rút gọn phân số A=\(\dfrac{1.98+1.97+...+98.1}{1.2+2.3+...+98.99}\)
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AL
18 tháng 3 2021
Đặt A=1.98+2.97+3.96+...+96.3+97.2+98.1
B=1.2+2,3+3.4+...+96.97+97.98+98.99
Ta có: A=1+(1+2)+...+(1+2+3+...+97+98)
=\(\dfrac{1.2}{2}+\dfrac{2.3}{2}+...+\dfrac{98.99}{3}\)
=\(\dfrac{1.2+2.3+3.4+4.5+...+98.99}{2}\)=\(\dfrac{B}{2}\)
=>E=\(\dfrac{B}{2}\):2=\(\dfrac{1}{2}\)
OM
0
LN
18 tháng 10 2017
Tử số của E = 1 + ( 1 + 2 ) + ( 1 + 2 + 3 ) + ........ + ( 1 + 2 + 3 + .... + 98 )
= \(\dfrac{1.2}{2}+\dfrac{2.3}{2}+\dfrac{3.4}{2}+......+\dfrac{98.99}{2}\)
\(=\left(1.2+2.3+.........+98.99\right):2\)
\(\Rightarrow E=\dfrac{1}{2}\left(đpcm\right)\)
DT
4
S
22 tháng 7 2016
\(\text{Đặt C = 1.2 + 2.3 + 3.4 + ..... +98.99 }\)
\(\text{ Và A = 1.98 + 2.97 + 3.96 + .... + 98.1 }\)
\(\text{Khi đó : }A=1+\left(1+2\right)+....+\left(1+2+...+98\right)\)
\(=\frac{1.2}{2}+\frac{2.3}{2}+\frac{3.4}{2}+....+\frac{98.99}{2}\)
\(=\frac{1.2+2.3+3.4+....+98.99}{2}=\frac{C}{2}\)
\(\Rightarrow B=\frac{C}{\frac{2}{C}}=\frac{1}{2}\)
NR
1 tháng 1 2019
Bải giải
B=1.98+2.97+3.96+...+98.11.2+2.3+3.4+...+98.99
B=1.(100−2)+2.(100−3)+3.(100−4)+...+98.(100−99)1.2+2.3+3.4+...+98.991.(100−2)+2.(100−3)+3.(100−4)+...+98.(100−99)1.2+2.3+3.4+...+98.99
B=100.(1+2+3+...+98)−(1.2+2.3+3.4+...+98.99)1.2+2.3+3.4+...+98.99100.(1+2+3+...+98)−(1.2+2.3+3.4+...+98.99)1.2+2.3+3.4+...+98.99
B=100.(1
DV
1
O
2
17 tháng 3 2016
B=\(\frac{1.\left(100-2\right)+2.\left(100-3\right)+3.\left(100-4\right)+...+98.\left(100-99\right)}{1.2+2.3+3.4+...+98.99}\)
B=\(\frac{100.\left(1+2+3+...+98\right)-\left(1.2+2.3+3.4+...+98.99\right)}{1.2+2.3+3.4+...+98.99}\)
B=\(\frac{100.\left(1+98\right).98:2}{1.2+2.3+3.4+...+98.99}-\frac{1.2+2.3+3.4+...+98.99}{1.2+2.3+3.4+...+98.99}\)
B=\(\frac{50.98.99}{1.2+2.3+3.4+...+98.99}\)
Đặt M = 1.2+2.3+3.4+....+98.99
=> 3M=3.(1.2+2.3+3.4+...+98.99)
=> 3M = 1.2.3+2.3.(4-1)+...+098.99.(100-97)
3M= 1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+98.99.100-97.98.100
3M=98.99.100
=> M = 98.33.100
=> B = \(\frac{50.98.99}{98.33.100}-1=\frac{3}{2}-1=\frac{1}{2}\)
RM
0
Sửa đề: \(A=\frac{1\cdot98+2\cdot97+\cdots+98\cdot1}{1\cdot2+2\cdot3+\cdots+98\cdot99}\)
Đặt \(B=1\cdot98+2\cdot97+\cdots+98\cdot1\)
\(=2\left(1\cdot98+2\cdot97+\cdots+49\cdot50\right)\)
\(=2\cdot\left\lbrack1\left(99-1\right)+2\left(99-2\right)+\cdots+49\left(99-49\right)\right\rbrack\)
\(=2\cdot\left\lbrack99\left(1+2+\cdots+49\right)-\left(1^2+2^2+\cdots+49^2\right)\right\rbrack\)
\(=2\cdot\left\lbrack99\cdot\frac{49\cdot50}{2}-\frac{49\left(49+1\right)\left(2\cdot49+1\right)}{6}\right\rbrack=2\cdot\left\lbrack99\cdot49\cdot25-49\cdot25\cdot33\right\rbrack\)
\(=2\cdot25\cdot49\cdot33\left(3-1\right)=50\cdot49\cdot33\cdot2=100\cdot49\cdot33\)
Đặt \(C=1\cdot2+2\cdot3+\cdots+98\cdot99\)
\(=1\left(1+1\right)+2\left(2+1\right)+\cdots+98\left(98+1\right)\)
\(=\left(1^2+2^2+\cdots+98^2\right)+\left(1+2+\cdots+98\right)\)
\(=\frac{98\left(98+1\right)\left(2\cdot98+1\right)}{6}+\frac{98\cdot99}{2}=49\cdot33\cdot197+49\cdot99=49\cdot33\cdot\left(197+3\right)=49\cdot33\cdot200\)
Ta có: \(A=\frac{1\cdot98+2\cdot97+\cdots+98\cdot1}{1\cdot2+2\cdot3+\cdots+98\cdot99}\)
\(=\frac{100\cdot49\cdot33}{49\cdot33\cdot200}=\frac{100}{200}=\frac12\)