Lời giải:
Đặt \(\underbrace{111....1}_{n}=a\Rightarrow 9a+1=1\underbrace{00....0}_{n-1}=10^{n}\)
Khi đó:
\(\underbrace{33....3^2}_{n}+\underbrace{5...5}_{n-1}\underbrace{444...4^2}_{n}\)
\(=(\underbrace{333....3}_{n})^2+(\underbrace{55...5}_{n-1}.10^n+\underbrace{4444....4}_{n})^2\)
\(=(\underbrace{333....3}_{n})^2+\left(\frac{\underbrace{55...5}_{n}-5}{10}.10^n+\underbrace{4444....4}_{n}\right)^2\)
\(=(3a)^2+(\frac{5a-5}{10}.(9a+1)+4a)^2\)
\(=(3a)^2+(\frac{9a^2-1}{2})^2=9a^2+\frac{81a^4+1-18a^2}{4}\)
\(=\frac{81a^4+1+18a^2}{4}=\frac{(9a^2+1)^2}{4}=\left(\frac{9a^2+1}{2}\right)^2\) là số chính phương vì \(\frac{9a^2+1}{2}\in\mathbb{Z}\) )
Ta có đpcm.
Ribi Nkok Ngok, Khôi Bùi , Phùng Tuệ Minh, Nguyễn Thành Trương
Nguyen, Nguyễn Ngô Minh Trí, Akai Haruma
Help me!
Đặt ẩn phụ 1111111...11(n) chữ số 1=t ý,giải tiếp :3