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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) \(x^2+y^2-2x-2y+2xy+2\)
\(=\left(x+y\right)^2-2\left(x+y\right)+1+1\)
\(=\left(x+y-1\right)^2+1>0\left(\forall x,y\right)\)
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
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: \(\left(x-\dfrac{1}{2}\right)^2\ge0\forall x\Rightarrow\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\left(đpcm\right)\)
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
Ta có : x2 - x + 1
=.\(x^2+2x\frac{1}{2}+\frac{1}{4}+\frac{3}{4}\)
\(=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\)
Mà \(\left(x+\frac{1}{2}\right)^2\ge0\forall x\)
Nên : \(\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\forall x\)
Hay \(\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\forall x\)
Vậy giá trị của biểu thức luôn luôn dương với mọi x
Ta có : x2 - 8x + 17
= x2 - 2.x.4 + 16 + 1
= (x - 4)2 + 1
Mà (x - 4)2 \(\ge0\forall x\)
Nên : (x - 4)2 + 1 \(\ge1\forall x\)
Hay (x - 4)2 + 1 \(>0\forall x\)\(>0\forall x\)
Vậy giá trị của biểu thức luôn luôn dương với mọi x
Mấy câu trên dễ
\(M=4a^2-6a+12\)
\(M=\left(2a\right)^2-2\cdot2a\cdot\frac{3}{2}+\left(\frac{3}{2}\right)^2+\frac{39}{4}\)
\(M=\left(2a-\frac{3}{2}\right)^2+\frac{39}{4}\ge\frac{39}{4}\forall x\left(đpcm\right)\)
1. a) 2x2y - 3xy2 - 6x + 9y = 2x( xy - 3 ) - 3y ( xy - 3) = ( 2x - 3y)(xy - 3)
b) x2 - 2x + 8 = x2 - 2x + 12 - 1 + 9 = ( x - 1 )2 + 32 ( xem lại đề bài )
2. a) ( 2x - 1) 2 - (2x-1)(2x+3) = 5
(2x-1)(2x-1-2x-3) = 5
-4(2x-1) = 5
2x - 1 = -1,25
2x = -0,25
x= -0,125
b) x(x-9 ) = 0
x= 0 hoặc x = 9
c, ko hiểu
3, M = (2a)2 - 2.2a.1,5 + ( 1,5)2 + 9,75
M= ( 2a - 1,5)2 + 9,75
Vì ( 2a - 1,5 )2 \(\ge\)0 \(\forall x\)
\(\Rightarrow\)( 2a - 1,5)2 + 9,75 \(\ge9,75\forall x\)
Vậy biểu thức trên luôn dương
a)
\(A=x^2-4x+18=\left(x^2-4x+4\right)+14=\left(x-2\right)^2+14\ge14>0\)
\(B=x^2-x+2=\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{7}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{7}{4}\ge\dfrac{7}{4}>0\)
\(C=x^2-2xy+2y^2-2y+15\)
\(C=\left(x^2-2xy+y^2\right)+\left(y^2-2y+1\right)+14\)
\(C=\left(x-y\right)^2+\left(y-1\right)^2+14\ge14>0\)
A=(x-3)(x-5)+2=x^2-5x-3x+15+2=x^2-8x+17=x^2-8x+16+1=(x-4)^2+1>0
B=x^2-5x+7=x^2-5/2*2x+(5/2)^2-(5/2)^2+7=(x-5/2)^2+3/4>0
C=x^2-xy+y^2=x^2-1/2*2xy+1/4y^2-1/4y^2+y^2=(x-1/2y)^2+3/4y^2>0
a: Ta có: \(a^2+b^2+c^2=ab+bc+ac\)
\(\Leftrightarrow2a^2+2b^2+2c^2=2ab+2bc+2ac\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(a^2-2ac+c^2\right)+\left(b^2-2bc+c^2\right)=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\)
=>a=b=c
b: ta có: \(x^2+x+1\)
\(=x^2+2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}\)
\(=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\forall x\)
Ta có: \(x^2-x+1\)
\(=x^2-2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}\)
\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\forall x\)
GIÚP MIK VS NHA MN
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