Question

[Nguyễn Tùng]

Chứng minh rằng \(1+999...9^2+\left(\frac{999...9}{10^n}\right)^2\) (n chữ số 9) là số chính phương

Answer 1

[Mr Lazy]

\(A=1+99..9^2+0,99..9^2=1+\left(10^n-1\right)^2+\left(\frac{10^n-1}{10^n}\right)^2\)

\(=\frac{10^{2n}+10^{2n}\left(10^n-1\right)^2+\left(10^n-1\right)^2}{10^{2n}}\)

\(=\frac{10^{4n}-2.10^{2n}.10^n+3.10^{2n}-2.10^n+1}{10^{2n}}\)

\(=\frac{10^{4n}+10^{2n}+1-2.10^{2n}.10^n+2.10^{2n}.1-2.10^n.1}{10^{2n}}\)

\(=\frac{\left(10^{2n}-10^n+1\right)^2}{10^{2n}}\)\(=\left(\frac{10^{2n}-10^n+1}{10^n}\right)^2\)

Answer 2

[Mr Lazy]

Thay số vào thấy đề sai