nczl.net
当前位置:首页 >> 1 x 2 y 2xy >>

1 x 2 y 2xy

化为: [(1+x²)y'-2xy]/(1+x²)²=1 [y/(1+x²)]'=1 积分得: y/(1+x²)=x+C 得y=(1+x²)(x+C)

解: 可以用主元法的思想来解这道题。 把x作为主元,整理,得 x²+2yx+2y²-1=0 这个方程有实数根(否则x取不到值) Δ=4y²-4(2y²-1) =4y²-8y²+4 =4-4y²≥0 y²≤1 所以-1≤y≤1 y的极小值是-1,当x=1时取得。 ...

因为x^2+2xy+4y^2=1 即(x+y)^2=1-3y^2 因为x,y都是正数,所以当y变小接近零,x变大接近1时 (x+y)^2

(xy-1)²+(x+y-2)(x+y-2xy) =(xy)²-2xy+1+(x+y)²-2(x+y)-2xy(x+y)+4xy =(xy+1))²+(x+y)²-2(x+y)(xy+1) =(xy+1-x-y)²

2xy = 1 - x^2, y = (1-x^2)/(2x), x^2 + y^2 = x^2 + (1-x^2)^2/(4x^2) = x^2 + (1-2x^2 + x^4)/(4x^2) = x^2 + x^2/4 + 1/(4x^2) - 1/2 = 5x^2/4 + 1/(4x^2) - 1/2 >= 2[(5/4)(1/4)]^(1/2) - 1/2 = 2*5^(1/2)/4 - 1/2 = [5^(1/2) -1]/2 x^2 + ...

令y =sin a ; x = cos a sin b ; z = cos a cos b,a,b∈[0,π/2] 2xy+yz=y(2x+z) = sin a ( 2 cos a sin b + cos a cos b ) = sin a cosa ( 2sin b + cos b ) =(√5/2)sin 2a * sin(b+arctan1/2)

答案:(x-y-1)^2 解析:这道题涉及到完全平方公式的应用,完全平方公式如下 (a+b)²=a²﹢2ab+b² ﹙a-b﹚²=a²﹣2ab+b² 所以x^2-2xy+y^2是完全平方式 所以可变成(x-y)^2 然后 -2x+2y可变成-2(x-y) 所以原式...

(1)原式=y(x-y) 2 (2)原式=(x-y)(x-y)=(x-y) 2 (3)原式= 13 55 (18.9+37.1-1)= 13 55 ×55=13(4)原式= 2 =4.

(x+y-2xy)(x+y-2)+(xy-1)² =(x+y)²-(2xy+2)(x+y)+4xy+x²y²-2xy+1 =(x+y)²-2(xy+1)(x+y)+(x²y²+2xy+1) =(x+y)²-2(xy+1)(x+y)+(xy+1)² =[(x+y)-(xy+1)]² =(x+y-xy-1)²

网站首页 | 网站地图
All rights reserved Powered by www.nczl.net
copyright ©right 2010-2021。
内容来自网络,如有侵犯请联系客服。zhit325@qq.com