Question

Câu 5 nha các bạn và cả câu 6  nữa nhé!!!!

Answer 1

5) Ta có:

a + b + 1 = 111...1 + 444...4 + 1

                  (2n c/s 1)(n c/s 4)

= 111...1000...0 + 111...1 + 111...1 x 4 + 1

 (n c/s 1)(n c/s 0) (n c/s 1)   (n c/s 1)

= 111...1 x 1000...0 + 111...1 + 111...1 x 4 + 1

(n c/s 1)    (n c/s 0)    (n c/s 1) (n c/s 1)

= 111...1 x 1000..05 + 1

 (n c/s 1)   (n-1 c/s 0)

= 111...1 x 3 x 333...35 + 1

  (n c/s 1)      (n-1 c/s 3)

= 333...3 x 333...34 + 333...34

(n c/s 3)   (n-1 c/s 3)(n-1 c/s 3)

= 333...34² là số chính phương (đpcm)

(n-1 c/s 3)

6) (x - 2)⁶ = (x - 2)⁸

=> (x - 2)⁸ - (x - 2)⁶ = 0

=> (x - 2)⁶.[(x - 2)² - 1] = 0

=> \(\left[\begin{array}{nghiempt}\left(x-2\right)^6=0\\\left(x-2\right)^2-1=0\end{array}\right.\)=> \(\left[\begin{array}{nghiempt}x-2=0\\\left(x-2\right)^2=1\end{array}\right.\)=> \(\left[\begin{array}{nghiempt}x-2=0\\x-2=1\\x-2=-1\end{array}\right.\)=> \(\left[\begin{array}{nghiempt}x=2\\x=3\\x=1\end{array}\right.\)

Vậy \(x\in\left\{1;2;3\right\}\)