1/2.3 + 1/3.4 + 1/4.5 +...+ 1/89.90
bài 1:Tính nhanh
\(\frac{1}{3.4}+\frac{1}{4.5}+.....+\frac{1}{89.90}\)
Ta có : \(\frac{1}{3.4}+\frac{1}{4.5}+......+\frac{1}{89.90}\)
\(=\frac{1}{3}+\frac{1}{4}-\frac{1}{4}+.......+\frac{1}{89}-\frac{1}{90}\)
\(=\frac{1}{3}-\frac{1}{90}=\frac{30}{90}-\frac{1}{90}=\frac{29}{90}\)
\(\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{89.90}\)
\(=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{89}-\frac{1}{90}\)
\(=\frac{1}{3}-\frac{1}{90}\)
\(=\frac{29}{90}\)
Đặt \(A=\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{89.90}\)
\(A=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{89}-\frac{1}{90}\)
\(A=\frac{1}{3}-\frac{1}{90}\)
\(A=\frac{29}{90}\)
1/2.3+1/3.4+1/4.5+1/5.6+1/6.7
\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}\\ =\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}\\ =\dfrac{1}{2}-\dfrac{1}{7}\\ =\dfrac{5}{14}\)
1/1.2+1/2.3+1/3.4+1/4.5+...+1/99.100
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}=1-\dfrac{1}{100}=\dfrac{99}{100}\)
1/2.3 + 1/3.4 + 1/4.5 + ... + 1/18.19 + 1/19.20
\(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\cdot\cdot\cdot+\dfrac{1}{18\cdot19}+\dfrac{1}{19\cdot20}\)
\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\cdot\cdot\cdot+\dfrac{1}{18}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{20}\)
\(=\dfrac{1}{2}-\dfrac{1}{20}\)
\(=\dfrac{9}{20}\)
#\(Urushi\)☕
A= 1/2.3 + 1/3.4 + 1/4.5 + .... + 1/2014.2015
A=1/2.3+1/3.4+1/4.5+..+1/2014.2015
A=1/2-1/3+1/3-1/4+1/4-1/5+...+1/2014-1/2015
A=1/2-1/2015
A=2013/4030
tính 1/2.3+1/3.4+1/4.5+......+1/99.100
1/2.3+1/3.4+1/4.5+...+1/99.100
=\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{99}-\frac{1}{100}\)
=\(\frac{1}{2}-\frac{1}{100}=\frac{50}{100}-\frac{1}{100}=\frac{49}{100}\)
=1/2-1/3+1/3-1/4+1/4-1/5+......+1/99-1/100
=1/2-1/100
=49/100
tính 1/2.3+1/3.4+1/4.5+..+1/99.100
1/2*3+1/3*4+...+1/99*100
=1/2-1/3+1/3-1/4+...+1/99-1/100
=50/100-1/100=49/100
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}\)
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)\(=>\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(=>\frac{1}{2}-\frac{1}{100}=>\frac{50}{100}-\frac{1}{100}=\frac{49}{100}\)
A=1/2.3+1/3.4+1/4.5+...+1/99.100
Ta có: 1/2-1/3+1/3-1/4+1/4-1/5+...+1/99-1/100
= 1/2-1/100
= 50/100-1/100
= 49/100
A = 1/2 - 1/3 + 1/3 -1/4 + 1/4 -1/5 + ...+ 1/98 - 1/99 + 1/99 - 1/100
Ta thấy đoạn giữa sẽ bị trừ lẫn nhau nên bằng 0
A = 1/2 - 1/100 = 49/100
tích nha
D=1/2.3+1/3.4+1/4.5+...+1/19.20
\(D=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{19.20}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{19}-\frac{1}{20}\)
\(=\frac{1}{2}-\frac{1}{20}\)
\(=\frac{9}{20}\)
\(D=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{19.20}\)
\(D=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{19}-\frac{1}{20}\)
\(D=\frac{1}{2}-\frac{1}{20}\)
\(D=\frac{9}{20}\)
Vậy : . . .
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