chứng tỏ rằng giá trị của đa thức: P= -3xy mũ 3 + 5y mũ 2 - 3/2 xy + 2x mũ 2 tay y= -x luôn không âm
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x^2 -6x +10 = x^2 -2.x.3 +3^2 +1 = (x-3)^2 +1
Ma (x-3)^2 >=0 <=> (x-3)^2 +1 >=1>0 (voi moi x)
b) 4x - x^2 -5 = -(x^2 -4x +5) =-[(x^2 -4x +4)+1] = -[(x-2)^2 +1]
Ma (x+2)^2 >=0 <=> (x-2)^2 +1 >=1 <=> -[(x-2)^2 +1] <=-1 => -[(x-2)^2 +1] <0
2) a) P= x^2 -2x +5 = x^2 -2x +1 +4 = (x-1)^2 +4
Ta co: (x-1)^2 >=0 <=> (x-1)^2 +4 >=4
Vay gia tri nho nhat P=4 khi x=1
b) Q= 2x^2 -6x = 2(x^2 -3x) = 2(x^2 - 2.x.3/2 + 9/4 -9/4)= 2[(x-3/2)^2 -9/4]
Ta co: (x-3/2)^2 >=0 <=>(x-3/2)^2 -9/4 >= -9/4 <=> 2[(x-3/2)^2 -9/4] >= -9/2
Vay gia tri nho nhat Q= -9/2 khi x= 3/2
c) M= x^2 +y^2 -x +6y +10 = (x^2 -2.x.1/2 + 1/4) +(y^2 +2.y.3+9)+3/4
= ( x-1/2)^2 + (y+3)^2 +3/4
M>= 3/4
Vay GTNN cua M = 3/4 khi x=1/2 va y=-3
3)a) A= 4x - x^2 +3 = -(x^2 -4x -3) = -( x^2 -4x+4 -7) =-[(x-2)^2 -7]
Ta co: (x-2)^2>=0 <=> (x-2)^2 -7 >=-7 <=> -[(x-2)^2 -7] <=7
Vay GTLN A=7 khi x=2
b) B= x-x^2 = -(x^2 -2.x.1/2+1/4-1/4) = -[(x-1/2)^2 -1/4]
GTLN B= 1/4 khi x=1/2
c) N= 2x - 2x^2 -5 =-2( x^2 -x+5/2) = -2(x^2 - 2.x.1/2 +1/4 +9/4)
= -2[(x-1/2)^2 +9/4]
GTLN N= -9/2 khi x=1/2
a) \(P\left(x\right)=3x^3-2x+2x^2+7x+8-x^4)\)
\(P\left(x\right)=3x^3(-2x+7x)+2x^2+8-x^4)\)
\(P\left(x\right)=3x^3+5x+2x^2+8-x^4)\)
\(P\left(x\right)=-x^4+3x^3+2x^2+5x+8\)
\(Q\left(x\right)=2x^2-3x^3+3x^2-5x^4\)
\(Q\left(x\right)=(2x^2+3x^2)-3x^3-5x^4\)
\(Q\left(x\right)=5x^2-3x^3-5x^4\)
\(Q\left(x\right)=-5x^4-3x^2+5x^2\)
b)
\(P\left(x\right)+Q\left(x\right)=(3x^3-2x+2x^2+7x+8-x^4)+\left(2x^2-3x^3+3x^2-5x^4\right)\)
\(P\left(x\right)+Q\left(x\right)=3x^3-2x+2x^2+7x+8-x^4+2x^2-3x^3+3x^2-5x^4\)
\(P\left(x\right)+Q\left(x\right)=\left(3x^3-3x^3\right)+\left(-2x+7x\right)+\left(2x^2+2x^2+3x^2\right)+8+\left(-x^4-5x^4\right)\)\(P\left(x\right)+Q\left(x\right)=5x+7x^2+8-6x^4\)
Vậy: \(R\left(x\right)\) \(=5x+7x^2+8-6x^4\)
c. \(R\left(x\right)\) \(=5x+7x^2+8-6x^4\)
\(=5x+7x^2+4+4-6x^4\)
\(=\) \((12x-4)^2+4\ge4-6x^4\)
Câu c MIK KHÔNG CHẮC LÀ ĐÚNG
B1 : a, M = x3-3xy(x-y)-y3-x2+2xy-y2
= ( x3-y3)-3xy(x-y) -(x2-2xy+y2)
= (x-y)(x2+xy+y2)-3xy(x-y)-(x-y)2
= (x-y) [(x2+xy+y2-3xy-(x-y)]
= (x-y)[(x2-2xy+y2)-(x-y)
= (x-y)[(x-y)2-(x-y)]
= (x-y)(x-y)(x-y-1)
= (x-y)2(x-y-1)
= 72(7-1) = 49 . 6= 294
N = x2(x+1)-y2(y-1)+xy-3xy(x-y+1)-95
= x3+x2-(y3-y2)+xy-(3x2y-3xy2+3xy)-95
= x3+x2-y3+y2+xy-3x2y+3xy2-3xy-95
= (x3-y3)+(x2-2xy+y2)-(3x2y+y2)-(3x2y-3xy2)-95
=(x-y)(x2+xy+y2)+(x-y)2-3xy(x-y)-95
= (x-y)(x2+xy+y2+x-y-3xy)-95
= (x-y)[(x2-2xy+y2)+(x-y)]-95
= (x-y)[(x-y)2+(x-y)]-95
=(x-y)(x-y)(x-y+1)-95
= (x-y)2(x-y+1)-95
= 72(7+1)-95=297
a)x²−2x−4y²−4ya)x²-2x-4y²-4y
=x²−2x−4y²−4y+2xy−2xy=x²-2x-4y²-4y+2xy-2xy
=(x²−2xy−2x)+(2xy−4y²−4y)=(x²-2xy-2x)+(2xy-4y²-4y)
=x(x−2y−2)+2y(x−2y−2)=x(x-2y-2)+2y(x-2y-2)
=(x+2y)(x−2y−2)=(x+2y)(x-2y-2)
b)x4+2x³−4x−4b)x4+2x³-4x-4
=x4+2x³+2x²−2x²−4x−4=x4+2x³+2x²-2x²-4x-4
=(x4+2x³+2x²)−(2x²+4x+4)=(x4+2x³+2x²)-(2x²+4x+4)
=x²(x²+2x+2)−2(x²+2x+2)=x²(x²+2x+2)-2(x²+2x+2)
=(x²−2)(x²+2x+2)=(x²-2)(x²+2x+2)
c)x³+2x²y−x−2yc)x³+2x²y-x-2y
=x²(x+2y)−(x+2y)=x²(x+2y)-(x+2y)
=(x²−1)(x+2y)=(x²-1)(x+2y)
=(x+1)(x−1)(x+2y)=(x+1)(x-1)(x+2y)
d)3x²−3y²−2(x−y)²d)3x²-3y²-2(x-y)²
=3(x²−y²)−2(x−y)²=3(x²-y²)-2(x-y)²
=3(x+y)(x−y)−2(x−y)²=3(x+y)(x-y)-2(x-y)²
=(x−y)[3(x+y)−2(x−y)]=(x-y)[3(x+y)-2(x-y)]
=(x−y)(3x+3y−2x+2y)=(x-y)(3x+3y-2x+2y)
=(x−y)(x+5y)=(x-y)(x+5y)
e)x³−4x²−9x+36e)x³-4x²-9x+36
=(x³−4x²)−(9x−36)=(x³-4x²)-(9x-36)
=x²(x−4)−9(x−4)=x²(x-4)-9(x-4)
=(x−4)(x²−9)=(x-4)(x²-9)
=(x−4)(x²−3²)=(x-4)(x²-3²)
=(x−4)(x+3)(x−3)=(x-4)(x+3)(x-3)
f)x²−y²−2x−2yf)x²-y²-2x-2y
=(x²−y²)−(2x+2y)=(x²-y²)-(2x+2y)
=(x+y)(x−y)−2(x+y)=(x+y)(x-y)-2(x+y)
=(x+y)(x−y−2)
hok tốt nhé
k đi
a: \(A=\dfrac{2}{3}x^2y\cdot\dfrac{3}{4}x^4y^3=\dfrac{1}{2}x^6y^4\)
\(B=\dfrac{-1}{2}xy^2\cdot4x^5y^2=-2x^6y^4\)
b: \(C=A-B=\dfrac{-3}{2}x^6y^4\)
Bậc là 10
c: A-B nhận được giá trị âm với mọi x,y