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z x 3y 2 3xy 3 xy 1

z=x^2+3xy+y^2 ∂z/∂x = 2x+3y ∂z/∂x|(1,0) = 2+0 =2 ------ ∂z/∂y = 3x +2y ∂z/∂y|(1,0) = 3 +0 =3

z=x³+y³-3xy(x-y) ∂z/∂x=3x²-3y(x-y)-3xy=3x²-6xy+3y² ∂z/∂y=3y²-3x(x-y)+3xy=3y²+6xy-3x² ∂²z/∂x²=6x-6y ∂²z/∂x∂y=6y+6x W...

1=(x²+4y²-3xy)/z≥(4xy-3xy)/z, 故则当xy/z取得最大值时x²=4y²,x=2y,z=x²+4y²-3xy=2y² 2/x+1/y-2/z=2/y-1/y²=1-(1/y-1)²≤1 即当y=1,x=2,z=2时取最大值1

把式中含w的项合并,得 M=xw+2yw+3xy+3zw+4xz+5yz=(x+2y+3z)w+3xy+4xz+5yz 将w=1-x-y-z代入上式,整理得M=(x+2y+3z)-(x^2+2y^2+3z^2) =(x-x^2)+2(y-y^2)+3(z-z^) =1/4-(x-1/2)^2+2/4-2(y-1/2)^2+3/4-(z-1/2)^2 =6/4-(x-1/2)^2-2(y-1/2)^2-(z-1/...

(1)原式=(2x+5y)(2x-5y); (2)原式=y(x2-1)=y(x+1)(x-1); (3)原式=(2x+y-z)(2x-y+z); (4)原式=(5a-3b)(3a-5b);(5)原式=-3xy(y2-9)=-3xy(y+3x)(y-3x); (6)原式=4a2(x2-4y2 )=4a2(x+2y)(x-2y); (...

(1)原式=3xy(4x2-yz);(2)原式=x(x2-25)=x(x+5)(x-5);(3)原式=4a2-12ab+9b2=(2a-3b)2;(4)原式=m2(m-n)-(m-n)=(m-n)(m2-1)=(m-n)(m+1)(m-1).

∵x 2 -3xy+4y 2 -z=0,∴z=x 2 -3xy+4y 2 ,又x,y,z均为正实数,∴ xy z = xy x 2 -3xy+ 4y 2 = 1 x y + 4y x -3 ≤ 1 2 x y × 4y x -3 =1(当且仅当x=2y时劝=”),∴ ( xy z ) max =1,此时,x=2y.∴z=x 2 -3xy+4y 2 =(2y) 2 -3×2y×y+4y 2 =2y...

由正实数x,y,z满足x2-3xy+4y2-z=0,∴z=x2-3xy+4y2.∴xyz=xyx2?3xy+4y2=1xy+4yx?3≤12xy?4yx?3=1,当且仅当x=2y>0时取等号,此时z=2y2.∴2x+1y-2z=22y+1y?22y2=?(1y?1)2+1≤1,当且仅当y=1时取等号,即2x+1y-2z的最大值是1.故答案为1.

解:已知x=-2,y=-3,z=1, 得3xy-[2xy-(2xyz-xz)-4xz]-xyz =3xy-2xy+(2xyz-xz)+4xz-xyz =xy+2xyz-xz+4xz-xyz =xy+xyz+3xz =x×(y+yz+3z) =-2×[-3+(-3)×1+3×1] =-2×(-3-3+3) =-2×(-3) =6 即已知x-2,y=-3,z=1,则3xy-[2xy-(2xyz-xz)...

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