Câu hỏi của Lê Thanh Nhàn

Question

Tìm GT nguyên của x để A = $\frac{x^4+x^2+x+2}{x^4+3x^3+7x^2+3x+6}$ có giá trị nguyên

Nguyễn Việt Lâm · Accepted Answer

$x^4+x^2+x+2=x^4+\left(x+\frac{1}{2}\right)^2+\frac{7}{4}>0$ $x^4+3x^3+7x^2+3x+6=\left(x^2+\frac{3x}{2}\right)^2+\frac{19}{4}\left(x+\frac{6}{19}\right)^2+\frac{105}{19}>0$ $\Rightarrow A>0$ $2-A=\frac{x^4+6x^3+13x^2+5x+10}{x^4+3x^3+7x^2+3x+6}=\frac{\left(x^2+3x\right)^2+4\left(x+\frac{5}{8}\right)^2+\frac{135}{16}}{x^4+3x^3+7x^2+3x+6}>0$ $\Rightarrow A< 2\Rightarrow0< A< 2$ $\Rightarrow A=1$ $\Rightarrow x^4+3x^3+7x^2+3x+6=x^4+x^2+x+2$ $\Leftrightarrow3x^3+6x^2+2x+4=0$ $\Leftrightarrow\left(x+2\right)\left(3x^2+2\right)=0\Rightarrow x=-2$Câu trả lời: $x^4+x^2+x+2=x^4+\left(x+\frac{1}{2}\right)^2+\frac{7}{4}>0$ $x^4+3x^3+7x^2+3x+6=\left(x^2+\frac{3x}{2}\right)^2+\frac{19}{4}\left(x+\frac{6}{19}\right)^2+\frac{105}{19}>0$ $\Rightarrow A>0$ $2-A=\frac{x^4+6x^3+13x^2+5x+10}{x^4+3x^3+7x^2+3x+6}=\frac{\left(x^2+3x\right)^2+4\left(x+\frac{5}{8}\right)^2+\frac{135}{16}}{x^4+3x^3+7x^2+3x+6}>0$ $\Rightarrow A< 2\Rightarrow0< A< 2$ $\Rightarrow A=1$ $\Rightarrow x^4+3x^3+7x^2+3x+6=x^4+x^2+x+2$ $\Leftrightarrow3x^3+6x^2+2x+4=0$ $\Leftrightarrow\left(x+2\right)\left(3x^2+2\right)=0\Rightarrow x=-2$

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