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z=x-y的概率密度公式 (y+z-2x)^2+(z+x-2y)^2+(x+y-2z)^2-3(y-z)^2-3(z-x)^2-3(x-y)^2

火烧 2022-10-11 21:31:28 1142
y+z-2x ^2+ z+x-2y ^2+ x+y-2z ^2-3 y-z ^2-3 z-x ^2-3 x-y ^2 y+z-2x ^2+ z+x-2y ^2+ x+y-2z ^2-3 y-z ^2

(y+z-2x)^2+(z+x-2y)^2+(x+y-2z)^2-3(y-z)^2-3(z-x)^2-3(x-y)^2  

(y+z-2x)^2+(z+x-2y)^2+(x+y-2z)^2-3(y-z)^2-3(z-x)^2-3(x-y)^2

令a=y-z
b=z-x
c=x-y
原式=(b-c)²+(c-a)²+(a-b)²-3a²-3b²-3c²
=b²-2bc+c²+c²-2ac+a ²+a²-2ab+b²-3a²-3b²-3c²
-(a²+b²+c²+2ab+2bc+2ca)
=-(a+b+c)²
=-(y-z+z-x+x-y)²
=0

已知sinα-cosβ=m,sinβ+cosα=n,其中m2+n2≤2,则sin(α-β)=2?(m2+n2)22?(m2+n2)2

将等式sinα-cosβ=m两边平方,
有:sin2α+cos2β-2sinαcosβ=m2;
将等式sinβ+cosα=n两边平方,
有:sin2β+cos2α+2sinβcosα=n2;
将上面两个等式相加,
有2-2(sinαcosβ-sinβcosα)=m2+n2,
故sinαcosβ-sinβcosα=

2?(m2+n2) 2


即sin(α-β)=

2?(m2+n2) 2


故答案为:

2?(m2+n2) 2

z=x-y的概率密度公式 (y+z-2x)^2+(z+x-2y)^2+(x+y-2z)^2-3(y-z)^2-3(z-x)^2-3(x-y)^2

计算:(1)3(y-z)2-(2y+z)(-z+2y)(2)(2m2n-2)2?3m-3n3÷4n-2

(1)3(y-z)2-(2y+z)(-z+2y)
=3(y2-2yz+z2)-(4y2-z2)
=-y2-6yz+4z2;

(2)(2m2n-2)2?3m-3n3÷4n-2
=4m4n-4?3m-3n3÷4n-2
=3m4-3n-4+3-(-2)
=3mn

sin^2θ-cos^2θ-2sinθcosθ/sin^2θ-2sinθcosθ+cos^2θ【提示:sin^2θ-cos^2θ-2sinθcosθ全都是分

上下除以cos²θ
由sinθ/cosθ=tanθ就得到了

已知a=m/2+1,b= m/2+2,c= m/2+3,求a^2+2ab+b^2-2ac-abc+c^2的值

a^2+2ab+b^2-2ac-abc+c^2
=(a+b)^2-2c(a+b)+c^2+2bc-abc
=(a+b-c)^2+bc(2-a)
=(m/2+1+m/2+2-m/2-3)^2+(m/2+2)(m/2+3)(2-m/2-1)
=m^2/4+(m^2/4+5m/2+6)(1-m/2)
=(m^2)/4+(m^2)/4+5m/2+6-(m^3)/8-)(5m^2)/4-3m
=-(m^3)/8-(3m^2)/4-m/2+6

求(2+1)(2^2+1)(2^4+1)。。。。。。(2^32+1)+1

(2+1)(2^2+1)(2^4+1)。。。。。。(2^32+1)+1
=(2-1)(2+1)(2^2+1)(2^4+1)。。。。。。(2^32+1)+1
=2^64-1+1
=2^64

计算(1)-1+2×3; (2) (-3 ) 2 ÷ 3 2 -(-2 ) 3 ; &nb...

(1)原式=-1+2×3
=5;

(2)原式=9×

2 3

+8
=14;

(3)原式=

1 4

-

3 4

=-

1 2

(4)90°-45°58′=44°2′;

(5)38°36′+72.5°=38.6°+72.5°=111.1°.

2m2n2+8mn+8+x(mn+2)2

2m^2n^2+8mn+8+x(mn+2)^2
=2(m^2n^2+4mn+4)+x(mn+2)^2
=2(mn+2)^2+x(mn+2)^2
=(mn+2)^2(x+2)

s=1+2+2^2+2^3+...+2^1999求s的值

这个是等比数列求和,运用公式就好,Sn=a1(1-q^n)/(1-q),此题a1=1.n=2000,q=2,结果等于2的2000次方-1,希望采纳

若x^2+xy=2、y^2+xy=7,则x^2-y^2=

X^2-Y^2=X^2+XY-Y^2-XY=2-7=-5

  
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