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Question

1. Cho phân thức $P=\frac{5x^2}{x^6+x^5-x^3-5x^2-4x+1}$ Chứng minh rằng có 1 đa thức Q(x) với hệ số nguyên sao cho $Q\left(x_0\right)=P\left(x_0\right)$ với mọi $x_0$ là nghiệm của đa thức \(R\l...

Nguyễn Việt Lâm · Accepted Answer

1.

$x^8-4x^4+1=0\Rightarrow x^8-4x^4=-1$

Đồng thời: $x^8-4x^4+1=0\Leftrightarrow x^8+2x^4+1=6x^4\Leftrightarrow\left(x^4+1\right)^2=6x^4$

$\Leftrightarrow\left(x^4+1\right)^2-x^4=5x^4\Leftrightarrow\left(x^4-x^2+1\right)\left(x^4+x^2+1\right)=5x^4$

$\Leftrightarrow\frac{5x^4}{x^4-x^2+1}=x^4+x^2+1$

Ta có: $P\left(x\right)=\frac{5x^5}{x^9+x^8-x^6-5x^5-4x^4+x^3}=\frac{5x^5}{x^9-x^6-5x^5+x^3-1}=\frac{5x^5}{x\left(x^8-4x^4\right)-x^6-x^5+x^3-1}$

$=\frac{5x^5}{-x-x^6-x^5+x^3-1}=\frac{-5x^5}{x^6+x^5-x^3+x+1}=\frac{-5x^5}{\left(x^2+x+1\right)\left(x^4-x^2+1\right)}$

$=\frac{5x^4}{x^4-x^2+1}.\left(\frac{-x}{x^2+x+1}\right)=\frac{-x\left(x^4+x^2+1\right)}{x^2+x+1}=\frac{-x\left(x^4+2x^2+1-x^2\right)}{x^2+x+1}$

$=\frac{-x\left[\left(x^2+1\right)^2-x^2\right]}{x^2+x+1}=\frac{-x\left(x^2+x+1\right)\left(x^2-x+1\right)}{x^2+x+1}$

$=-x\left(x^2-x+1\right)=-x^3+x^2-x$

Vậy luôn tồn tại $Q\left(x\right)=-x^3+x^2-x$ thỏa mãn điều kiện đề bàiCâu trả lời: 1.

$x^8-4x^4+1=0\Rightarrow x^8-4x^4=-1$

Đồng thời: $x^8-4x^4+1=0\Leftrightarrow x^8+2x^4+1=6x^4\Leftrightarrow\left(x^4+1\right)^2=6x^4$

$\Leftrightarrow\left(x^4+1\right)^2-x^4=5x^4\Leftrightarrow\left(x^4-x^2+1\right)\left(x^4+x^2+1\right)=5x^4$

$\Leftrightarrow\frac{5x^4}{x^4-x^2+1}=x^4+x^2+1$

Ta có: $P\left(x\right)=\frac{5x^5}{x^9+x^8-x^6-5x^5-4x^4+x^3}=\frac{5x^5}{x^9-x^6-5x^5+x^3-1}=\frac{5x^5}{x\left(x^8-4x^4\right)-x^6-x^5+x^3-1}$

$=\frac{5x^5}{-x-x^6-x^5+x^3-1}=\frac{-5x^5}{x^6+x^5-x^3+x+1}=\frac{-5x^5}{\left(x^2+x+1\right)\left(x^4-x^2+1\right)}$

$=\frac{5x^4}{x^4-x^2+1}.\left(\frac{-x}{x^2+x+1}\right)=\frac{-x\left(x^4+x^2+1\right)}{x^2+x+1}=\frac{-x\left(x^4+2x^2+1-x^2\right)}{x^2+x+1}$

$=\frac{-x\left[\left(x^2+1\right)^2-x^2\right]}{x^2+x+1}=\frac{-x\left(x^2+x+1\right)\left(x^2-x+1\right)}{x^2+x+1}$

$=-x\left(x^2-x+1\right)=-x^3+x^2-x$

Vậy luôn tồn tại $Q\left(x\right)=-x^3+x^2-x$ thỏa mãn điều kiện đề bài

Nguyễn Việt Lâm · Answer

2.

$\left(3a+2b\right)\left(3a+2c\right)=16bc$

$\Leftrightarrow9a^2+6ab+6ac+4bc=16bc$

$\Leftrightarrow3a^2+2a\left(b+c\right)=4bc\le\left(b+c\right)^2$

$\Leftrightarrow3+2\frac{b+c}{a}\le\left(\frac{b+c}{a}\right)^2$

$\Leftrightarrow\left(\frac{b+c}{a}\right)^2-\frac{2\left(b+c\right)}{a}-3\ge0$

$\Leftrightarrow\left(\frac{b+c}{a}+1\right)\left(\frac{b+c}{a}-3\right)\ge0$

$\Leftrightarrow\frac{b+c}{a}-3\ge0\Leftrightarrow b+c\ge3a$

b. Đặt $\frac{b+c}{a}=x\ge3$

$VT=\frac{1}{x}+x=\frac{1}{x}+\frac{x}{9}+\frac{8x}{9}\ge2\sqrt{\frac{x}{9x}}+\frac{8}{9}.3=\frac{10}{3}$

Dấu "=" xảy ra khi $x=3$ hay $b+c=3a$Câu trả lời: 2.