a) Ta có:

\(x^2+4x+5\)

\(=x^2+2.x.2+4+1\)

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

Vì \(\left(x+2\right)^2\ge0\forall x\)

\(\Rightarrow\left(x+2\right)^2+1>0\forall x\)

\(\Rightarrow x^2+4x+5>0\forall x\)

b) Ta có:

\(x^2-x+1\)

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

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

Vì \(\left(x-\dfrac{1}{2}\right)^2\ge0\forall x\)

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

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

c) Ta có:

\(12x-4x^2-10\)

\(=-\left(4x^2-12x+10\right)\)

\(=-\left[\left(2x\right)^2-2.2x.3+9+1\right]\)

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

Vì \(-\left(2x-3\right)^2\le0\forall x\)

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

\(\Rightarrow12x-4x^2-10< -1\)

a: Ta có: \(x^2-8x+20\)

\(=x^2-8x+16+4\)

\(=\left(x-4\right)^2+4>0\forall x\)

b: Ta có: \(-x^2+6x-19\)

\(=-\left(x^2-6x+19\right)\)

\(=-\left(x^2-6x+9+10\right)\)

\(=-\left(x-3\right)^2-10< 0\forall x\)

1/

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

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

2/

a, \(P=x^2-2x+5=x^2-2x+1+4=\left(x-1\right)^2+4\ge4\)

Dấu "=" xảy ra khi x-1=0 <=> x=1

Vậy Pmax = 4 khi x = 1

b, \(M=x^2+y^2-x+6y+10=\left(x^2-x+\dfrac{1}{4}\right)^2+\left(y^2+6y+9\right)+\dfrac{3}{4}=\left(x-\dfrac{1}{2}\right)^2+\left(y+3\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}x-\dfrac{1}{2}=0\\y+3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=-3\end{matrix}\right.\)

Vậy Mmax = 3/4 khi x = 1/2, y = -3

Phủ định của mệnh đề A là mệnh đề “\(\exists x \in \mathbb{R},{x^2} + 4x + 5 = 0\)”

Phủ định của mệnh đề B là mệnh đề “\(\exists x \in \mathbb{R},{x^2} + x < 1\)”

Phủ định của mệnh đề C là mệnh đề “\(\forall x \in \mathbb{Z},2{x^2} + 3x - 2 \ne 0\)”

Phủ định của mệnh đề D là mệnh đề “\(\forall x \in \mathbb{Z},{x^2} \ge x\)”

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

b: \(x-2\cdot\sqrt{x}\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\)

c: \(=x^2-2\cdot x\cdot\dfrac{1}{2}y+\dfrac{1}{4}y^2+\dfrac{3}{4}y^2=\left(x-\dfrac{1}{2}y\right)^2+\dfrac{3}{4}y^2>0\forall x,y\ne0\)

1. a,\(A=x^2-2x+5=x^2-2.x.1+1^2-1+5\)

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

Do \(\left(x-1\right)^2\ge0\) với \(\forall x\) \((\)dấu "=" xảy ra \(\Leftrightarrow x=1)\)

\(\Rightarrow\left(x-1\right)^2+4\ge4\) hay \(A\ge4\) \((\) dấu "=" xảy ra \(\Leftrightarrow x=1)\)

Vậy Min A=4 tại x=1

b,\(B=2x^2-6x=2\left(x^2-3x\right)\)

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

\(=2.\left[\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{4}\right]\)

\(=2.\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{2}\)

Do \(2.\left(x-\dfrac{3}{2}\right)^2\ge0\) với mọi x (dấu "=" xảy ra <=> x=\(\dfrac{3}{2}\))

\(\Rightarrow2.\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{2}\ge-\dfrac{9}{2}\) hay \(B\ge-\dfrac{9}{2}\)

(dấu "=" xảy ra <=> x=\(\dfrac{3}{2}\))

Vậy Min B = \(-\dfrac{9}{2}\) tại x=\(\dfrac{3}{2}\)

Bài 2

a,\(A=6x-x^2+3=-\left(x^2-6x-3\right)\)

\(=-\left(x^2-2.x.3+3^2-9-3\right)\)

\(=-\left[\left(x-3\right)^2-12\right]\)

\(=-\left(x-3\right)^2+12\)

Do \(-\left(x-3\right)^2\le0\) với mọi x (dấu "=" xảy ra <=> x=3)

\(\Rightarrow-\left(x-3\right)^2+12\le12\) hay \(A\le12\) (dấu "=" xảy ra <=> x=3)

Vậy Max A =12 tại x=3

b,\(B=x-x^2+2=-\left(x^2-x-2\right)\)

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

\(=-\left[\left(x-\dfrac{1}{2}\right)^2-\dfrac{9}{4}\right]\)

\(=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{9}{4}\)

Do \(-\left(x-\dfrac{1}{2}\right)^2\le0\) với mọi x (dấu "=" xảy ra \(\Leftrightarrow x=\dfrac{1}{2}\))

\(\Rightarrow\left(x-\dfrac{1}{2}\right)^2+\dfrac{9}{4}\le\dfrac{9}{4}\) hay \(B\le\dfrac{9}{4}\) (dấu "=" xảy ra \(\Leftrightarrow x=\dfrac{1}{2}\))

Vậy Max B=\(\dfrac{9}{4}\) tại x=\(\dfrac{1}{2}\)

c,\(C=5x-x^2-5=-\left(x^2-5x+5\right)\)

\(=-\left[x^2-2.x.\dfrac{5}{2}+\left(\dfrac{5}{2}\right)^2-\dfrac{25}{4}+5\right]\)

\(=-\left[\left(x-\dfrac{5}{2}\right)^2-\dfrac{5}{4}\right]\)

\(=-\left(x-\dfrac{5}{2}\right)^2+\dfrac{5}{4}\)

Do \(-\left(x-\dfrac{5}{2}\right)^2\le0\) với mọi x (dấu "=" xảy ra <=> x=\(\dfrac{5}{2}\))

\(\Rightarrow-\left(x-\dfrac{5}{2}\right)^2+\dfrac{5}{4}\le\dfrac{5}{4}\) hay \(C\le\dfrac{5}{4}\) (dấu ''='' xảy ra <=> x=\(\dfrac{5}{2}\))

Vậy Max C=\(\dfrac{5}{4}\) tại x=\(\dfrac{5}{2}\)

Mình làm tiếp phần của Dũng Nguyễn nha.

b) \(4x-x^2-5\)

\(=-\left(x^2-4x+5\right)\)

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

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

Vì \(-\left(x-2\right)^2\le0\) với mọi x

\(\Rightarrow-\left(x-2\right)^2-1\le-1\)

\(\Rightarrow-\left(x-2\right)^2-1< 0\) với mọi x

Vậy \(4x-x^2-5< 0\) với mọi x

c) \(x^2-x+1\)

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

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

Vì \(\left(x-\dfrac{1}{2}\right)^2\ge0\)

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

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

Vậy \(x^2-x+1>0\) với mọi x

d) \(-x^2+2x-4\)

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

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

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

Vì \(-\left(x-1\right)^2\le0\) với mọi x

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

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

Vậy \(-x^2+2x-4< 0\) với mọi x

a,\(x^2-6x+10=x^2-2.x.3+3^2-9+10\)

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

Do \(\left(x-3\right)^2\ge0\) với mọi x

\(\Rightarrow\left(x-3\right)^2+1\ge1>0\)(đpcm)

Yukru làm mấy câu còn lại hộ tớ!

\(x^2+4x+8=x^2+2.2x+4+4=\left(x+2\right)^2+4\\ \left(x+2\right)^2\ge0\forall x\\ =>\left(x+2\right)^2+4>4\left(>0\right)\forall x\\ =>x^2+4x+8>0\left(\forall x\right)\)

\(Ta\) \(có:\) \(x^2+4x+8\)

\(=x^2+4x+4+4\)

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

\(mà:\) \(\left(x+2\right)^2\ge0\)

\(\Rightarrow\left(x+2\right)^2+4>0\) \(hay\) \(x^2+4x+8>0\) với mọi x

cũng đc đấy nhưng mình chỉ xem trình các bạn thế nào thôi !

\(B=4x^2+y^2+12x-4xy-6y+16\)

\(=\left(4x^2+y^2+9-4xy-6y+12x\right)+7\)

\(=\left[\left(2x\right)^2+y^2+3^2-2.2x.y-2.y.3+2.2x.3\right]+7\)

\(=\left(2x-y+3\right)^2+7\)

Ta có :

\(\left(2x-y+3\right)^2\ge0\forall x,y\)

\(\Rightarrow\left(2x-y+3\right)^2+7\ge7>0\forall x,y\)

Hay B > 0 với mọi x,y

Ta có : \(B=\left(2x\right)^2-2.2x\left(y-3\right)+\left(y-3\right)^2-\left(y-3\right)^2+y^2-6y+16\)

\(=\left(2x-y+3\right)^2-y^2+6y-9+y^2-6y+16\)

\(=\left(2x-y+3\right)^2+7\)

Vì \(\left(2x-y+3\right)^2\ge0\forall x,y\Rightarrow B\ge7\)

hay B > 0 với mọi x,y

a) \(x^2+x+2=\left(x^2+x+\frac{1}{4}\right)+\frac{7}{4}=\left(x+\frac{1}{2}\right)^2+\frac{7}{4}\ge\frac{7}{4}>0\)đúng \(\forall x\in R\)

b) \(x^2-4x+10=\left(x^2-4x+4\right)+6=\left(x-2\right)^2+6\ge6>0\)đúng \(\forall x\in R\)

c) \(x\left(x-4\right)+10=x^2-4x+10\)(giải như câu b)

d) \(x\left(2-x\right)-4=-\left(x^2-2x+1\right)-3=-\left(x-1\right)^2-3\le-3< 0\)đúng ​\(\forall x\in R\)​

e) \(x^2-5x+2017=\left(x^2-5x+\frac{25}{4}\right)+\frac{8043}{4}=\left(x-\frac{5}{2}\right)^2+\frac{8043}{4}\ge\frac{8043}{4}>0\)đúng \(\forall x\in R\)