a, x2+x+1
=(x2+2.\(\frac{1}{2}\)x+\(\frac{1}{4}\))+1-\(\frac{1}{4}\)
=(x+\(\frac{1}{2}\))2+\(\frac{3}{4}\)
Ta có :\(\left\{{}\begin{matrix}\left(x+\frac{1}{2}\right)^2\ge0với\forall x\\\frac{3}{4}>0\end{matrix}\right.\)
\(\Rightarrow\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\)với \(\forall x\)
b, 2x2+y2+2x+2xy+2
=x2+x2+y2+2x+2xy+1+1
=(x2+2xy+y2)+(x2+2x+1)+1
=(x+y)2+(x+1)2+1
Ta có :
\(\left\{{}\begin{matrix}\left(x+y\right)^2\ge0với\forall x,y\\\left(x+1\right)^2\ge0với\forall x\\1>0\end{matrix}\right.\)
\(\Rightarrow\)(x+y)2+(x+1)2+1>0 với \(\forall\)x,y
a) \(x^2+x+1\)
\(=x^2+2\cdot\frac{1}{2}\cdot x+\frac{1}{4}+\frac{3}{4}\)
\(=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\)
ta có \(\left(x+\frac{1}{2}\right)^2\ge0\) với \(\forall x\)
\(\Rightarrow\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\) với \(\forall x\)
hay \(x^2+x+1>0\) với \(\forall x\)
b)\(2x^2+y^2+2x+2xy+2\)
\(=\left(x^2+2xy+y^2\right)+\left(x^2+2x+1\right)+1\)
\(=\left(x+y\right)^2+\left(x+1\right)^2+1\)
ta có \(\left(x+y\right)^2\ge0\) với \(\forall x\),
\(\left(x+1\right)^2\ge0\) với \(\forall x\)
\(\Rightarrow\left(x+y\right)^2+\left(x+1\right)^2\ge0\) với \(\forall x\)
\(\Rightarrow\left(x+y\right)^2+\left(x+1\right)^2+1>0\) với \(\forall x\)
hay \(2x^2+y^2+2x+2xy+2>0\) với \(\forall x\)
