Đặt A= x² + 6x + 10

=> A= x² + 2.3x + 3² +1

A = (x+3)² +1 ≥ 1

=> A > 0 với mọi x (đpcm)

x² - 6x + 10 

= x² - 2.x.3 + 3² + 1

= ( x - 3 )² + 1

Vì \(\left(x-3\right)^2\ge0\forall x\)

1 > 0

=> \(\left(x-3\right)^2+1\ge0\forall x\)         ( đpcm )

Study well 

Ta có: x² – 6x + 10 = x² – 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 x² – 6x + 10 > 0 với mọi x.

Ta có: \(x^2-6x+10=x^2-2.x.3+9+1\)

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

Vì: \(\left(x-3\right)^2\ge0\)\(\forall x\)

Nên \(\left(x-3\right)^2+1\ge1>0\)\(\forall x\)

\(\Rightarrowđpcm\)

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!

a: Ta có: \(x^2-8x+20\)

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

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

b: Ta có: \(-x^2+6x-19\)

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

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

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

a, Sửa đề:

-x²-2x-2

=-(x²+2x+2)

=-(x²+2x+1+1)

=-[(x+1)²+1]<0\(\forall\)x

b, -x²-6x-11

=-(x²+6x+11)

=-(x²+2.x.3+3²+2)

=-[(x+3)²+2]<0\(\forall\)x

Đúng tick nha,

a, -x - 2x - 2

= -(x+2x+1)-1

= -(x+1)² -1

Có (x + 1)² ≥0 ⇒- (x + 1) ≤ 0 ⇒ -(x + 1)² - 1≤ -1

Do đó - x - 2x - 2 < 0 ∀ x

b, -x² - 6x - 11

= -(x² + 2.3.x+ 3²)-2

= -(x+3)² - 2

Có (x + 3)² ≥0 ⇒- (x + 3) ≤ 0 ⇒ -(x + 3)² - 2 ≤ -2

Do đó -x² - 6x - 11 <0 ∀ x

a, x²-2x+2>0

⇔(x²-2x+1)+1>0(luôn đúng)

a. x² - 2x + 2 > 0

⇔ (x² - 2x + 1) + 1 > 0

b,6x-x²-10

=-x²+6x-9-1

=-(x-6x+9)-1

=-(x-3)²-1

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

\(A=9x^2-6x+2=\left(3x\right)^2-2.3x+1+1=\left(3x-1\right)^2+1>0\forall x\)

Vậy ta có đpcm 

\(B=x^2-2xy+y^2+1=\left(x-y\right)^2+1>0\forall x;y\)

Vậy ta có đpcm 

Trả lời:

\(A=9x^2-6x+2=\left(3x\right)^2-2.3x.1+1+1=\left(3x-1\right)^2+1\ge1>0\forall x\)

Vậy A > 0 với mọi x 

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

Vậy B > 0 với mọi x;y