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已知2x y 10

(x²+y²-x²-y²+2xy+2xy-2y²)÷2y (4xy-2y²)÷2y 2x-y=10, 结果10 如果不对请追问 帮到您请好评 谢谢

[(x 2 +y 2 )-(x-y) 2 +2y(x-y)]÷4y,=[x 2 +y 2 -x 2 +2xy-y 2 +2xy-2y 2 ]÷4y,=[4xy-2y 2 ]÷4y,= 1 2 (2x-y),∵2x-y=10,∴原式= 1 2 ×10=5.

解:∵[(x^2+y^2)-(x-y)^2+2y(x-y)]/y =[x^2+y^2-(x^2-2xy+y^2)+2y(x-y)]/y =[[x^2+y^2-x^2+2xy-y^2)+2y(x-y)]/y =[2xy+2y(x-y)]/y =y[2x+2(x-y)]/y =2x+2x-2y =4x-2y =2(2x-y) 又∵已知2x-y=10 ∴2(2x-y)=2*10=20 ∴[(x^2+y^2)-(x-y)^2+2y(x-y)]/y=...

(1)原式=y2-4-y2-4y+5=1-4y;(2)∵2x-y=10,∴[(x2+y2)-(x-y)2+2y(x-y)]÷y=(x2+y2-x2+2xy-y2+2xy-2y2)÷y=(4xy-2y2)÷y=4x-2y=2(2x-y)=20.

首先x=(y+10)/2 [x^2-(x-y)^2+y^2+2y(x-y)]/2y =[(x+x-y)(x-x+y)+y^2+2y(x-y)]/2y =[(2x-y)y+y^2+2y(x-y)]/2y 代入x=(y+10)/2,经过简单的化简计算 答案 10

已知2x-y=10, 式子[(x^2+y^2)-(x-y)^2+2y(x-y)]/4y =[x^2+y^2-x^2+2xy-y^2+2xy-2y^2]/4y =[4xy-2y^2]/4y =2y(2x-y)/4y =2y*10/4y =5

因为随机变量X~P(2),所以EX=2,DX=2.因为Y=2X-10,故利用数学期望的线性性质可得,EY=2EX-10=-6,利用方差的性质可得,DY=22DX=8.故答案为:8.

2x - y = 10 [(x^2 + y^2) - (x - y)^2 + 2y(x - y)]/(4y) = [(x^2 + y^2) - (x^2 - 2xy + y^2) + 2y(x - y)]/(4y) = [2xy + 2y(x - y)]/(4y) = [2y(2x - y)]/(4y) = (2x - y)/2 = 5

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