Question

Tìm x nguyên để biểu thức sau có giá trị nguyên  

1) A=(2x^3+x^2+2x+4)/(2x+1) 

2) B=(3x^2-8x+1)/(x-3)

3) C=(x^3+2x^2+5x)/(x^2+4x+4)

Answer 1

1.

\(A=\frac{2x^3+x^2+2x+4}{2x+1}=\frac{x^2(2x+1)+(2x+1)+3}{2x+1}=x^2+1+\frac{3}{2x+1}\)

Với $x$ nguyên, để $A$ nguyên thì $3\vdots 2x+1$

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

$\Rightarrow x\in \left\{0; -1; 1; -2\right\}$

Answer 2

2.

\(B=\frac{3x^2-8x+1}{x-3}=\frac{3x(x-3)+x+1}{x-3}=\frac{3x(x-3)+(x-3)+4}{x-3}=3x+1+\frac{4}{x-3}\)

Với $x$ nguyên, để $B$ nguyên thì $4\vdots x-3$

$\Rightarrow x-3\in \left\{\pm 1; \pm 2; \pm 4\right\}$

$\Rightarrow x\in \left\{2; 4; 5; 1; 7; -1\right\}$

Answer 3

3.

\(C=\frac{x^3+2x^2+5x}{x^2+4x+4}=\frac{x(x^2+4x+4)-2(x^2+4x+4)+9x+8}{x^2+4x+4}\)

\(=x-2+\frac{9x+8}{x^2+4x+4}\)

Với $x$ nguyên, để $C$ nguyên thì:

$9x+8\vdots x^2+4x+4$

$\Rightarrow 9x+8\vdots (x+2)^2$

$\Rightarrow 9x+8\vdots x+2$

$\Rightarrow 9(x+2)-10\vdots x+2$

$\Rightarrow 10\vdots x+2$

$\Rightarrow x+2\in \left\{\pm 1; \pm 2; \pm 5; \pm 10\right\}$

$\Rightarrow x\in \left\{-1; -3; 0; -4; 3; -7; 8; -12\right\}$
Thử lại thấy $x\in \left\{-1; -3; 0; -4; -12\right\}$