CMR:
\(B=\dfrac{1.98+2.97+3.96+...+96.3+97.2+98.1}{1.2+2.3+3.4+...+96.97+97.98+98+99}=\dfrac{1}{2}\)
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CMR:
\(B=\dfrac{1.98+2.97+3.96+...+96.3+97.2+98.1}{1.2+2.3+3.4+...+96.97+97.98+98+99}=\dfrac{1}{2}\)
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=\(\dfrac{1}{2}\)
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
ta có
b2=ac\(\Rightarrow\dfrac{a}{b}=\dfrac{b}{c}\) (1)
c2=bd\(\Rightarrow\dfrac{b}{c}=\dfrac{c}{d}\) (2)
từ(1),(2)\(\Rightarrow\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\)
\(\Rightarrow\dfrac{a^3}{b^3}=\dfrac{b^3}{c^3}=\dfrac{c^3}{d^3}\)
áp dung tính chấ t dăy tỉ số bằng nhau ta có
\(\Rightarrow\dfrac{a^3}{b^3}=\dfrac{b^3}{c^3}=\dfrac{c^3}{d^3}=\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}\) (ĐPCM)