Question

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

Answer 1

Ta có: \(A=x^2+x+10\)

\(=x^2+2.x.\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{4}+10\)

\(=\left(x+\dfrac{1}{2}\right)^2+\dfrac{39}{4}>0\)

-> ĐPCM.

Answer 2

ta có:

A=x²+x+10

=x² +2x.\(\dfrac{1}{2}\)+\(\dfrac{1}{4}-\dfrac{1}{4}+10\)

=\(\left(x+\dfrac{1}{2}\right)^2+\dfrac{39}{4}>0\)

\(\RightarrowĐPCM\)

Answer 3

Vì \(A=x^2+x+10\)

\(\Leftrightarrow x^2+2x.\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{4}+10\)

\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{39}{4}>0\)

Tương tự ta có: ĐPCM