Question

[BHQV]

Chứng minh rằng đa thức \(\dfrac{x^2+x+1}{-2x^2+2x-2}\) luôn nhận giá trị âm với mọi x

Answer 1

[Nguyễn Lê Phước Thịnh]

Đặt \(A=\dfrac{x^2+x+1}{-2x^2+2x-2}\)

\(x^2+x+1=x^2+2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>=\dfrac{3}{4}>0\forall x\)

\(-2x^2+2x-2\)

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

\(=-2\left(x^2-x+\dfrac{1}{4}+\dfrac{3}{4}\right)\)

\(=-2\left[\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\right]\)

\(=-2\left(x-\dfrac{1}{2}\right)^2-\dfrac{3}{2}< =-\dfrac{3}{2}< 0\forall x\)

Do đó: \(A=\dfrac{x^2+x+1}{-2x^2+2x-2}< 0\forall x\)

Answer 2

[Toru]

\(\dfrac{x^2+x+1}{-2x^2+2x-2}=\dfrac{x^2+x+1}{-2\left(x^2-x+1\right)}\)

Ta thấy:

\(x^2+x+1\\=x^2+2\cdot x\cdot\dfrac12+\left(\dfrac12\right)^2-\left(\dfrac12\right)^2+1\\=\left(x+\dfrac12\right)^2+\dfrac34\)

Vì \(\left(x+\dfrac{1}{2}\right)^2\ge0\forall x\)

\(\Rightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\forall x\)

hay \(x^2+x+1>0\forall x\) (1)

Lại có:

\(x^2-x+1\\=x^2-2\cdot x\cdot\dfrac12+\left(\dfrac12\right)^2-\left(\dfrac12\right)^2+1\\=\left(x-\dfrac12\right)^2+\dfrac34\)

Vì \(\left(x-\dfrac{1}{2}\right)^2\ge0\forall x\)

\(\Rightarrow\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\forall x\)

hay \(x^2-x+1>0\forall x\) (2)

Từ (1) và (2) \(\Rightarrow\dfrac{x^2+x+1}{x^2-x+1}>0\forall x\)

\(\Rightarrow\dfrac{x^2+x+1}{-2\left(x^2-x+1\right)}< 0\forall x\)

hay đa thức \(\dfrac{x^2+x+1}{-2x^2+2x-2}< 0\forall x\)

\(\text{#}Toru\)