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\(B=\frac{9}{1\cdot2}-\frac{9}{2\cdot3}-.....-\frac{9}{99\cdot100}\)
\(=\frac{9}{2}-9\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+....+\frac{1}{99\cdot100}\right)\)
\(=\frac{9}{2}-9\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{99}-\frac{1}{100}\right)\)
\(=\frac{9}{2}-9\left(1-\frac{1}{100}\right)\)
\(=\frac{9}{2}-\frac{891}{100}\)
tự tính nốt.
\(B=\frac{9}{1.2}-\frac{9}{2.3}-.....-\frac{9}{99.100}\)
\(B=\frac{9}{1.2}+\frac{-9}{2.3}+\frac{-9}{3.4}+....+\frac{-9}{99.100}\)
\(B=\frac{9}{2}+\left(-9\right)\left(\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{99.100}\right)\)
\(B=\frac{9}{2}+\left(-9\right).\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}\right)\)
\(B=\frac{9}{2}+\left(-9\right).\left(1-\frac{1}{100}\right)\)
\(B=\frac{9}{2}+\left(-9\right).\frac{99}{100}\)
\(B=\frac{9}{2}+\frac{-891}{100}\)
\(B=\frac{450}{100}+\frac{-891}{100}\)
\(B=\frac{-441}{100}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
A=9/1.2+ 9/2.3+ 9/3.4+ .... +9/98.99 + 9/99/100
=9(1- 1/2 + 1/2 -1/3+...+1/99 -1/100)
=9.(1- 1/100)
=9.99/100
=891/100
A=9/1.2+9/2.3+...+9/99.100
A/9=1/1.2+1/2.3+....+1/99.100
A/9=1-1/2+1/2-1/3+....+1/99-1/100
A/9=1+(-1/2+1/2)+(-1/3+1/3)+....+(-1/99+1/99)-1/100
A/9=1-1/100
A/9=99/100
A=99/100.9=891/100
Vậy A=891/100
mik ko biết đúng hay sai mn góp ý giúp mik nha
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{9}{1.2}+\frac{9}{2.3}+\frac{9}{3.4}+...+\frac{9}{99.100}\)
=\(9.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)
=\(9.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
=\(9.\left(\frac{1}{1}-\frac{1}{100}\right)\)
=\(9.\frac{99}{100}\)
=\(\frac{891}{100}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=1-\dfrac{1}{100}=\dfrac{99}{100}\)
b, \(\left(1-\dfrac{1}{100}\right)\left(1-\dfrac{1}{99}\right)...\left(1-\dfrac{1}{2}\right)=\dfrac{99.98...1}{100.99...2}=\dfrac{1}{100}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a/ A= 1-3+5-7+9-11+......+97-99
= -2+(-2)+(-2)+......+(-2)
= (-2).25=-50
b/B=-1-2-3-4-...-100
=-(1+2+3+4+...+100)
=-5050
c/C=1-2+3-4+5-6+......+99-100
= -1+(-1)+(-1)+.............+(-1)
=(-1).50=-50
d/D=1-2-3+4+5-6-7+8+9-....+94-95
= (1-2-3+4)+(5-6-7+8)+.......+(92-93-94+95)
= 0+0+0+...+0=0
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=9\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)=9\left(1-\dfrac{1}{100}\right)=\dfrac{891}{100}\)
B=9(1/1.2-1/2.3-...-1/99.100)
B = \(\frac{9}{1.2}-\frac{9}{2.3}-...-\frac{9}{99.100}\)
B = \(\frac{9}{2}-\left(\frac{9}{2.3}+\frac{9}{3.4}+...+\frac{9}{99.100}\right)\)
B = \(\frac{9}{2}-9.\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)
B = \(\frac{9}{2}-9.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
B = \(\frac{9}{2}-9.\left(\frac{1}{2}-\frac{1}{100}\right)\)
B = \(\frac{9}{2}-9.\frac{49}{100}\)
B = \(\frac{9}{2}-\frac{441}{100}=\frac{9}{100}\)