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Ta có
Z = 1/1.2.3 + 1/2.3.4 + 1/3.4.5 + ... + 1/98.99.100
2Z = 2/1.2.3 + 2/2.3.4 + 2/3.4.5 + ... + 2/98.99.100
2Z = 1/1.2 - 1/2.3 + 1/2.3 - 1/3.4 + 1/3.4 - 1/4.5 + ... + 1/98.99 - 1/99.100
2Z = 1/1.2 - 1/99.100
2Z = 4949/9900
=> Z = 4949/19800
=> 4949/19800 . x = 49/200
x = 49/200 : 4949/19800
x = 99/101
Vậy x = 99/101
Ủng hộ nha
T/c:A=1/1*2*3+1/2*3*4+1/3*4*5+1/4*5*6+...+1/97*98*99+1/98*99*100
2A=2/1*2*3+2/2*3*4+2/3*4*5+2/4*5*6+...+2/97*98*99+1/98*99*100
2A=(1/1*2-1/2*3)+(1/2*3-1/3*4)+(1/3*4-1/4*5)+.....+(1/97*98-1/98*99)+(1/98*99-1/99*100)
2A=1/2+1/99*100
A=tự tính nha
\(B=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{98.99.100}\)
\(=>2B=\frac{2}{1.2.3}+\frac{2}{2.3.4}+....+\frac{2}{98.99.100}\)
Dễ dàng CM đẳng thức phụ sau : \(\frac{2}{\left(n-1\right)n\left(n+1\right)}=\frac{1}{\left(n-1\right)n}-\frac{1}{n\left(n+1\right)}\)
Áp dụng vào tính 2B,ta có:
\(2B=\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+....+\left(\frac{1}{98.99}-\frac{1}{99.100}\right)\)
\(=\frac{1}{1.2}-\frac{1}{99.100}=\frac{4949}{9900}=>B=\frac{4949}{9900}:2=\frac{4949}{19800}\)
Vậy.....
1/1.2.3 + 1/2.3.4 + .... + 1/98.99.100
= 1/2(1/1.2-1/2.3) + 1/2(1/2.3-1/3.4) + ..... + 1/2(1/98.99-1/99.100)
= 1/2(1/1.2-1/2.3+1/2.3-....+1/98.99-1/99.100)
= 1/2(1/2 - 1/9900)
= 1/2(4950/9900 - 1/9900)
= 1/2. 4949/9900
= 4949/19800
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{98.99.100}-3x=\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+...+\frac{1}{27.28.29.30}\)
\(\Leftrightarrow\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)-3x=\frac{1}{3}.\left(\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+...+\frac{1}{27.28.29}-\frac{1}{28.29.30}\right)\)
\(\Leftrightarrow\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{99.100}\right)-3x=\frac{1}{3}.\left(\frac{1}{1.2.3}-\frac{1}{28.29.30}\right)\)
\(\Leftrightarrow\frac{4949}{19800}-3x=\frac{451}{8120}\)
\(\Leftrightarrow x\approx0,0648\)
=1+\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{1}{2}\) -\(\frac{1}{3}\) -\(\frac{1}{4}\)+\(\frac{1}{3}\) - \(\frac{1}{4}\)-\(\frac{1}{5}\)+.....+\(\frac{1}{99}\)-\(\frac{1}{100}\)-\(\frac{1}{101}\)
=1+\(\frac{1}{101}\)
=\(\frac{102}{101}\)
1/1.2.3 = 1/2 .[1/1.2 - 1 / 2.3]
1/2.3.4 = 1/2[ 1/2- 1/3 ]
...................
1/99.100.101 = 1/2[ 1/99. 100 - 1/100.101]
=> A= 1/2 [ 1/1.2- 1/2.3 + 1/2.3 - 1/3.4 + 1/3.4 - 1/ 4.5 +.........+ 1/99 .100 - 1/100. 101]
A = 1/2 . [1/1.2 -1/100 .101]
A= 1/2 . 5049 /10100 = 5049 / 20200.
Mình nghĩ là vậy đó.
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{19.20.21}\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{19.20}-\frac{1}{20.21}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{20.21}\right)\)
\(=\frac{1}{2}.\frac{209}{420}\)
\(=\frac{209}{840}\)
\(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{19\cdot20\cdot21}\)
\(=\frac{1}{2}\left(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+...+\frac{2}{19\cdot20\cdot21}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+...+\frac{1}{19\cdot20}-\frac{1}{20\cdot21}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{19\cdot21}\right)\)
bn tự lm tp
A=1/2 *(1/1*2-1/2*3+1/2*3-1/3*4+........+1/98*99-1/99*100)
=1/2*(1/2-1/99*100)
=1/2*(4950-1/9900)
=4950/19800
\(A=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{98\cdot99\cdot100}\)
\(A=\frac{1}{2}\left[\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+...+\frac{2}{98\cdot99\cdot100}\right]\)
\(A=\frac{1}{2}\left[\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+....+\frac{1}{98\cdot99}-\frac{1}{99\cdot100}\right]\)
\(A=\frac{1}{2}\left[\frac{1}{2}-\frac{1}{99\cdot100}\right]=\frac{1}{2}\cdot\frac{4949}{9900}=\frac{4949}{19800}\)