a) \(A=x^2-2x+2=\left(x-1\right)^2+1>0\forall x\inℝ\)

b) \(x-x^2-3=-\left(x^2-x+3\right)\)

\(=-\left(x^2-x+\frac{1}{4}+\frac{11}{4}\right)\)

\(=-\left[\left(x-\frac{1}{2}\right)^2+\frac{11}{4}\right]\)

\(=-\left[\left(x-\frac{1}{2}\right)^2\right]-\frac{11}{4}\le\frac{-11}{4}< 0\forall x\inℝ\)

\(x^2-x+1>0\)

Ta có:

\(x^2-x+1\)

=\(\left(x\right)^2-2\left(\frac{1}{2}\right)\left(x\right)+\left(\frac{1}{2}\right)^2-\frac{1}{4}+1\)

=\(\left(x-\frac{1}{2}\right)^2+\frac{3}{4}>0\)\(\forall x\in R\)

a) \(x^2+8x+17=\left(x^2+8x+16\right)+1=\left(x+4\right)^2+1\ge1>0\)

\(x^2-x+1=\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{3}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)

giải giúp mik với

a) \(x^2+8x+17>0\) với mọi x

Ta có: \(x^2+8x+17=x^2+8x+16+1\)

\(=\left(x+4\right)^2+1>0\) với mọi x

Vậy \(x^2+8x+17>0\) với mọi x

b) \(x^2-x+1\ge\dfrac{3}{4}\) với mọi x

Ta có \(x^2-x+1=x^2-2.\dfrac{1}{2}x+\dfrac{1}{4}+\dfrac{3}{4}\)

\(=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\) với mọi x

Vậy \(x^2-x+1\ge\dfrac{3}{4}\) với mọi x

M làm được r

Ko cần nx