1) CMR : \(\dfrac{1.98+2.97+...+97.2+98.1}{1.2+2.3+...+98.99}\)= \(\dfrac{1}{2}\)
Chứng minh rằng :\(\dfrac{1.98+2.97+3.96+...+96.3+97.2+98.1}{1.2+2.3+3.4+...+96.97+97.98+98.99}=\dfrac{1}{2}\)
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=\(\dfrac{1}{2}\)
tính:
\(\dfrac{1.98+2.97+3.96+....+96.3+97.2+98.1}{1.2+2.3+3.4+4.5+.....+98.99}\)
Bải giải
B=
B=1.(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+...+98.99
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+...+98.99
B=50.98.991.2+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=12
CMR:
\(B=\dfrac{1.98+2.97+3.96+...+96.3+97.2+98.1}{1.2+2.3+3.4+...+96.97+97.98+98+99}=\dfrac{1}{2}\)
CMR:
\(B=\dfrac{1.98+2.97+3.96+...+96.3+97.2+98.1}{1.2+2.3+3.4+...+96.97+97.98+98+99}=\dfrac{1}{2}\)
Chứng minh rằng: B = \(\frac{1.98+2.97+3.96+...+96.3+97.2+98.1}{1.2+2.3+3.4+...+96.97+97.98+98.99}=\frac{1}{2}\)
tính
\(\frac{1.98+2.97+3.96+...+96.3+97.2+98.1}{1.2+2.3+3.4+...+98.99}\)
Tính \(A=\frac{1.98+2.97+3.96+...+97.2+98.1}{1.2+2.3+3.4+...+97.98+98.99}\)
Chứng minh rằng \(\frac{1.98+2.97+3.96+...+96.3+97.2+98.1}{1.2+2.3+3.4+...+96.97+97.98+98.99}=\frac{1}{2}\)
Chứng minh rằng \(\frac{1.98+2.97+3.96+...+96.3+97.2+98.1}{1.2+2.3+3.4+...+96.97+97.98+98.99}\)=1/2