m n=x^2-3m=3x-3等于是多少-搜答案-神马作文网

∵M、N都是单项式∴3xM也是单项式∵3xM-15x^2=6x^2y^2+N又∵-15x^2≠6x^2y^2∴3xM=6x^2y^2,N=-15x^2

m+n=3-2kmn=k^23-2k=k^2得到k=1或-3所求=mn(m+n-1)=0或72

（x＋y）²-3x-3y+2=（x＋y）²-3x+y）+2=203m-2mn＋3n分之m＋2mn＋n=4mn分之4mn=1

3x^2-5x/2+11=x-m/2-3x+n-m/2-3x=-5x/2m=-xn=3x^2+113x^2-(5x+11)/(5x-2)=m/(x-2)-n/(3x+1)(15x^3-6x^2-5x-

1.X=3,其实y轴相当于x=0的图形,只要保持如果直线上每个点的X值都不变,那么他就平行于y轴,对本题来说,就是X=3这条直线.2.对于坐标系内的点,他到x轴的距离,就是|y|,到y轴的距离就是|x

3x(M-5x)=3xM-15x²=6x²y³+n3xM=6x²y³-15x²=NM=2xy³MN=-30x³y

把x=3代入原方程,得：9-6m+2m²-4mn+4n²=0(m²-6m+9)+(m²-4mn+4n²)=0(m-3)²+(m-2n)

X^2+2(M+1)X+3M^2+4MN+4N^2+2=(X+M+1)^2-M^2-2M-1+3M^2+4MN+4N^2+2=(X+M+1)^2+M^2-2M+1+M^2+4MN+4N^2=(X+M+

m三次方-4m=m(m²-4）=m(m+2)(m-2).m三次方n-9mn=mn(m²-9)=mn(m+3)(m-3).3m(2x-y)平方-3mn二次方=3m[(2x-y)

2x（M+3x）=6x²y²+N2Mx+6x²=6x²y²+N∴2Mx=6x²y²N=6x²即：M=3xy²N

2x(M+3x)=6x²y²+N2xM+6x²=6x²y²+N由于6x²=6x²y²不恒等,根据对应项相等,得2xM=6

(3x^2-2m^2+mn)^2+(3m^2-mn+2n^2-12x)^2+4=4x-x^2所以(3x^2-2m^2+mn)^2+(3m^2-mn+2n^2-12x)^2+4+x^2-4x=0所以(3

f(x)>0的解集为(1,2)1

1、减去-2x+3x²等于-3-5x+4x²的代数式=-3-5x+4x²+(-2x+3x²)=7x²-7x-32、M+N=-2,是不是m+n=-2?2

由韦达定理m+n=-（2k-3）,mn=k^2所以-（2k-3）=mn=k^2解得k=1或k=-3当k=1时,方程变为x²-x+1=0,此时方程得耳塔小于0,无解,k=1舍去当k=-3时,方

1、3m(2x-y)²-3mn²

原式=3m[(2x-y)²-n²]=3m(2x-y+n)(2x-y-n)原式=4x²+4xy+y²=(2x+y)²

奇函数f（o）=0所以n=2f（-x）=-f（x）m=1

1.4x-5x=-x2.-1/3ab^2+ab^2=2/3ab^23.4a-5a+8a=7a4.-2x+3y+7x-1/2y=5x+5/2y5.5mn-2m+4mn+6m-2=mn+4m-26.剩下的

2x^2+(3m-n)*x-2m^2+3mn-n^2=(2x-(m-n))*(x+2m-n)=0x1=(m-n)/2x2=n-2m

根据一元二次方程根式：x=(-b+/-(b^2-4ac)^(1/2))/(2a)得x=(3m+/-(9m^2-4(2m^2-mn-n^2))^(1/2))/2=(3m+/-(m-2n))/2得x1=2