A = \(\left(x^{9999}-x^9\right)+\left(x^{8888}-x^8\right)+...+\left(x^{1111}-x\right)+\left(x^9+x^8+....+x+1\right)\)

Ta có

\(x^{9999}-x^9=x^9\left(x^{9990}-1\right)\)

Mà \(x^{9990}-1⋮x^{10}-1\)

\(x^{10}-1=\left(x-1\right)\left(x^9+x^8+...+x+1\right)\)

\(\Rightarrow x^{9999}-x^9⋮x^9+x^8+...+x+1\)

CMTT có

\(x^{8888}-x^8;x^{7777}-x^7;...x^{1111}-x\) đều chia hết cho

\(x^9+x^8+...+x+1\)

Mặt khác

\(x^9+x^8+x^7+...+x+1⋮x^9+x^8+x^7+..+x+1\)

\(\Rightarrow A⋮B\left(ĐPCM\right)\)

khó quá khó

9999 x 9999 - 9999 + 9999 * 9999=199950003

8888 x 8888 - 8888 + 8888 * 8888 =78987656

7777 x 7777 - 7777 + 7777 * 7777 =120955681

9999 x 9999 - 9999 + 9999 * 9999=199950003

8888 x 8888 - 8888 + 8888 * 8888 =157984200

7777 x 7777 - 7777 + 7777 * 7777 =120955681

\(x+14⋮7\)

Ta có: \(14⋮7\) => \(x⋮7\) => \(x=7n\) (x < 1200)

\(x-16⋮8\)

Ta có: \(16⋮8\) => \(x⋮8\) => \(x=8n\) (x < 1200)

\(54-x⋮9\)

Ta có \(54⋮9\) => \(x⋮9\) => \(x=9n\) (x < 1200)

a: 450 chia hết cho x

396 chia hết cho x

=>\(x\inƯC\left(450;396\right)\)

=>\(x\inƯ\left(18\right)\)(Vì ƯCLN(450;396)=18)

mà x>12

nên x=18

b: 285+x chia hết cho x

=>285 chia hết cho x(1)

306-x chia hết cho x

=>306 chia hết cho x(2)

Từ (1), (2) suy ra \(x\inƯC\left(285;306\right)\)

=>\(x\inƯ\left(3\right)\)

mà x>=3

nên x=3

c: x chia 8;12;16 đều dư 1

=>x-1 chia hết cho 8;12;16

=>\(x-1\in B\left(48\right)\)

mà 40<x<100

nên x-1=48 hoặc x-1=96

=>x=49 hoặc x=97

Bài 1 : Tìm x, biết :

\(\left(x-2\right)\left(x^2+2x+7\right)+2\left(x^2-4\right)-5\left(x-2\right)=0\)

\(\Rightarrow\left(x-2\right)\left(x^2+2x+7\right)+2\left(x-2\right)\left(x+2\right)-5\left(x-2\right)=0\) \(\Rightarrow\left(x-2\right)\left(x^2+2x+7\right)+\left(x-2\right)\left(2\left(x+2\right)-5\right)=0\)

\(\Rightarrow\left(x-2\right)\left(x^2+2x+7+2\left(x+2\right)-5\right)=0\)

\(\Rightarrow\left(x-2\right)\left(x^2+2x+7+2x+4-5\right)=0\)

\(\Rightarrow\left(x-2\right)\left(x^2+4x+6\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-2=0\\x^2+4x+6=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\\left(x+2\right)^2+2>0\end{matrix}\right.\Rightarrow x=2\)

5) Ta có ( x + 7 + 1 ) chia hết cho ( x+7 )

=> có biểu thức A=(x+7+1) : (x+7)

A= 1- 7 chia hết [(1-7)+ 7]

=> x = (1-7) : [(-6) + 7]

=> x= (-6) : 1

=> x = -6