如果x y=1,x的2次方 y的2次方=3
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看明白了,m=-1.四次三项式的话,x的Im-1I次方乘以y的2次方是四次,m-3不等于0(等于0了就变成二项式了),所以Im-1I*2=4,m=-1或者m=3,又因为m不能等于3,只能是m=-1
x²-2xy+y²/x²-xy÷(xy的-1次方-x的-1次方y)=(x-y)²/x(x-y)÷(x/y-y/x)=(x-y)/x÷(x²-y
-xy(x的2次方y的5次方-xy的3次方-y)=-xy²(x²y的4次方-xy²-1)=-(xy²)[(xy²)²-(xy²)-
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
x^2+xy=3xy+y^2=2相加,左边和左边相加,右边和右边相加x^2+xy+xy+y^2=3+2所以x^2+2xy+y^2=5
分子分母同时除以y^2,有[(x/y)^2-x/y+3]/[((x/y)^2)+x/y+6]=5/12再问:能除y的2次方么,至少也要分解好吧,不然你把详细过程写下再答:不要分解,直接同时除以y^2,
由x²-2xy+y²=12x²-5xy-3y²=0得(x-y)²=1(x-3*y)*(2*x+y)=0所以x-y=1或者-1x=3y或者2x+y=0分
1/x+1/y=5x+y=5xy(2x-xy+2y)/(x+2xy+y)=[2(x+y)-xy]/[(x+y)+2xy]=9xy/7xy=9/7再问:9xy/7xy是怎么得出来的?
原式=xy(x²-2xy+y²)=xy(x-y)²=3×(-1)²=3
原式=-x²y³+3x²y³=2x²y³
是不是求(7x的2次方-y的2次方)÷(x的2次方-2xy+3y的2次方)再问:是的根据前面的条件求出答案再答:令x/2=y/3=a,则x=2a,y=3a即(7x²-y²)/(x&
(x^2-2xy+y^2)+(x^2+2x+1)=0(x-y)^2+(x+1)^2=0x=-1y=-1xy=1(xy)^2006=1
原式=3x的2次方y/(-1/2xy)-xy的2次方/(-1/2xy)+1/2xy/(-1/2xy)=-6x+2y-1
原式=(3/2x²-5xy+y²)-[-3xy+2(1/4x²-xy)+2/3y²]=3/2x²-5xy+y²-(-3xy+1/2x
原式=2x的2次方-y的2次方-3xy-x的2次方+2y的2次方-3xy=x的2次方+y的2次方-6xy=2-6×(-1/2)=2+3=5
解x²+y²=7xy=-15x²-(3xy+4y²)-(11xy-2y²)-7x²=(5x²-7x²)+(-3xy-11
=2x²y²-2x³y-4xy³
原式可化为(y-x)^2+2(y-x)+1=(y-x-1)^2
A=3X^2-2XYB=Y^2+XYC=X^2-XY+Y^2=(x-y)^2-xy题目不完整再问:不用了