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=$\frac{1.98+2.97+3.96+...+98.1}{1.2+2.3+3.4+...+}$

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=$\frac{\mathrm{50.98.99}}{1.2+2.3+3.4+...+98.99}$

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 = $\frac{\mathrm{50.98.99}}{\mathrm{98.33.100}}-1=\frac{3}{2}-1=\frac{1}{2}$

\(2\)

or\(\frac{1}{2}\)

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)