Question

Chứng tỏ nếu a thuộc Z thì A=x^2+x+10>0

Answer 1

Ta có: \(A=x^2+x+10=x^2+x+\dfrac{1}{4}+\dfrac{39}{4}=x^2+\dfrac{1}{2}x+\dfrac{1}{2}x+\dfrac{1}{4}+\dfrac{39}{4}=x\left(x+\dfrac{1}{2}\right)+\dfrac{1}{2}\left(x+\dfrac{1}{2}\right)+\dfrac{39}{4}=\left(x+\dfrac{1}{2}\right)^2+\dfrac{39}{4}\)

Vì \(\left(x+\dfrac{1}{2}\right)^2\ge0\forall x\Rightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{39}{4}>0\)

\(\Rightarrow A=x^2+x+10>0\)

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