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:  x 2  – 6x + 10 =  x 2  – 2.x.3 + 9 + 1 = x - 3 2  + 1

Vì  x - 3 2  ≥ 0 với mọi x nên  x - 3 2  + 1 > 0 mọi x

Vậy  x 2  – 6x + 10 > 0 với mọi x.(đpcm)

Ta có:

\(x^2-x+1\)

\(=x^2-2\cdot\dfrac{1}{2}\cdot x+\dfrac{1}{4}+\dfrac{3}{4}\)

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

Mà: \(\left(x-\dfrac{1}{2}\right)^2\ge0\forall x\) và \(\dfrac{3}{4}>0\)

Nên: \(x^2-x+1>0\)

\(x^2-x+1\)

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

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

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

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

\(x^2-x+1=x^2-2.x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\dfrac{3}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\\ Mà:\left(x-\dfrac{1}{2}\right)^2>0\forall x\in R\\ Vậy:\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\in R\\ Vậy:x^2-x+1>0\forall x\in R\)

b: \(=\left(x-5\right)^2-9y^2\)

\(=\left(x-5-3y\right)\left(x-5+3y\right)\)

Bài 1: 

b: \(=\left(x-5\right)^2-9y^2\)

\(=\left(x-5-3y\right)\left(x-5+3y\right)\)

\(1,\\ a,=3x\left(x-3y\right)\\ b,=\left(x-5\right)^2-9y^2=\left(x-3y-5\right)\left(x+3y-5\right)\\ c,=3x\left(x-y\right)-2\left(x-y\right)=\left(3x-2\right)\left(x-y\right)\\ 2,\\ Sửa:x^2-6x+10=\left(x-3\right)^2+1\ge1>0,\forall x\)

1, =3x (2x -3y)

c, = 3x(x-y) -2(x-y)

= (3x-2)(x-y)

2, Ta có: x² -6x+10= (x-3)² +11

Nhận xét: (x-3)² >= 0 với mọi số thực x

=> (x-3)² +1 >= 1 >0 (đpcm)

c: \(=\left(x+1\right)^2+1>0\forall x\)

Trả lời:

a, \(x^2-6x+11=x^2-6x+9+2=\left(x-3\right)^2+2\ge2\forall x\)

Dấu "=" xảy ra khi x - 3 = 0 <=> x = 3

Vậy GTNN của biểu thức bằng 2 khi x = 3

b, \(-x^2+6x-11=-\left(x^2-6x+11\right)=-\left(x^2-6x+9+2\right)=-\left[\left(x-3\right)^2+2\right]\)

\(=-\left(x-3\right)^2-2\le-2\forall x\)

Dấu "=" xảy ra khi x - 3 = 0 <=> x = 3

Vậy GTLN của biểu thức bằng - 2 khi x = 3

c, \(x^2+2x+2=x^2+2x+1+1=\left(x+1\right)^2+1\ge1>0\forall x\inℤ\)  (đpcm)

Dấu "=" xảy ra khi x + 1 = 0 <=> x = - 1

a) \(x^2-6x+10=\left(x^2-6x+9\right)+1=\left(x-3\right)^2+1\ge1>0\forall x\)

b) \(4x-x^2-5=-\left(x^2-4x+4\right)-1=-\left(x-2\right)^2-1\le-1< 0\forall x\)