E= x2+x+1
<=>E=x2+2.x.\(\dfrac{1}{2}\)+\(\dfrac{1}{4}\)-\(\dfrac{1}{4}\)+1
<=>E=\(\left(x^2+2x\cdot\dfrac{1}{2}+\dfrac{1}{4}\right)\)+\(\dfrac{3}{4}\)
<=>E=\(\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
Do \(\left(x+\dfrac{1}{2}\right)^2\ge0\Rightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\)
=>E>0
\(E=x^2+x+1=x^2+2.x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2-\dfrac{1}{4}+1\)
\(=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
Do \(\left(x+\dfrac{1}{2}\right)^2\ge0\) với mọi x
\(\Rightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\) hay \(E>0\) với mọi x
Vậy biểu thức \(E=x^2+x+1\) luôn dương với mọi x
\(E=x^2+x+1=x^2+2x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2-\left(\dfrac{1}{2}\right)^2+1=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
Có \(\left(x+\dfrac{1}{2}\right)^2\ge0\forall x\&\dfrac{3}{4}>0\Rightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\)
\(\Rightarrow dpcm\)
