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Ta có : x2 + 2x + 2
= x2 + 2x + 1 + 1
= (x + 1)2 + 1
Mà : (x + 1)2 \(\ge0\forall x\)
Nên : (x + 1)2 + 1 \(\ge1\forall x\)
Vậy x2 + 2x + 2 luôn dương
-(x2-8x+16)-(y2-4y+4)= -(x-4)2-(y-2)2
Ta có : -(x-4)2<= 0
suy ra: -(x-4)2-(y-2)2<=0 (dpcm)
a) x2 + x + 1 = ( x2 + x + 1/4 ) + 3/4 = ( x + 1/2 )2 + 3/4 ≥ 3/4 > 0 ∀ x ( đpcm )
b) 4x2 - 2x + 1 = 4( x2 - 1/2x + 1/16 ) + 3/4 = 4( x - 1/4 )2 + 3/4 ≥ 3/4 > 0 ∀ x ( đpcm )
c) x4 - 3x2 + 9 (*)
Đặt t = x2
(*) <=> t2 - 3t + 9 = ( t2 - 3t + 9/4 ) + 27/4 = ( t - 3/2 )2 + 27/4 = ( x2 - 3/2 )2 + 27/4 ≥ 27/4 > 0 ∀ x ( đpcm )
d) x2 + y2 - 2x - 4y + 6 = ( x2 - 2x + 1 ) + ( y2 - 4y + 4 ) + 1 = ( x - 1 )2 + ( y - 2 )2 + 1 ≥ 1 > 0 ∀ x, y ( đpcm )
e) x2 + y2 - 2x - 2y + 2xy + 2 = ( x2 + 2xy + y2 - 2x - 2y + 1 ) + 1
= [ ( x2 + 2xy + y2 ) - ( 2x + 2y ) + 1 ] + 1
= [ ( x + y )2 - 2( x + y ) + 12 ] + 1
= ( x + y - 1 )2 + 1 ≥ 1 > 0 ∀ x, y ( đpcm )
a) \(x^2+x+1=\left(x^2+x+\frac{1}{4}\right)+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\left(\forall x\right)\)
b) \(4x^2-2x+1=4\left(x^2-\frac{x}{2}+\frac{1}{16}\right)+\frac{3}{4}=4\left(x-\frac{1}{4}\right)^2+\frac{3}{4}>0\left(\forall x\right)\)
c) \(x^4-3x^2+9=\left(x^4-3x^2+\frac{9}{4}\right)+\frac{27}{4}=\left(x^2-\frac{3}{2}\right)^2+\frac{27}{4}>0\left(\forall x\right)\)
d) \(x^2+y^2-2x-4y+6\)
\(=\left(x^2-2x+1\right)+\left(y^2-4y+4\right)+1\)
\(=\left(x-1\right)^2+\left(y-2\right)^2+1>0\left(\forall x,y\right)\)
e) \(x^2+y^2-2x-2y+2xy+2\)
\(=\left(x+y\right)^2-2\left(x+y\right)+1+1\)
\(=\left(x+y-1\right)^2+1>0\left(\forall x,y\right)\)
Rút gọn
\(\left(2x+1\right)\left(4x^2-3x+1\right)+\left(2x-1\right)\left(4x^2+3x+1\right)\)
\(=8x^3-12x^2+2x+4x^2-3x+1+8x^3+12x^2+2x-4x^2-3x-1\)
\(=16x^3-2x\)
Phân tích đa thức thnahf nhân tử
\(4y^2+16y-x^2-8x\)
\(=\left(4y^2-x^2\right)+\left(16y-8x\right)\)
\(=\left(2y-x\right)\left(2y+x\right)+8\left(2y-x\right)\)
\(=\left(2y-x\right)\left(2y+x+8\right)\)
Chứng minh .............
Có: \(x^2+x+1=\left(x^2+2\cdot x\cdot\frac{1}{2}+\frac{1}{4}\right)+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\)
Vì: \(\left(x+\frac{1}{2}\right)^2\ge0\)
=> \(\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\)
Kết luận......
a) \(x^2 +x +1 = x^2 +x +1/4 +3/4 = (x+1/2)^2 +3/4\)
các câu khác dùng phương pháp tương tự
a) x^2 + x +1 = x^2 + x + 1/4 + 3/4 = ( x+ 1/2)^2 + 3/4
Vì (x+1/2)^2 >= 0 => (x+1/2)^2 + 3/4>=3/4 > 0
b) 4x^2 - 2x + 1 = (2x)^2 - 2x + 1/4 + 3/4 = (2x +1/2)^2 + 3/4
Vì (2x +1/2)^2 >=0 => (2x +1/2)^2 + 3/4 >= 3/4 > 0
c) x^4 -3x^2 + 9 = x^4 - 3x^2 + 9/4 + 25/4 = ( x^2+ 3/2)^2 + 9/4
Vì ( x^2+ 3/2)^2 >= 0 => ( x^2+ 3/2)^2 + 9/4 >=9/4 >0
d) x^2 + y^2 -2x-2y + 2xy +1
= ( x^2 + 2xy + y^2) - 2( x+y) +1
= ( x+y)^2 -2(x+y) +1
= (x +y +1)^2 >=0
g) x^2+y^2+2(x-2y)+6
= (x^2 + 2x +1) + (y^2 -4y+4) +1
= ( x+1)^2 + (y-2)^2 +1
Vì (x+1)^2; (y-2)^2 >= 0 => ( x+1)^2 + (y-2)^2 +1>=1>0
Bài 24: Chứng minh các đa thức sau luôn dương với mọi x,y
a) x2+ 2x +2
b) 4x2 - 12x +11
c) x2 - x +1
a/ \(x^2+2x+2=x^2+2x+1+1=\left(x+1\right)^2+1\)
vì: \(\left(x+1\right)^2\ge0\forall x\Rightarrow\left(x+1\right)^2+1\ge1>0\left(đpcm\right)\)
b/ \(4x^2-12x+11=\left(4x^2-2\cdot2x\cdot3+9\right)+2=\left(2x-3\right)^2+2\)
vì: \(\left(2x-3\right)^2\ge0\forall x\Rightarrow\left(2x-3\right)^2+2\ge2>0\left(đpcm\right)\)
c/ \(x^2-x+1=x^2-2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
Vì: \(\left(x-\dfrac{1}{2}\right)^2\ge0\forall x\Rightarrow\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\left(đpcm\right)\)
\(4x^2+2x+1\)
\(=\left[\left(2x\right)^2+2.2x.\frac{1}{2}+\left(\frac{1}{2}\right)^2\right]-\left(\frac{1}{2}\right)^2+1\)
\(=\left(2x+\frac{1}{2}\right)^2+\frac{3}{4}\)
\(Có:\left(2x+\frac{1}{2}\right)^2\ge0\)\(\text{với mọi x}\)
\(\Rightarrow\left(2x+\frac{1}{2}\right)^2+\frac{3}{4}\ge0+\frac{3}{4}=\frac{3}{4}>0\)\(\text{với mọi x}\)
\(\text{Vậy 4x^2}+2x+1\)\(\text{luôn dương với mọi x}\)
a) 4x2 - 12x + 11=4x2-12x+9+2=(2x-3)2+2
vì (2x-3)2\(\ge\)0
nên (2x-3)2+2 dương với mọi x
=>4x2 - 12x + 11luôn luôn dương với mọi x
b) x2 - 2x + y2 + 4y + 6
=x2-2x+1+y2+4y+4+1
=(x-1)2+(y+2)2+1
vì (x-1)2\(\ge\)0 ; (y+2)2\(\ge\)0
nên (x-1)2+(y+2)2+1 dương với mọi x;y
=>x2 - 2x + y2 + 4y + 6 luôn dương với mọi x;y
a/B=x2+2x+2013