tính:
\(\dfrac{1.98+2.97+3.96+....+96.3+97.2+98.1}{1.2+2.3+3.4+4.5+.....+98.99}\)
K
Khách
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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=\(\dfrac{1}{2}\)
BT
2
LV
CMR:
\(B=\dfrac{1.98+2.97+3.96+...+96.3+97.2+98.1}{1.2+2.3+3.4+...+96.97+97.98+98+99}=\dfrac{1}{2}\)
0
HT
0
WR
0
LV
0
DT
4
S
22 tháng 7 2016
\(\text{Đặt C = 1.2 + 2.3 + 3.4 + ..... +98.99 }\)
\(\text{ Và A = 1.98 + 2.97 + 3.96 + .... + 98.1 }\)
\(\text{Khi đó : }A=1+\left(1+2\right)+....+\left(1+2+...+98\right)\)
\(=\frac{1.2}{2}+\frac{2.3}{2}+\frac{3.4}{2}+....+\frac{98.99}{2}\)
\(=\frac{1.2+2.3+3.4+....+98.99}{2}=\frac{C}{2}\)
\(\Rightarrow B=\frac{C}{\frac{2}{C}}=\frac{1}{2}\)
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+98).98:21.2+2.3+3.4+...+98.99−1.2+2.3+3.4+...+98.991.2+2.3+3.4+...+98.99100.(1+98).98:21.2+2.3+3.4+...+98.99−1.2+2.3+3.4+...+98.991.2+2.3+3.4+...+98.99
B=50.98.991.2+2.3+3.4+...+98.9950.98.991.2+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 = 50.98.9998.33.100−1=32−1=1250.98.9998.33.100−1=32−1=12