如果(x y) (y-1)i
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x+y=3xy可得(x+y)/xy=3,得1/y+1/x=3答案就是3
x^2+xy+y=14y^2+xy+x=28两式相加x^2+y^2+2xy+x+y=42(x+y)^2+(x+y)-42=0(x+y-6)(x+y+7)=0x+y=6或x+y=-7
xy/(x+y)上下同除以xy,得到1/(1/y+1/x)=5,则(1/x)+(1/y)=1/5
xy-1+x-y=XY+X-Y-1(加法交换律)=(XY+X)-(Y+1)=X(Y+1)-(Y+1)(提取公因式X)=(Y+1)(X-1)(提取公因式Y+1)这样可以么?
(x-y)^2=x^2-2xy+y^2=9x^2+2xy+y^2=9+4=13(x+y)^2=13
(1)因为x^2+xy=2,y^2+xy=7所以x^2+2xy+y^2=(x^2+xy)+(y^2+xy)=2+7=9(2)因为a^2-ab=3,b^2+ab=2所以a^2+b^2=(a^2-ab)+
如果有理数xy满足丨x-1丨+(xy-2)²=0求x和y的值丨x-1丨+(xy-2)²=0丨x-1丨=0x=1(xy-2)²=0xy=2y=2
你传错了吧,怎么无解啊~
∵x2+y2=1,∴x=sinθ,y=cosθ,∴(1-xy)(1+xy)=1-x2y2=1-(sinθcosθ)2=1-(12sin2θ)2=1-14sin22θ,当sin2θ=0时,1-14sin
x²y-3xy²+kxy-1/3xy-4=x²y-3xy²+(k-1/3)xy-4不含xy的项,说明含xy的项(k-1/3)xy的系数(k-1/3)等于0k-1
(1)(-2)*4=(-2)x4+1=-7(2)(-1*3)*(2)=[(-1)x3+1]*2=(-2)*2=(-2)x2+1=13(3)任意选择两个有理数,分别填入下列□和○内,并比较两个运算结果,
1/x-1/y=(y-x)/xy=3所以y-x=3xy代入后面的公式得(-5xy)/(-9xy)=5/9
观察到sin²θ+cos²θ=1,则可做三角代换令x=sinθ,y=cosθ(1-xy)(1+xy)=1-(xy)²=1-(sinθcosθ)²=1-(sin2
设x=sint,y=cost,那么原式就等于(1+sintcost)(1-sintcost)=1-sint2cost2=1-sin2t2/4.因为sin2t的平方最大值为1最小值为0所以原式的最大值为
1/x-1/y=2(y-x)/xy=2y-x=2xy(3x+5xy-3y)/(x-xy-y)=[5xy-3(y-x)]/[-xy-(y-x)]=(5xy-6xy)/(-xy-2xy)=-xy/-3xy
∵x>y>0,∴y-x<0,y+1>0,∴x+1y+1-xy=y(x+1)y(y+1)-x(y+1)y(y+1)=xy+y−xy−xy(y+1)=y−xy(y+1)<0.∴x+1y+1-xy的结果是负
2x+3y-3xy+4xy-3x-4y+7=(2x-3x+3y-4y)+4xy-3xy+7=(-x-y)+xy+7=-(x+y)+xy+7=-4-1+7=2
0.5x²+xy+0.5y²=0.5(x²+2xy+y²)=0.5(x+y)²=0.5
已知xy+x+y=6则xy+x+y+1=(x+1)(y+1)=7=7×1因为x,y都是自然数(非负整数)所以x+1,y+1≥1所以只能x+1=7y+1=1即x=6,y=0或x+1=1,y+1=7即x=
Ix-2I+(y+1)²=0,绝对值与平方均为非负数,则有x=2,y=-15x²+3xy-5y²-2xy+4y²-4x²=x^2-y^2+xy=4-1