a , Ta có \(x^2+x+1=x^2+2x\frac{1}{2}+\left(\frac{1}{2}\right)^2+\)\(\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\) \(\ge\frac{3}{4}>0\left(đpcm\right)\)
b , Ta có : \(4x^2-2x+3\)= \(\left(2x\right)^2-2.2x.1+1^2+2\) = \(\left(2x-1\right)^2+2\ge2>0\left(đpcm\right)\)
c , Ta có \(3x^2+2x+1=x^2-\frac{2x}{3}+\frac{1}{9}+2x^2+\frac{8x}{3}+\frac{8}{9}\)
= \(\left(x-\frac{1}{3}\right)^2+2\left(x^2+\frac{4x}{3}+\frac{4}{9}\right)=\left(x-\frac{1}{3}\right)^2+2\left(x+\frac{2}{3}\right)^2\ge0\)
Vì Dấu "=" không thể xảy ra , do đó \(3x^2+2x+1>0\left(đpcm\right)\)
a) \(A=x^2+x+1=x^2+2.x\cdot\frac{1}{2}+\frac{1}{4}+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0,\forall x\)
b)\(B=4x^2-2x+2=\left(2x\right)^2-2.2x\cdot\frac{1}{2}+\frac{1}{4}+\frac{11}{4}=\left(2x-\frac{1}{2}\right)^2+\frac{11}{4}>0,\forall x\)
c)\(C=3x^2+2x+1=3\left(x^2+\frac{2}{3}x+\frac{1}{3}\right)=3\left(x^2+2x\cdot\frac{1}{3}+\frac{1}{9}+\frac{2}{9}\right)=3\left(x+\frac{1}{3}\right)^2+\frac{2}{3}>0,\forall x\)
tại x = 18; y = 4 
