limarctan1/x-y

x趋近于0时limarctan1/x^2的极限x趋近于0时limarctan1/x^2的极限x趋近于0时limarctan1/x^2的极限limarctan1/x²=arctan(+∞)=&

{(x,y)|x|+|y|{(x,y)|x|+|y|{(x,y)|x|+|y|区域是一个正方形

x>y?x:yx>y?x:yx>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-y^/y-x化简.x^/y-x-y^/y-x化简.x^/y-x-y^/y-x化简.原式=(x²-y²)/(y-x)=(x+y)(x-y)/[-(x-y)]=-(x+y

化简x(y-x)-y(x-y)化简x(y-x)-y(x-y)化简x(y-x)-y(x-y)y^2-x^2你觉得怎么样

行列式计算xyx+yyx+yxx+yxy|xyx+y||yx+yx||x+yxy|行列式计算xyx+yyx+yxx+yxy|xyx+y||yx+yx||x+yxy|行列式计算xyx+yyx+yxx+y

y/(x+y)-x/(x-y)y/(x+y)-x/(x-y)y/(x+y)-x/(x-y)y/(x+y)-x/(x-y)=y(x-y)/(x+y)(x-y)-x(x+y)/(x+y)(x-y)=[y(

x-y/x-x+y/y-(x+y)(x-y)/y²y/x=2x-y/x-x+y/y-(x+y)(x-y)/y²y/x=2x-y/x-x+y/y-(x+y)(x-y)/y²

xyx+yyx+yxx+yxyxyx+yyx+yxx+yxyxyx+yyx+yxx+yxy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）

(3x-y)(3y+x)-(x-y)(x+y)(3x-y)(3y+x)-(x-y)(x+y)(3x-y)(3y+x)-(x-y)(x+y)(3x-y)(3y+x)-(x-y)(x+y)=9xy+3x&

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

化简x/y(x+y)-y/x(x+y)=化简x/y(x+y)-y/x(x+y)=化简x/y(x+y)-y/x(x+y)=x^2-y^2为分子,xy(x+y)为分母

化简:(x+y)(x-y)+(x+y)+(x-y)化简:(x+y)(x-y)+(x+y)+(x-y)化简:(x+y)(x-y)+(x+y)+(x-y)(x+y)(x-y)+(x+y)+(x-y)=(x

x+yx+yx+y2*(3x+2y)

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

|x+y||x+y||x+y|证明：由|x+y|又已知|2x-y|（1）+（2）得-5/6亦即|y|（注意正负范围值、大于小于是可以直接相加的）可以得出这两个式子-1/3＜x+y＜1/3-1/6＜2x

xyx+yyx+yxx+yxyxyx+yyx+yxx+yxyxyx+yyx+yxx+yxyc1+c2+c3第2,3列加到第1列2(x+y)yx+y2(x+y)x+yx2(x+y)xyr2-r1,r3-

化简y(x+y)+(x+y)(x-y)-x^2化简y(x+y)+(x+y)(x-y)-x^2化简y(x+y)+(x+y)(x-y)-x^2y(x+y)+(x+y)(x-y)-x^2=(x+y)(y+x

|x|+|y||x|+|y||x|+|y||x|+|y|

x*x-y*y-(x+y)(x+y),因式分解x*x-y*y-(x+y)(x+y),因式分解x*x-y*y-(x+y)(x+y),因式分解原式＝(x+y)(x-y)-(x+y)^2=(x+y)(x-y