Question

Chứng minh các đa thức sau luôn có giá trị dương
a/\(9x^2-6x+2\)
b/\(x^2+x+1\)
c/\(2x^2+2x+1\)

Answer 1

a) 9x² - 6x + 2 = (3x)² - 2.3x.1 + 1² + 1 = (3x - 1)² + 1 mà\(\left(3x+1\right)^2\ge0\Rightarrow\left(3x+1\right)^2+1\ge1>0\)

b) x² + x + 1 = x² + 2.x.\(\frac{1}{2}+\left(\frac{1}{2}\right)^2+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\)mà\(\left(x+\frac{1}{2}\right)^2\ge0\Rightarrow\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\)

c) 2x² + 2x + 1 =\(\left(\sqrt{2}x\right)^2+2\sqrt{2}x.\frac{1}{\sqrt{2}}+\left(\frac{1}{\sqrt{2}}\right)^2+\frac{1}{2}=\left(\sqrt{2}x+\frac{1}{\sqrt{2}}\right)^2+\frac{1}{2}\ge\frac{1}{2}>0\)

Answer 2

a)    \(9x^2-6x+2=\left(\left(3x\right)^2-2.3x.1+1\right)+1=\left(3x-1\right)^2+1>0\)

b)   .\(x^2+x+1=\left(\left(x^2\right)+2.x.\frac{1}{2}+\frac{1}{4}\right)-\frac{1}{4}+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\)

c)    \(2x^2+2x+1=x^2+\left(x^2+2x+1\right)=x^2+\left(x+1\right)^2>0\)