- Với \(x=0\Rightarrow144>0\) (đúng)

- Với \(x\ne0\)

\(VT=\left(x-2\right)\left(x-6\right)\left(x+3\right)\left(x+4\right)+57x^2\)

\(=\left(x^2+12-8x\right)\left(x^2+12+7x\right)+57x^2\)

\(=x^2\left[\left(x+\frac{12}{x}-8\right)\left(x+\frac{12}{x}+7\right)+57\right]\)

\(=x^2\left[\left(x+\frac{12}{x}-8\right)^2+15\left(x+\frac{12}{x}-8\right)+57\right]\)

\(=x^2\left[\left(x+\frac{12}{x}-8+\frac{15}{2}\right)^2+\frac{3}{4}\right]>0;\forall x\ne0\)

Vậy...

a) Ta có: \(x^2-x+1=x^2-2\cdot x\cdot\frac{1}{2}+\frac{1}{4}+\frac{3}{4}=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\)

Ta có: \(\left(x-\frac{1}{2}\right)^2\ge0\forall x\)

\(\Rightarrow\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\forall x\)

hay \(x^2-x+1>0\forall x\)(đpcm)

b) Ta có: \(-x^2+2x-4=-\left(x^2-2x+4\right)=-\left(x^2-2x+1+3\right)=-\left(x-1\right)^2-3\)

Ta có: \(\left(x-1\right)^2\ge0\forall x\)

\(\Rightarrow-\left(x-1\right)^2\le0\forall x\)

\(\Rightarrow-\left(x-1\right)^2-3\le-3< 0\forall x\)

hay \(-x^2+2x-4< 0\forall x\)(đpcm)

a) x² – x + 1 

=(x² – x + 1/4 )+3/4

=(x-1/2)²+3/4

ta có (x-1/2)²>=0

(x-1/2)²​+3/4>=​+3/4>0

vậy (x-1/2)²​+3/4>0 với mọi số thực x

b)  -x²+2x -4

= -x²+2x -1-3

=-(x²-2x +1)-3

=-(x-2)²​-3

ta có (x-2)²>=0

=>-(x-2)²=<0

=>-(x-2)²​-3=<​-3<0

vậy -(x-2)²​-3<0 với mọi số thực x

a) Đề sai thì phải.Phải là CM: \(x^2-x+1>0\) với mọi x

Ta có:

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

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

Vì \(\left(x-\frac{1}{2}\right)^2\ge0\) nên \(\left(x-\frac{1}{2}\right)^2+\frac{3}{4}>0\)

Vậy \(x^2-x+1>0\) với mọi \(x\in R\)

b)Ta có:

\(-x^2+2x-4=-\left(x^2-2x+1\right)-3\)

\(=-\left(x-1\right)^2-3\)

Vì \(-\left(x-1\right)^2\le0\) với mọi x nên \(-\left(x-1\right)^2-3< 0\)

Vậy \(-x^2+2x-4< 0\) với mọi \(x\in R\)

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

Ta có: \(4x^2+x+1\)

\(=\left(2x\right)^2+2\cdot2x\cdot\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{15}{16}\)

\(=\left(2x+\dfrac{1}{4}\right)^2+\dfrac{15}{16}>0\forall x\)

a ) Đề sai

b ) \(x^2-x+1=x^2-x+\dfrac{1}{4}+\dfrac{3}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\forall x\left(đpcm\right)\)

c ) \(x-x^2-2=-\left(x^2-x+\dfrac{1}{4}\right)-\dfrac{7}{4}=-\left(x-\dfrac{1}{2}\right)^2-\dfrac{7}{4}\le-\dfrac{7}{4}< 0\forall x\left(đpcm\right)\)