`A=3/4+8/9+.............+9999/10000`
`=1-1/4+1-1/9+,,,,,,,,,,+1-1/10000`
`=99-(1/4+1/9+.........+1/10000)<99-0=99`
`=>A<99`
Giải:
\(A=\dfrac{3}{4}+\dfrac{8}{9}+\dfrac{15}{16}+...+\dfrac{9999}{10000}\)
\(A=\left(1-\dfrac{1}{4}\right)+\left(1-\dfrac{8}{9}\right)+\left(1-\dfrac{1}{16}\right)+...+\left(1-\dfrac{1}{10000}\right)\)
\(A=99-\left(\dfrac{1}{4}+\dfrac{1}{9}+\dfrac{1}{19}+...+\dfrac{1}{10000}\right)< 99\)
\(\Rightarrow A< 99\left(đpcm\right)\)
Chúc bạn học tốt!
Ta có :A=3/4+8/9+15/16+...+9999/10000
=>A=(1-1/4)+(1-1/9)+(1-1/16)+...+(1-/10000)
=>A=(1-1/2^2)+(1-1/3^2)+(1-1/4^2)+...+(1-1/100^2)
=>A=(1+1+1+....+1)-(1/2^2+1/3^2+1/4^2+...+1/100^2)
=>A=99-(1/2^2+1/3^2+1/4^2+...+1/100^2)<99>
=>A<99>
Ta có B=1/2^2+1/3^2+1/4^2+...+1/100^2
=>B<1>
=>B<1>
=>B<1>
=>A>99-(1-1/100)
=>A>99-1+1/100
=>A>98+1/100>98
=>A>98(2)
Từ (1),(2)=>99>A>98