............+ 1/99x100 và 1
\(A=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+....+\dfrac{1}{99\cdot100}\)
\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(A=1+\left(-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}\right)-\dfrac{1}{100}\)
\(A=1+0-\dfrac{1}{100}\)
\(A=1-\dfrac{1}{100}< 1\)
\(\Rightarrow A< 1\)
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A=1/1x2+1/2x3+1/3x4+......+1/99x100
\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
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\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
=\(1-\frac{1}{100}\)
=\(\frac{99}{100}\)
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A = 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/99.100
A = 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + .... + 1/99 - 1/100
A = 1 - 1/100
A = 99/100
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Xem thêm câu trả lời
A= 1/1x2 + 1/2x3 + 1/3x4 + .........+1/99x100
A=1/1x2+1/2x3+...+1/99x100
A=1-1/2+1/2-1/3+1/3-...+1/99-1/00
A=1-1/100
A=99/100
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Tính A= 1/1x2+1/2x3+1/3x4+........+1/98x99+1/99x100
tìm x biết
a, (1/1x2+1/2x3+1/5x4+...+1/99x100) X=1/1x2+2x3+3x4+...+98x99
b, X/1x3+X/3x5+X/5x7+...+X/2013x2015=4/2015
c, X+1/2015+X+2/2016=X+3/2017+X+4/2018
b) \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2013.2015}\)
\(=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2013.2015}\right)\)
\(=\frac{1}{2}\left(\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{2015-2013}{2013.2015}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2013}-\frac{1}{2015}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{2015}\right)=\frac{1007}{2015}\)
Phương trình tương đương với:
\(\frac{1007X}{2015}=\frac{4}{2015}\Leftrightarrow X=\frac{4}{1007}\)
c) \(\frac{x+1}{2015}+\frac{x+2}{2016}=\frac{x+3}{2017}+\frac{x+4}{2018}\)
\(\Leftrightarrow\frac{x+1}{2015}-1+\frac{x+2}{2016}-1=\frac{x+3}{2017}-1+\frac{x+4}{2018}-1\)
\(\Leftrightarrow\frac{x-2014}{2015}+\frac{x-2014}{2016}=\frac{x-2014}{2017}+\frac{x-2014}{2018}\)
\(\Leftrightarrow x-2014=0\)
\(\Leftrightarrow x=2014\)