1) \(A=6x^3+15x^2-4x-3=3x^2\left(2x+5\right)-2\left(2x+5\right)+7\)
- Để \(A⋮\left(6x+5\right)\) thì: \(7⋮\left(2x+5\right)\)
\(\Rightarrow2x+5\inƯ\left(7\right)\)
\(\Rightarrow2x+5\in\left\{1;-1;7;-7\right\}\)
| \(2x+5\) | \(1\) | \(-1\) | \(7\) | \(-7\) |
| \(x\) | \(-2\) | \(-3\) | \(1\) | \(-6\) |
2) \(A=10x^4-13x^3-9x^2+x+17=5x^3\left(2x-3\right)+x^2\left(2x-3\right)-3x\left(2x-3\right)-4\left(2x-3\right)+5\)
- Để \(A⋮\left(2x-3\right)\) thì: \(5⋮\left(2x-3\right)\)
\(\Rightarrow2x-3\inƯ\left(5\right)\)
\(\Rightarrow2x-3\in\left\{1;-1;5;-5\right\}\)
| \(2x-3\) | \(1\) | \(-1\) | \(5\) | \(-5\) |
| \(x\) | \(2\) | \(1\) | \(4\) | \(-1\) |
Lời giải:
1.
$A=6x^3+15x^2-4x-3=3x^2(2x+5)-2(2x+5)+7$
$=(2x+5)(3x^2-2)+7$
Để $A$ chia hết cho $2x+5$ thì $7\vdots 2x+5$
$\Rightarrow 2x+5\in\left\{\pm 1; \pm 7\right\}$
$\Rightarrow x\in \left\{-2; -3; 1; -6\right\}$
2.
$A=10x^4-13x^3-9x^2+x+17$
$=5x^3(2x-3)+x^2(2x-3)-3x(2x-3)-4(2x-3)+5$
$=(2x-3)(5x^3+x^2-3x-4)+5$
Để $A$ chia hết cho $2x-3$ thì $5\vdots 2x-3$
$\Rightarrow 2x-3\in\left\{\pm 1; \pm 5\right\}$
$\Rightarrow x\in\left\{2; 1; 4; -1\right\}$
