1-cos(x2 y2)
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![1-cos(x2 y2)](/uploads/image/f/41599-55-9.jpg?t=1-cos%28x2+y2%29)
y(1+x2y2)dx=xdy设xy=u,则y=u/x,dy=d(u/x)=(xdu-udx)/x^2方程化为u/x(1+u^2)dx=x*(xdu-udx)/x^2化简得u(1+u^2)dx=xdu
sin
xy/x+y=1/3x+y=3xyx2y2/x2+y2=1/5(xy)²/[(x+y)²-2xy]=1/5(xy)²/[(3xy)²-2xy]=1/5(xy)&
也可以考虑,分子分母同时乘以1-cosx,被积函数化为:(1-cosx)/sin²xI=∫(1-cosx)/sin²xdx=∫[csc²x-cscxcotx]dx=-co
(1)4x2m+1y的系数是4,次数是2m+2;-5x2y2的系数是-5,次数是4;-31x5y的系数是-31,次数是6;(2)由(1)可得2m+2=8,解得m=3.
第一式是六次四项式,最高次为6,式中已知的最高次是:--4x3y2这一项5次,所以--1/5x2ym这一项的次数必须是6次,既有2+m=6,m=4;第二式次数与第一是相同,则有2n+(5--m)=6,
x3次方y-2x2y2+xy3=xy(x²-2xy+y²)=xy(x-y)²=3x3²=27如果本题有什么不明白可以追问,再问:=xy(x2-2xy+y2)=x
(x+1)(x+3)(x-2)(x-4)+24=[(x+1)(x-2)][(x+3)(x-4)]+24=(x^2-x-2)(x^2-x-12)+24=(x^2-x-7+5)(x^2-x-7-5)+24
原式=x4+x3y+4x3y+x2y+4x2y2+4x2y2+xy2+4xy3+xy3+y4,=x3(x+y)+4x2y(x+y)+xy(x+y)+4xy2(x+y)+y3(x+y),=-x3-4x2
(1)x2(a-b)+4(b-a),=x2(a-b)-4(a-b),=(a-b)(x2-4),=(a-b)(x+2)(x-2);(2)x2(2x+y)2-4x2y2,=x2[(2x+y)2-4y2],
(2x4-4x3y-x2y2)-2(x4-2x3y-y3)+x2y2=2x4-4x3y-x2y2-2x4+4x3y+2y3+x2y2=2y3,因为化简的结果中不含x,所以原式的值与x值无关.
原式=(x^4-2x²y²+y^4)+6xy(x²+2xy+y²)-2xy(x+y)=(x²-y²)²+6xy(x+y)²
原式=2x2y-2xy2-[-3x2y2+3x2y+3x2y2-3xy2]=2x2y-2xy2+3x2y2-3x2y-3x2y2+3xy2=2x2y-3x2y-2xy2+3xy2+3x2y2-3x2y
原式=[x3y2-x2y-x2y+x3y2]÷3x2y=(2x3y2-2x2y)÷3x2y=23xy-23;当x=3,y=-1时,原式=23×3×(-1)-23=-83.
(2X²-2y²)-3(X²y²+X²)+3(X²y²+y²)=2x²-2y²-3x²y&
因为x²+4y²+x²y²-6xy+1=0(x²-4xy+4y²)+(x²y²-2xy+1)=0(x-2y)²