Đáp án: $1+999\cdots99^2+0.9999\cdots99^2=( 10^n+ 10^{-n}-1)^2$
Giải thích các bước giải:
Ta có:
$1+999\cdots99^2+0.9999\cdots99^2$
$=1+(10^n-1)^2+(1-10^{-n})^2$
$=1+(10^n)^2-2\cdot 10^n+1 +1-2\cdot 10^{-n}+(10^{-n})^2$
$=((10^n)^2+2+(10^{-n})^2)-(2\cdot 10^n+2\cdot 10^{-n})+1$
$=((10^n)^2+2\cdot 10^n\cdot 10^{-n}+(10^{-n})^2)-2\cdot ( 10^n+ 10^{-n})+1$
$=( 10^n+ 10^{-n})^2-2\cdot ( 10^n+ 10^{-n})+1$
$=( 10^n+ 10^{-n}-1)^2$