#define DIV(x,y) x y;DIV(6 9,3)

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#define DIV(x,y) x y;DIV(6 9,3)
x(x-y)+4(y-x)因式分解

x(x-y)+4(y-x)=x(x-y)-4(x-y)=(x-y)(x-4)=(x-y)(X+2)(X-2)

x^2-y^2/x+y-4x(x-y)+y^2/2x-y,

(x^2-y^2)/(x+y)-(4x(x-y)+y^2)/(2x-y)=(x-y)(x+y)/(x+y)-(4x^2-4xy+y^2)/(2x-y)=(x-y)-(2x-y)^2/(2x-y)=(x

1、x(x-y)(x+y)-x(x+y)^2

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)^2+(x+y)(x-y)】除以 2x

【(x-y)^2+(x+y)(x-y)】除以2x=(x-y)*(x-y+x+y)/2x=(x-y)*2x/2x=x-y

[(-x-y)(-x+y)-(x+y)^2-x(y-y^2)}÷1/2y

[(-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

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

#define max(x,y) x>y?x:

if(x>y)returnx;elsereturny;在一句完整的语句后面需要用到;比如一开始的定义自变量inta;赋值时要用到a=1;各种结构在执行完要处理的语句时也要用到.但是切记,各种结构只处理

化简:(x-3y)(x+y)-(x-2y)(x+2y)-(x-y)平方

(x-3y)(x+y)-(x-2y)(x+2y)-(x-y)平方=x²-2xy-3y²-x²+4y²-x²+2xy-y²=(x²-

(1)(x^2/x)-y-x-y

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

{(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)

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

若X-Y>X,X+Y

由X-Y>X可得Y<0由X+Y

因式分解 x方(x-y)(y-x)

您好:x方(x-y)+(y-x)=x方(x-y)-(x-y)=(x²-1)(x-y)=(x+1)(x-1)(x-y)如果本题有什么不明白可以追问,如果满意记得采纳如果有其他问题请采纳本题后另

因式分解:x²(x-y)+(y-x)

原式=x²(x-y)-(x-y)=(x²-1)(x-y)=(x+1)(x-1)(x-y)

# define ABS_MOD(x,y) (((x) < 0) ((((x) % (y)) + (y)) % (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-