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

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.`

Ta có: \(2x^2+4y^2+4xy-6x+10\)\(=x^2+4xy+4y^2+x^2-6x+9+1\)\(=\left(x+2y\right)^2+\left(x-3\right)^2+1\)

Vì \(\left(x+2y\right)^2\ge0;\left(x-3\right)^2\ge0\)\(\Rightarrow\left(x+2y\right)^2+\left(x-3\right)^2\ge0\)\(\Leftrightarrow\left(x+2y\right)^2+\left(x-3\right)^2+1\ge1>0\)\(2x^2+4y^2+4xy-6x+10>0\left(đpcm\right)\)

\(\left(x+y\right)^2\ge4xy\)

\(\Leftrightarrow x^2+2xy+y^2\ge4xy\)

\(\Leftrightarrow x^2-2xy+y^2\ge0\)

\(\Leftrightarrow\left(x-y\right)^2\ge0\) (luôn đúng với \(\forall x,y\))

-Vậy BĐT đã được c/m.

-Dấu "=" xảy ra khi \(x=y\)

ta co

vt (x+y)²=x2+y2+2xy

=x2-2xy+y2+4xy≥ 4xy (dpcm)

ta có 4x² + 4y² + 6x + 3 ≥ 4xy

<=> (x² - 4xy + 4y²) + 3(x² + 2x + 1) ≥ 0

<=> (x - 2y)² + 3(x +1)² ≥ 0 (luôn đúng với mọi x,y

vậy với mọi x,y ta có 4x² + 4y² +6x + 3 ≥ 4xy

Theo bài ra :4x²+4y²+6x+3≥4xy

⇔4x²+4y²+6x +3 -4xy≥0 ⇔ [x²-4xy+(2y)²] +3x²+6x+3≥0

⇔(x-2y)²+3(x²+2x+1)≥0  ⇔ (x-2y)² +3(x+1)²  ≥0 ,∀ x,y

vậy 4x²+4y²+6x+3≥4xy

$4x^2+4y^2+4xy>6y-4\left(1\right)$

$\iff 4x^2+4y^2+4xy-6y+4>0\left(2\right)$

$\iff 4x^2+4xy+y^2+3y^2-6y+3+1>0$

$\iff {(2x+y)}^2+3(y^2-2y+1)+1>0$

$\iff {(2x+y)}^2+3{(y-1)}^2+1>0$

$+){(}^{}$

a) \(-\left(x^2-6x+10\right)=-\left(x^2-6x+9+1\right)=-\left[\left(x-3\right)^2+1\right]\le-1< 0\forall x\)

BĐT đúng

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

BĐT đúng

c)Dấu "=" ko xảy ra???

\(=\left(4x^2+2.2x.y+y^2\right)+2\left(2x+y\right)+1+2\)

\(=\left(2x+y\right)^2+2.\left(2x+y\right).1+1+1\)

\(=\left(2x+y+1\right)^2+1\ge1>0\) (đpcm)

a. −x² + 6x - 10

= x² − 6x) − 10

= x² − 2.x.3 + 3² − 9) − 10

= x − 3)² + 9 − 10

= x − 3)² −1

Vì x − 3)² ≥ 0 ∀ x ⇒ x − 3)² ≤ 0 ⇒ x − 3)² −1 ≤ −1

Vậy x − 3)² −1 < 0 ⇒ −x² + 6x - 10 luôn âm với mọi x

b. x² + x + 1

= x² + 2.x.\(\frac{1}{2}\)+ (\(\frac{1}{2}\))²

)² + \(\frac{3}{4}\)

Vì (x + \(\frac{1}{2}\))² ≥ 0 ∀ x ⇒ (x + \(\frac{1}{2}\))² + \(\frac{3}{4}\) ≥ \(\frac{3}{4}\) ∀ x

Vậy (x + \(\frac{1}{2}\))² + \(\frac{3}{4}\) ≥ 0 hay x² + x + 1 > 0 ∀ x.

a) Xét \(x^2-4x+4=\left(x-2\right)^2\ge0\)

<=> \(x^2-4x\ge-4>-5\)

b) \(2x^2+4y^2-4x-4xy+5\)

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

= \(\left(x-2\right)^2+\left(x-2y\right)^2+1\ge1>0\)