a: \(x^2+x+1=x^2+x+\dfrac{1}{4}+\dfrac{3}{4}\)

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

b: \(4y^2+2y+1\)

\(=4\left(y^2+\dfrac{1}{2}y+\dfrac{1}{4}\right)\)

\(=4\left(y^2+2\cdot y\cdot\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{3}{16}\right)\)

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

c: \(-2x^2+6x-10\)

\(=-2\left(x^2-3x+5\right)\)

\(=-2\left(x^2-3x+\dfrac{9}{4}+\dfrac{11}{4}\right)\)

\(=-2\left(x-\dfrac{3}{2}\right)^2-\dfrac{11}{2}< =-\dfrac{11}{2}< 0\forall x\)

`#3107.101107`

a)

`x^2 + x + 1`

`= (x^2 + 2*x*1/2 + 1/4) + 3/4`

`= (x + 1/2)^2 + 3/4`

Vì `(x + 1/2)^2 \ge 0` `AA` `x`

`=> (x + 1/2)^2 + 3/4 \ge 3/4` `AA` `x`

Vậy, `x^2 + x + 1 > 0` `AA` `x`

b)

`4y^2 + 2y + 1`

`= [(2y)^2 + 2*2y*1/2 + 1/4] + 3/4`

`= (2y + 1/2)^2 + 3/4`

Vì `(2y + 1/2)^2 \ge 0` `AA` `y`

`=> (2y + 1/2)^2 + 3/4 \ge 3/4` `AA` `y`

Vậy, `4y^2 + 2y + 1 > 0` `AA` `y`

c)

`-2x^2 + 6x - 10`

`= -(2x^2 - 6x + 10)`

`= -2(x^2 - 3x + 5)`

`= -2[ (x^2 - 2*x*3/2 + 9/4) + 11/4]`

`= -2[ (x - 3/2)^2 + 11/4]`

`= -2(x - 3/2)^2 - 11/2`

Vì `-2(x - 3/2)^2 \le 0` `AA` `x`

`=> -2(x - 3/2)^2 - 11/2 \le 11/2` `AA` `x`

Vậy, `-2x^2 + 6x - 10 < 0` `AA `x.`

\(2x^2+4y^2+4xy-6x+100=\left(x^2+4xy+4y^2\right)+\left(x^2-6x+9\right)+91=\left(x+2y\right)^2+\left(x-3\right)^2+91\ge91>0\)

Có : \(2x^2+3x+2\)

\(\Leftrightarrow\) \(\left(x^2+2x+1^2\right)+\left(x^2+x+1^2\right)\)

\(\Leftrightarrow\) \(\left(x^2+2.x.1+1^2\right)\) + \(\left(x^2+2.x.\frac{1}{2}+\left(\frac{1}{2}\right)^2+\frac{3}{4}\right)\)

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

Vì \(\left(x+1\right)^2\ge0và\left(x+\frac{1}{2}\right)^2\ge0\)

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

Vậy \(2x^2+3x+2>0\left(\forall_x\right)\)

Ta có: \(-x^2+3x-4\)

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

\(=-\left(x^2-2\cdot x\cdot\dfrac{3}{2}+\dfrac{9}{4}+\dfrac{7}{4}\right)\)

\(=-\left(x-\dfrac{3}{2}\right)^2-\dfrac{7}{4}< 0\forall x\)

$-x^2+3x-4\\=-x^2+2.x.\dfrac{3}{2}-\dfrac{9}{4}-\dfrac{7}{4}\\=-(x-\dfrac{3}{2})^2-\dfrac{7}{4}<0$

=> ĐPCM

Ta có: \(-x^2-3x-4\)

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

\(=-\left(x^2+2\cdot x\cdot\dfrac{3}{2}+\dfrac{9}{4}+\dfrac{7}{4}\right)\)

\(=-\left(x+\dfrac{3}{2}\right)^2-\dfrac{7}{4}< 0\forall x\)

Bài 1:

\(a,A=2x^2+2x+1=\left(x^2+2x+1\right)+x^2=\left(x+1\right)^2+x^2\\ Mà:\left(x+1\right)^2\ge0\forall x\in R\\ \Rightarrow\left(x+1\right)^2+x^2>0\forall x\in R\\ Vậy:A>0\forall x\in R\)

2:

a: =-(x^2-3x+1)

=-(x^2-3x+9/4-5/4)

=-(x-3/2)^2+5/4 chưa chắc <0 đâu bạn

b: =-2(x^2+3/2x+3/2)

=-2(x^2+2*x*3/4+9/16+15/16)

=-2(x+3/4)^2-15/8<0 với mọi x

Bài 1:

\(B=4+x^2+x=\left(x^2+x+\dfrac{1}{4}\right)+\dfrac{15}{4}=\left(x+\dfrac{1}{2}\right)^2+\dfrac{15}{4}\ge\dfrac{15}{4}\forall x\in R\\ Vậy:B>0\forall x\in R\)

có  3x^2 + 2x + 4 = 2x^2 + x^2 + 2x +1 +3
                            = 2x^2 +3 + (x+1)^2
mà x^2 >=0 với mọi x
=> 2x^2 >=0 với mọi x
lại có (x+1)^2 >= 0 với mọi x
Suy ra 2x^2 + 3 + (x+1)^2 > 0 với mọi x ( đpcm )

                 \(3x^2+2x+4>0\)

\(\Leftrightarrow\)\(2x^2+x^2+2x+\frac{1}{4}+\frac{15}{4}>0\)

\(\Leftrightarrow\)\(\left(x^2+2x+\frac{1}{4}\right)+2x^2+\frac{15}{4}>0\)

\(\Leftrightarrow\) \(\left(x+\frac{1}{2}\right)^2+2x^2+\frac{15}{4}>0\)

BĐt cuối cùng luôn đúng nên ta có đpcm