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\(\frac{3}{4}+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.........+\frac{1}{99.100}\)
\(=\frac{3}{4}+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{3}{4}+1-\frac{1}{100}=\frac{87}{50}\)
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ta có 1+(1+2)+(1+2+3)+...+(1+2+3+...+100)
=4+(1+3).3/2+9+(1+4).4/2+...+(1+100).100/2
=1/2(1.2+2.3+.....+100.101)
=>1/2.100.101.102
con cái dưới thì bằng 99.100.101
=>F=51/99
ngu rua mà ko biet lam
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\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ =1-\dfrac{1}{100}=\dfrac{99}{100}\)
=1/1-1/2+1/2-1/3+1/3-1/4+....+1/99-1/100
=1-1/100
=99/100
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1/1.2 + 1/2.3 + 1/3.4 + ... + 1/99.100
= 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/99 - 1/100
= 1 - 1/100
= 99/100
Máy mình đang lỗi nên không gõ được công thức, xin lỗi bạn nhé! :'(
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Answer:
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)
\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=1-\dfrac{1}{100}\)
\(=\dfrac{100}{100}-\dfrac{1}{100}\)
\(=\dfrac{99}{100}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}=1-\dfrac{1}{100}=\dfrac{99}{100}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}=\frac{99}{100}\)
1/(1.2)+1/(2.3)+1/(3.4)+...+1/(99.100)
=1-1/2+1/2-1-1/3+1/3-1/4+...+1/99-1/100
=1-1/100
=99/100
tôi không chép bài giang ho đai ca đâu nha.
\(\frac{3}{4}+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}=\frac{3}{4}+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{3}{4}+1-\frac{1}{100}=\frac{3}{4}+\frac{99}{100}=\frac{174}{100}\)