2015(x-y) 2016(y-z) 2017(z-x)=0-搜答案-智慧作业帮

再问：谢谢，再答：不言谢！

(x-y)(z+y-x+y+2y)÷4y=(x-y)(z-x+4y)÷4y{(x+y)(x-y)-(x-y)的2次方+2y(x-y)}除以4y=(x-y)(x+y-x+y+2y)÷4y=(x-y)(4

若x−y+y

x−y+（y-2）2=0，∵x−y≥0，（y-2）2≥0，∴x-y=0，y-2=0，解得：y=2，x=2，∴xy=4．

求导 x*e^y*y'

设u=x×e^y×y'du/dx=y'e^y+x(y')²e^y+xy''e^y

已知x,y满足y

y＜√(x-1)+√(1-x)+1/2x-1≥0,1-x≥0,即x-1≤0∴x-1=0,x=1∴y＜√(x-1)+√(1-x)+1/2=1/2∴1-y＞0∴|1-y|/(y-1)=(1-y)/(y-1

1)x(x-y)(x+y)-x(x+y)^2=x((x-y)(x+y)-(x+y)^2)=x(x^2-y^2-x^2-2xy-y^2)=x(-2xy-2y^2)=-2xy(x+y)2)（2a+b)(2

(x+y)

x＞0,y＞0,a＞0　　a(x+y)≤√(x²+y²)　　a²(x+y)²≤x²+y²　　(1-a²)(x²+y

[(-x-y)(-x+y)-(x+y)^2-x(y-y^2)}÷1/2y=[x²-y²-x²-2xy-y²-xy+xy²]/(y/2)=[(x-2)y

求y'=y/(y-x)

∵令y=xt,则y'=xt'+t代入原方程,得xt'+t=t/(t-1)==>xt'=(2t-t^2)/(t-1)==>(t-1)dt/(2t-t^2)=dx/x==>2dx/x+[1/t+1/(t-

z/(x-y) × y/(x+y)

z/(x-y)×y/(x+y)=zy/(x-y)(x+y)=zy/(x²-y²)再问：还有两道题！麻烦你了！1.已知x-1/x=2，求x²/x四次方-x²+12

若x,y满足y

∵被开方数要大于等于0∴x+1≥0且-(x+1)²≥0∴x+1=0,x=-1∴y﹤0+0+5,y﹤5原式=√(y-5)²+|x+1|=|y-5|+|x+1|=-(y-5)+0=5-

{(x,y) |x|+|y|

区域是一个正方形

X、Y

解题思路：化简解答解题过程：varSWOC={};SWOC.tip=false;try{SWOCX2.OpenFile("http://dayi.prcedu.com/include/readq.ph

1.(x-y)(x+y)+(x-y)+(x+y)=x²-y²+x²-y²=2x²-2y²2.x(x+2)-(x+1)(x-1)=x²

|x|+|y|

讨论x,y与0的关系即可,即去掉绝对值：x0,y>0x+y

#define定义了一个宏.你可能需要这样用inta=-5,b=2;intc;c=ABS_MOD(a,b)然后编译器就帮你替换成c=a再问：有什么用，什么时候会用它再答：求模呀，只不过要这个要判断符号

x>y?x:y

判断X的数值是否大于Y的数值如果是则为真等式去X的值反之取Y的值

x y x+yy x+y xx+y x y

把所有列都加至第一列,第一列都是2x+2y将2x+2y提出,第一列剩下都是1,此时式外边有一因子（2x+2y）用2,3行加第1行负一倍得1yx+y0xy0x-y-x第一列展开得1*（-x^2-y(x-

) y=cos(x-y)

1.两边求导得：y'=-sin(x-y)(1-y')解得y'=sin(x-y)/[sin(x-y)-1]2.y'=-e^-xy''=e^-xy'"=-e^-x3.y'"=(e^2x)'"(sinx)+