llgd.net
当前位置:首页 >> 初中因式分解题目:若实数x,y,z满足(x%z)&#178... >>

初中因式分解题目:若实数x,y,z满足(x%z)&#178...

x+z=2y 解: (x-z)²-4(x-y)(y-z)=0 (x-y+y-z)²-4(x-y)(y-z)=0 (x-y)²+(y-z)²+2(x-y)(y-z)-4(x-y)(y-z)=0 (x-y)²+(y-z)²-2(x-y)(y-z)=0 [(x-y)-(y-z)]²=0 (x+z-2y)²=0 x+z-2y=0 x+z=2y

(x+y+z)²-(x-y-z)² =[(x+y+z)+(x-y-z)][(x+y+z)-(x-y-z)] =2x×(2y+2z) =4x(y+z)

1=x²+y²+z² =(x+y+z)²-2xy-2yz-2zx =-2yz-2x(y+z) =-2yz+2(y+z)² ≥-2yz+2(2√yz)² =-2yz+8yz =6yz, ∴yz≤1/6, 即所求yz最大值为1/6, 不存在yz最小值! 此时,x=-√6/3,y=z=√6/6。

x^2+y^2+z^2-2x+4y-6z+14= 0 (x^2-2x+1)+(y^2+4y+4)+(z^2-6z+9)= 0 ∴ (x-1)^2+(y+2)^2+(z-3)^2= 0 ∵ (x-1)^2≥0,(y+2)^2≥0,(z-3)^2≥0 ∴ x-1= y+2= z-3= 0 ∴ x=1,y= -2,z=3 ∴ (x-y-z)^2012 = (1+2-3)^2012 = 0 这类多元...

(x+y+z)³-x³-y³-z³ =(x+y+z-x)[(x+y+z)²+x(x+y+z)+x²]-(y+z)(y²-yz+z²) =(y+z)[x²+y²+z²+2xy+2xz+2yz+x²+xy+xz+x²]-(y+z)(y²-yz+z²) =(y+z)(3x²+3xy+3xz+3yz)...

(x+(y-z))(x-(y-z))=x²-(y-z)²=x²-y²+2yz-z²

求偏导数就把别的参数看作常数即可 f(x,y,z)=x²y+y²z+z²x 于是f'x=2xy+z²,f'y=2yz+x² 继求偏导得到 f''xx=2y,f''xy=2x,f''yz=2y,而f'''xxy=2 即f''xx(1,0,0)=0,f''xy(1,0,1)=2 f''yz(1,0,2)=0,f'''xxy=2

(1)“已知实数x,y,z满足4x²/(1+4x²)=y,4y²/(1+4y²)=z,4z²/(1+4z²)=x,求x,y,z的值。 (2):“已知实数x,y,z满足(1+4x²)/4x²=y,(1+4y²)/4y²=z,(1+4z²)/4z²=x,求x,y,z的...

没全看清你的表达。能写多少算多少。 f(z)=dF(x)/dx. F(z) = P(Z

由于积分区域Ω:x² + y² + z² = R²关于坐标三轴都对称 且被积函数中的x,y,z都是奇函数 若f(x,y,-z)=-f(x,y,z),则说f(x,y,z)关于z是奇函数 在对称区间上的奇函数的积分结果是0 所以用对称性可得∫∫∫ (x+y+z) dV = 0 剩下的∫∫...

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