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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<=>\(x=0\)
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
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,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
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 \(x=\dfrac{1}{2}\)
c) Ta có: \(C=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)\)
\(=\left(x^2+5x-6\right)\left(x^2+5x+6\right)\)
\(=\left(x^2+5x\right)^2-36\ge-36\forall x\)
Dấu '=' xảy ra khi x(x+5)=0
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
d) Ta có: \(x^2+5y^2-2xy+4y+3\)
\(=\left(x^2-2xy+y^2\right)+\left(4y^2+4y+1\right)+2\)
\(=\left(x-y\right)^2+\left(2y+1\right)^2+2\ge2\forall x,y\)
Dấu '=' xảy ra khi \(x=y=-\dfrac{1}{2}\)