Bài 1:

a: \(A=x^2+2x+4\)

\(=x^2+2x+1+3\)

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

Dấu '=' xảy ra khi x+1=0

=>x=-1

Vậy: \(A_{min}=3\) khi x=-1

b: \(B=x^2-20x+101\)

\(=x^2-20x+100+1\)

\(=\left(x-10\right)^2+1>=1\forall x\)

Dấu '=' xảy ra khi x-10=0

=>x=10

Vậy: \(B_{min}=1\) khi x=10

c: \(C=x^2-2x+y^2+4y+8\)

\(=x^2-2x+1+y^2+4y+4+3\)

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

Dấu '=' xảy ra khi x-1=0 và y+2=0

=>x=1 và y=-2

Vậy: \(C_{min}=3\) khi (x,y)=(1;-2)

Bài 2:

a: \(A=5-8x-x^2\)

\(=-\left(x^2+8x\right)+5\)

\(=-\left(x^2+8x+16-16\right)+5\)

\(=-\left(x+4\right)^2+16+5=-\left(x+4\right)^2+21< =21\forall x\)

Dấu '=' xảy ra khi x+4=0

=>x=-4

b: \(B=x-x^2\)

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

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

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

Dấu '=' xảy ra khi \(x-\dfrac{1}{2}=0\)

=>\(x=\dfrac{1}{2}\)

c: \(C=4x-x^2+3\)

\(=-x^2+4x-4+7\)

\(=-\left(x^2-4x+4\right)+7\)

\(=-\left(x-2\right)^2+7< =7\forall x\)

Dấu '=' xảy ra khi x-2=0

=>x=2

d: \(D=-x^2+6x-11\)

\(=-\left(x^2-6x+11\right)\)

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

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

Dấu '=' xảy ra khi x-3=0

=>x=3

Bài 6:

a) Ta có: \(A=-x^2+6x-11\)

\(=-\left(x^2-6x+11\right)\)

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

Dấu '=' xảy ra khi x=3

b) Ta có: \(B=-x^2-8x+5\)

\(=-\left(x^2+8x-5\right)\)

\(=-\left(x^2+8x+16-21\right)\)

\(=-\left(x+4\right)^2+21\le21\forall x\)

Dấu '=' xảy ra khi x=-4

c) Ta có: \(C=-x^2+4x+1\)

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

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

\(=-\left(x-2\right)^2+5\le5\forall x\)

Dấu '=' xảy ra khi x=2

Bài 7:

a) Ta có: \(x^2-6x+11\)

\(=x^2-6x+9+2\)

\(=\left(x-3\right)^2+2\ge2\forall x\)

Dấu '=' xảy ra khi x=3

a) \(A=x^2-4x+1=\left(x-2\right)^2-3\ge-3\)

\(minA=-3\Leftrightarrow x=2\)

b) \(B=-x^2-8x+5=-\left(x+4\right)^2+21\le21\)

\(maxB=21\Leftrightarrow x=-4\)

c) \(C=2x^2-8x+19=2\left(x-2\right)^2+11\ge11\)

\(minC=11\Leftrightarrow x=2\)

d) \(D=-3x^2-6x+1=-3\left(x+1\right)^2+4\le4\)

\(maxD=4\Leftrightarrow x=-1\)

a) A = (x-2)^2 - 3 >= -3

--> A nhỏ nhất bằng -3

 <=> x = 2

b) B = -(x+4)^2 + 21 <= 21

--> B lớn nhất bằng 21

<=> x = -4

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

Dấu \("="\Leftrightarrow x=\dfrac{1}{2}\)

\(b,=\left(x^2-2x+1\right)+\left(y^2+4y+4\right)+1=\left(x-1\right)^2+\left(y+2\right)^2+1\ge1\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)

\(c,=\left(x^2-2xy+y^2\right)+x^2+1=\left(x-y\right)^2+x^2+1\ge1\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x=y\\x=0\end{matrix}\right.\Leftrightarrow x=y=0\)

Bài 3: 

a) Ta có: \(A=25x^2-20x+7\)

\(=\left(5x\right)^2-2\cdot5x\cdot2+4+3\)

\(=\left(5x-2\right)^2+3>0\forall x\)(đpcm)

d) Ta có: \(D=x^2-2x+2\)

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

\(=\left(x-1\right)^2+1>0\forall x\)(đpcm)

Bài 1: 

a) Ta có: \(A=x^2-2x+5\)

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

\(=\left(x-1\right)^2+4\ge4\forall x\)

Dấu '=' xảy ra khi x=1

b) Ta có: \(B=x^2-x+1\)

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

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

Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)

\(A=15-8x-x^2=-\left(x+4\right)^2+31\)

Vì \(\left(x+4\right)^2\ge0\forall x\)\(\Rightarrow-\left(x+4\right)^2+31\le31\)

Dấu "=" xảy ra \(\Leftrightarrow-\left(x+4\right)^2=0\Leftrightarrow x=-4\)

Vậy maxA = 31 <=> x = - 4

\(B=4x-x^2+2=-\left(x-2\right)^2+6\)

Vì \(\left(x-2\right)^2\ge0\forall x\)\(\Rightarrow-\left(x-2\right)^2+6\le6\)

Dấu "=" xảy ra \(\Leftrightarrow-\left(x-2\right)^2=0\Leftrightarrow x=2\)

Vậy maxB = 6 <=> x = 2

a) \(A=15-8x-x^2=-\left(x^2+8x+16\right)-1\)

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

Dấu "=" xảy ra khi: \(-\left(x+4\right)=0\Rightarrow x=-4\)

b) \(B=4x-x^2+2=-\left(x^2-4x+4\right)+6\)

\(=-\left(x-2\right)^2+6\le6\left(\forall x\right)\)

Dấu "=" xảy ra khi: \(-\left(x-2\right)^2=0\Rightarrow x=2\)

c) Trang nghĩ nên sửa đề nhé:

\(C=-x^2-y^2+4x+4y+2\)

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

\(C=-\left(x-2\right)^2-\left(y-2\right)^2+10\le10\left(\forall x\right)\)

Dấu "=" xảy ra khi: \(\hept{\begin{cases}-\left(x-2\right)^2=0\\-\left(y-2\right)^2=0\end{cases}}\Rightarrow x=y=2\)

A = 15 - 8x - x²

= -( x² + 8x + 16 ) + 31

= -( x + 4 )² + 31 

-( x + 4 )² ≤ 0 ∀ x => -( x + 4 )² + 31 ≤ 31

Đẳng thức xảy ra <=> x + 4 = 0 => x = -4

=> MaxA = 31 <=> x = -4

B = 4x - x² + 2

= -( x² - 4x + 4 ) + 6

= -( x - 2 )² + 6

-( x - 2 )² ≤ 0 ∀ x => -( x - 2 )² + 6 ≤ 6

Đẳng thức xảy ra <=> x - 2 = 0 => x = 2

=> MaxB = 6 <=> x = 2

C = -x² - y² + 4x + 4y + 2 ( -x² mới ra :v )

= -( x² - 4x + 4 ) - ( y² - 4y + 4 ) + 10

= -( x - 2 )² - ( y - 2 )² + 10

\(\hept{\begin{cases}-\left(x-2\right)^2\le0\forall x\\-\left(y-2\right)^2\le0\forall y\end{cases}}\Rightarrow-\left(x-2\right)^2-\left(y-2\right)^2+10\le10\)

Đẳng thức xảy ra <=> \(\hept{\begin{cases}x-2=0\\y-2=0\end{cases}}\Leftrightarrow x=y=2\)

=> MaxC = 10 <=> x = y = 2

a.

\(A=x^2-4x+4+2=\left(x-2\right)^2+2\ge2\)

GTNN của A đạt 2 khi và chỉ khi \(x=2\)

b.

\(B=y^2-2.\dfrac{1}{2}y+\dfrac{1}{4}+\dfrac{3}{4}=\left(y-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)

GTNN của B đạt \(\dfrac{3}{4}\) khi và chỉ khi \(y=\dfrac{1}{2}\)

c.

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

GTNN của C đạt \(\dfrac{3}{4}\) khi và chỉ khi \(\left\{{}\begin{matrix}x=2\\y=\dfrac{1}{2}\end{matrix}\right.\)

a) \(A=x^2-4x+6\)

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

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

Mà: \(\left(x-2\right)^2\ge0\forall x\) nên \(A=\left(x-2\right)^2+2\ge2\forall x\)

Dấu "=" xảy ra:

\(\left(x-2\right)^2+2=2\Leftrightarrow x-2=0\)

\(\Leftrightarrow x=2\)

Vậy: \(A_{min}=2\) khi \(x=2\)

b) \(B=y^2-y+1\)

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

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

Mà: \(\left(y-\dfrac{1}{2}\right)^2\ge\forall x\) nên \(B=\left(y-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\)

Dấu "=" xảy ra:

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

\(\Leftrightarrow y=\dfrac{1}{2}\)

Vậy \(B_{min}=\dfrac{3}{4}\) khi \(y=\dfrac{1}{2}\)

c) \(C=x^2-4x+y^2-y+5\)

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

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

Mà: \(\left\{{}\begin{matrix}\left(x-2\right)^2\ge0\forall x\\\left(y-\dfrac{1}{2}\right)^2\ge0\forall x\end{matrix}\right.\) nên 

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

Dấu "=" xảy ra:

\(\left\{{}\begin{matrix}x-2=0\\y-\dfrac{1}{2}=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=\dfrac{1}{2}\end{matrix}\right.\)

Vậy: \(C_{min}=\dfrac{3}{4}\) khi \(\left\{{}\begin{matrix}x=2\\y=\dfrac{1}{2}\end{matrix}\right.\)

Bài 1:

a: \(M=x^2-10x+3\)

\(=x^2-10x+25-22\)

\(=\left(x^2-10x+25\right)-22\)

\(=\left(x-5\right)^2-22>=-22\forall x\)

Dấu '=' xảy ra khi x-5=0

=>x=5

b: \(N=x^2-x+2\)

\(=x^2-x+\dfrac{1}{4}+\dfrac{7}{4}\)

\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{7}{4}>=\dfrac{7}{4}\forall x\)

Dấu '=' xảy ra khi x-1/2=0

=>x=1/2

c: \(P=3x^2-12x\)

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

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

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

Dấu '=' xảy ra khi x-2=0

=>x=2

3. Tìm giá trị nhỏ nhất của các biểu thứca. A = 4x2  4x 11b. B = (x - 1) (x 2) (x 3) (x 6)c. C = x2 - 2x y2 - 4y 7Ai nha... - Hoc24