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已知:x-y=1,求证:x3-3xy-y3=

解:x^3+3xy+y^3=(x+y)(x^2-xy+y^2)+3xy, =(x^2-xy+y^2)+3xy, =(x+y)^2-3xy+3xy, =1.

已知x+y=1,求x³+y³+3xy的值 x³+y³+3xy =(x+y)(x²-xy+y²)+3xy =x²-xy+y²+3xy =x²+2xy+y² =(x+y)² =1

x^3+y^3+3xy = (x+y)(x^2-xy+y^2)+3xy = x^2-xy+y^2+3xy = x^2+2xy+y^2 = (x+y)^2 = 1 希望你能采纳,不懂可追问。谢谢。

y³+x³-3xy=1 在方程两端分别对 x 求导,得 3y²y'+3x²-3(y+xy')=0 y²y'+x²-y-xy'=0 从而 y'=(y-x²)/(y²-x)

A+B+C=(x3+3x2y-5xy2+6y3-1)+(y3+2xy2+x2y-2x3+2)+(x3-4x2y+3xy2-7y3+1)=(1+1-2)x3+(3+1-4)x2y+(-5+2+3)xy2+(6+1-7)y3-1+2+1=2.故A+B+C的值与x,y无关.

∵x=12+3=2-3,y=12?3=2+3,∴xy=1,x+y=4,∴x3+12xy+y3=12+x3+y3,=12+(x+y)(x2-xy+y2),=12+(x+y)[(x+y)2-3xy],=12+4×(42-3×1),=64.故本题答案为64.

z=x^3+y^3-3xy ∂z/∂x=3x²-3y ∂z/∂y=3y²-3x 驻点:(1,1) (0,0) ∂²z/∂x²=6x ∂²z/∂x∂y=3 ∂²z/∂y²=6y (1,1) A=6 B=3 C=6 B²-AC=9-360 ...

基本原理是:由上述的3个联立式子可以求得x, y, λ. 但是λ只是一个中间参数,不需要具体的求解,所以一般的做法是先将λ消去,只求x, y,的值; 有(1)式(2)式联立消去λ,得x/y=(3x^2-y)/(3y^2-x) ---->x^2-y^2+3xy(x-y)=0; 由此式可得,x-y=0, or x...

证明:(Ⅰ)x3+y3-x2y-xy2=(x-y)(x2-y2)=(x-y)2(x+y)…(2分)∵x>0,y>0,(x-y)2≥0,…(4分)∴x3+y3≥x2y+xy2;…(5分)(Ⅱ)由(Ⅰ)x3+y3≥x2y+xy2;同理z3+y3≥z2y+zy2;x3+z3≥x2z+xz2,…(6分)∴2(x3+y3+z3)≥x2(y+z)+y2(z+x)+z...

(x+y)3-x3-y3+3xy =x³+3x²y+3xy²+y³-x³-y³+3xy =3x²y+3xy²+3xy =3xy(x+y+1)

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