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Sửa đề: 98/97
\(=\dfrac{99}{98}-\dfrac{98}{97}+\dfrac{1}{97}-\dfrac{1}{98}\)
\(=1-1=0\)
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Ta có:(1+2+3+4+5+...+97+98+99)x100/2x50:2500
=[(99+1)x99:2)]x50x50:2500
=4950x2500:2500
=4950x1
=4950
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Cách 2 :
1+2+3+4+....+96+97+98+99
= (1 + 99) + (2 + 98) + ... + (59 + 51) + 50
= 100 + 100 + ... + 100 + 50 (có (99 - 1) : 2 = 49 số hạng 100)
= 100 x 49 + 50
= 4950
Theo công thức ta có :
1+2+3+4+....+96+97+98+99 = \(\frac{99.100}{2}=4950\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Tính nhanh :
100 - 99 + 98 - 97 + 96 - 95 + ...+ 2 -1
=(100-99)+(98 - 97)+(96 - 95)+...+(2-1)
=1+1+1+..,+1(50 số )
=5
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= \(\frac{99}{98}\)- \(\frac{98}{97}\)+ \(\frac{1}{9506}\)
= \(\frac{941094}{9506}\)- \(\frac{931588}{9506}\) + \(\frac{1}{9506}\)
= \(\frac{9506}{9506}\)+ \(\frac{1}{9506}\)
= 1 + \(\frac{1}{9506}\)
= \(\frac{1}{9506}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1+\frac{99}{98}-\frac{98}{97}+\frac{1}{97.98}\)
\(=1+1+\frac{1}{98}-\left(1+\frac{1}{97}\right)+\frac{1}{97}-\frac{1}{98}\)
\(=1+1+\frac{1}{98}-1-\frac{1}{97}+\frac{1}{97}-\frac{1}{98}\)
\(=1+1-1\)
\(=1\)
\(\frac{99}{98}-\frac{98}{97}+\frac{1}{97\times98}=\frac{99}{98}-\frac{98}{97}+\frac{98-97}{97\times98}\)
\(=\frac{99}{98}-\frac{98}{97}+\frac{1}{97}-\frac{1}{98}=\left(\frac{99}{98}-\frac{1}{98}\right)-\left(\frac{98}{97}-\frac{1}{97}\right)\)
\(=1-1=0\)