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 \(x=\dfrac{1}{2}\)

\(a,A=x^2-2x+2=\left(x-1\right)^2+1\ge1\)

dấu"=" xảy ra<=>x=1

\(b,B=2x^2-5x+2=2\left(x^2-\dfrac{5}{2}x+1\right)=2\left(x^2-2.\dfrac{5}{4}x+\dfrac{25}{16}-\dfrac{9}{16}\right)\)

\(=2\left[\left(x-\dfrac{5}{4}\right)^2-\dfrac{9}{16}\right]=2\left(x-\dfrac{5}{4}\right)^2-\dfrac{9}{8}\ge-\dfrac{9}{8}\)

dấu"=" xảy ra<=>x=5/4

c,\(C=x^2+2xy+4y^2+3=\left(x+y\right)^2+3\left(y^2+1\right)\ge3\)

dấu"=" xảy ra<=>x=y=0

d,\(D=\left|x-1\right|+|2x-1|=|1-x|+|2x-1|\ge|1-x+2x-1|\)

\(=|x|\ge0\)

dấu"=" xảy ra<=>\(x=0\)

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

Bài 1. 

a) A = -x² - 4x - 2 = -( x² + 4x + 4 ) + 2 = -( x + 2 )² + 2

\(-\left(x+2\right)^2\le0\forall x\Rightarrow-\left(x+2\right)^2+2\le2\)

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

=> MaxA = 2 <=> x = -2

b) B = -2x² - 3x + 5 = -2( x² + 3/2x + 9/16 ) + 49/8 = -2( x + 3/4 )² + 49/8

\(-2\left(x+\frac{3}{4}\right)^2\le0\forall x\Rightarrow-2\left(x+\frac{3}{4}\right)^2+\frac{49}{8}\le\frac{49}{8}\)

Đẳng thức xảy ra <=> x + 3/4 = 0 => x = -3/4

=> MaxB = 49/8 <=> x = -3/4

c) C = ( 2 - x )( x + 4 ) = -x² - 2x + 8 = -( x² + 2x + 1 ) + 9 = -( x + 1 )² + 9

\(-\left(x+1\right)^2\le0\forall x\Rightarrow-\left(x+1\right)^2+9\le9\)

Đẳng thức xảy ra <=> x + 1 = 0 => x = -1

=> MaxC = 9 <=> x = -1

d) D = -8x² + 4xy - y² + 3 = -( 4x² - 4xy + y² ) - 4x² + 3 = -( 2x - y )² - 4x² + 3

\(\hept{\begin{cases}-\left(2x-y\right)^2\le0\forall x,y\\-4x^2\le0\forall x\end{cases}}\Rightarrow-\left(2x-y\right)^2-4x^2+3\le3\)

Đẳng thức xảy ra <=> \(\hept{\begin{cases}2x-y=0\\4x=0\end{cases}}\Rightarrow x=y=0\)

=> MaxD = 3 <=> x = y = 0

Bài 2.

a) A = x² - 2x + 5 = ( x² - 2x + 1 ) + 4 = ( x - 1 )² + 4

\(\left(x-1\right)^2\ge0\forall x\Rightarrow\left(x-1\right)^2+4\ge4\)

Đẳng thức xảy ra <=> x - 1 = 0 => x = 1

=> MinA = 4 <=> x = 1

b) B = x² - x + 1 = ( x² - 2.1/2.x + 1/4 ) + 3/4 = ( x - 1/2 )² + 3/4

\(\left(x-\frac{1}{2}\right)^2\ge0\forall x\Rightarrow\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)

Đẳng thức xảy ra <=> x - 1/2 = 0 => x = 1/2

=> MinB = 3/4 <=> x = 1/2

c) C = ( x - 1 )( x + 2 )( x + 3 )( x + 6 )

C = [( x - 1 )( x + 6 )][( x + 2 )( x + 3)]

C = [ x² + 5x - 6 ][ x² + 5x + 6 ]

C = [ ( x² + 5x ) - 6 ][ ( x² + 5x ) + 6 ]

C = ( x² + 5x )² - 36

\(\left(x^2+5x\right)^2\ge0\forall x\Rightarrow\left(x^2+5x\right)^2-36\ge-36\)

Đẳng thức xảy ra <=> \(x^2+5x=0\Rightarrow x\left(x+5\right)=0\Rightarrow\orbr{\begin{cases}x=0\\x+5=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=-5\end{cases}}\)

=> MinC = -36 <=> x = 0 hoặc x = -5

d) D = x² + 5y² - 2xy + 4y + 3 

D = ( x² - 2xy + y² ) + ( 4y² + 4y + 1 ) + 2

D = ( x - y )² + ( 2y + 1 )² + 2

\(\hept{\begin{cases}\left(x-y\right)^2\ge0\forall x,y\\\left(2y+1\right)^2\ge0\forall y\end{cases}}\Rightarrow\left(x-y\right)^2+\left(2y+1\right)^2+2\ge2\)

Đẳng thức xảy ra <=> \(\hept{\begin{cases}x-y=0\\2y+1=0\end{cases}\Rightarrow}x=y=-\frac{1}{2}\)

=> MinD = 2 <=> x = y = -1/2

Giups mik vs

a: A=(x-1)(x-3)(x²-4x+5)

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

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

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

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

Dấu = xảy ra khi x-2=0

=>x=2

b: \(B=x^2-2xy+2y^2-2y+1\)

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

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

Dấu = xảy ra khi x-y=0 và y-1=0

=>x=y=1

c: \(C=5+\left(1-x\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)\)

\(=-\left(x-1\right)\left(x+6\right)\left(x+2\right)\left(x+3\right)+5\)

\(=-\left(x^2+5x-6\right)\left(x^2+5x+6\right)+5\)

\(=-\left[\left(x^2+5x\right)^2-36\right]+5\)

\(=-\left(x^2+5x\right)^2+36+5\)

\(=-\left(x^2+5x\right)^2+41< =41\)

Dấu = xảy ra khi \(x^2+5x=0\)

=>x(x+5)=0

=>\(\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)

a) -( x-y)² - (x-1)² -2 

GTLN = -2

a,A=x²+5y²-2xy+4y+3

=(x²-2xy+y²)+(4y²+4y+1)+2

=(x-y)²+(2y+1)²+2

Vì \(\left(x-y\right)^2\ge0;\left(2y+1\right)^2\ge0\)

\(\Rightarrow\left(x-y\right)^2+\left(2y+1\right)^2\ge0\)

\(\Rightarrow A=\left(x-y\right)^2+\left(2y+1\right)^2+2\ge2\)

Dấu "=" xảy ra khi x=y=-1/2

Vậy Amin=2 khi x=y=-1/2

b, B=(x²-2x)(x²-2x+2)

Đặt x²-2x+1=t, ta có:

B=(t-1)(t+1)=t²-1=(x²-2x+1)-1=(x-1)²-1 

Vì (x-1)²\(\ge\) 0

=>B=(x-1)²-1 \(\ge\)-1

Dấu "=" xảy ra khi x=1

Vậy Bmin =-1 khi x=1

c, C=(x+1)(x-2)(x-3)(x-6)

=(x+1)(x-6)(x-2)(x-3)

=(x²-6x+x-6)(x²-3x-2x+6)

=(x²-5x-6)(x²-5x+6)

Đặt x²-5x=t, ta có:

C=(t-6)(t+6)=t²-6²=t²-36=(x²-5x)²-36

Vì \(\left(x^2-5x\right)^2\ge0\Rightarrow C=\left(x^2-5x\right)^2-36\ge-36\)

Dấu "=" xảy ra khi x=0 hoặc x=5

Vậy Cmin=-36 khi x=0 hoặc x=5