c) x¹⁰ - 10x + 9

= x¹⁰ - x - 9x + 9

= x( x⁹ - 1) - 9( x - 1)

= x( x - 1)( x⁸ + x⁷ + x⁶ +...+ x + 1) - 9( x - 1)

= ( x - 1)[ x( x⁸ + x⁷ + x⁶ +...+ x + 1) - 9]

Do : ( x - 1) chia hết cho ( x- 1)( x - 1)

-->( x - 1)[ x( x⁸ + x⁷ + x⁶ +...+ x + 1) - 9] chia hết cho ( x - 1)²

Hay , x¹⁰ - 10x + 9 chia hết cho ( x - 1)² , đpcm

d) 8x⁹ - 9x⁸ + 1

= 8x⁹ - 8x⁸ - x⁸ + 1

= 8x⁸( x - 1) - ( x⁸ - 1)

= 8x⁸( x - 1) - ( x - 1)( x⁷ + x⁶ +...+ x + 1)

= ( x - 1)[ 8x⁸( - x⁷- x⁶ -...-x - 1) ]

Do : ( x - 1) chia hết cho ( x - 1)( x - 1)

--> ( x - 1)[ 8x⁸( - x⁷- x⁶ -...-x - 1) ] chia hết cho ( x - 1)( x - 1)

Hay , 8x⁹ - 9x⁸ + 1 chia hết cho ( x - 1)² , đpcm

Nhanh Nha

 (x^10-10x+9) chia cho (x^2-2x+1) 
=> (x^10-10x+9) = (x^2-2x+1)*(x^8 + 2x^7 + 3x^6 + 4x^5 + 5x^4 + 6x^3 + 7x^2 + 8x + 9) 
Vậy : (x^10-10x+9) chia hết cho (x^2-2x+1)

5.

$4x+3\vdots x-2$

$\Rightarrow 4(x-2)+11\vdots x-2$

$\Rightarrow 11\vdots x-2$

$\Rightarrow x-2\in \left\{1; -1; 11; -11\right\}$

$\Rightarrow x\in \left\{3; 1; 13; -9\right\}$

6.

$3x+9\vdots x+2$
$\Rightarrow 3(x+2)+3\vdots x+2$
$\Rightarrow 3\vdots x+2$

$\Rightarrow x+2\in \left\{1; -1; 3; -3\right\}$

$\Rightarrow x\in \left\{-1; -3; 1; -5\right\}$

7.

$3x+16\vdots x+1$

$\Rightarrow 3(x+1)+13\vdots x+1$

$\Rightarrow 13\vdots x+1$

$\Rightarrow x+1\in \left\{1; -1; 13; -13\right\}$

$\Rightarrow x\in\left\{0; -2; 12; -14\right\}$

8.

$4x+69\vdots x+5$

$\Rightarrow 4(x+5)+49\vdots x+5$

$\Rightarrow 49\vdots x+5$

$\Rightarrow x+5\in\left\{1; -1; 7; -7; 49; -49\right\}$

$\Rightarrow x\in \left\{-4; -6; 2; -12; 44; -54\right\}$

** Bổ sung điều kiện $x$ là số nguyên.

1. $x+9\vdots x+7$

$\Rightarrow (x+7)+2\vdots x+7$

$\Rightarrow 2\vdots x+7$

$\Rightarrow x+7\in \left\{1; -1; 2; -2\right\}$

$\Rightarrow x\in \left\{-6; -8; -5; -9\right\}$

2. Làm tương tự câu 1

$\Rightarrow 9\vdots x+1$

3. Làm tương tự câu 1

$\Rightarrow 17\vdots x+2$
4. Làm tương tự câu 1

$\Rightarrow 18\vdots x+2$

[toán 8] chứng minh chia hết | - Thím Hải