x3 -3x2 x 5=0分解因式
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1楼说实话答的不错,可是还是有些题...有些题和2楼答的一样x^3-x^2-x-2=x^3-2x^2+x^2-x-2=x^2(x-2)+(x-2)(x+1)=(x^2+x+2)(x-2)x^5+x+1
x3+3x2-2xy-kx-4yx3+2x2+x2-kx-2y(x+2)=x2(x+2)+x(x-k)-2y(x+2),若x3+3x2-2xy-ky-4y可分解为一次与二次因式之积,则x-k=x+2解
(1)x3+x2+x-3=x^3-x+x^2+2x-3=x(x-1)(x+1)+(x-1)(x+3)=(x-1)(x^2+2x+3)(2)x3-6x2+11x-6=x^3-x^2-(5x^2-11x+
原式=x4+2x3-x2-8x2-2x+8,=x2(x2+2x-8)-(x2+2x-8),=(x2+2x-8)(x2-1),=(x+4)(x-2)(x+1)(x-1).故答案为:(x+4)(x-2)(
∵代数式x3+y3+3x2y+axy2含有因式x-y,∴当x=y时,x3+y3+3x2y+axy2=0,∴令x=y,即x3+x3+3x3+ax3=0,则有5+a=0,解得a=-5.将a=-5代入x3+
原式=x(x2-5)=x(x+5)(x-5).故答案为x(x+5)(x-5).
设原式=(x2+ax+b)(x2+cx+d)=x4+(a+c)x3+(b+d+ac)x2+(ad+bc)x+bd,所以有a+c=2b+d+ac=3ad+bc=2bd=1,解得a=1b=1c=1d=1.
一式无法分解二式(x+1)*(x^2+x+1)*(x^2-x+1)三式无法分解
x3-x=x(x2-1)=x(x+1)(x-1);a2+4a+4=(a+2)2;4x2-25=(2x+5)(2x-5);9x2-y2-4y-4=9x2-(y+2)2=(3x+y+2)(3x-y-2).
^3代表3次方,^2代表平方X^3+X^2-Y^3-Y^2=(x^3-y^3)+(x^2-y^2)=(x-y)(x^2+xy+y^2)+(x-y)(x+y)=(x-y)(x^2+xy+y^2+x+y)
(a-2)(a-1)=0
(x-2)(3x+1)=0十字相乘法
x4+x3-3x2-4x-4=x4+x3+x2-4x2-4x-4=x2(x2+x+1)-4(x2+x+1)=(x2-4)(x2+x+1)=(x+2)(x-2)(x2+x+1).故答案为:(x+2)(x
原式=x^4(x+1)+x^2(x+1)+(x+1)=(x+1)[(x^4+2x^2+1)-x^2]=(x+1)[(x^2+1)^2-x^2]=(x+1)(x^2+x+1)(x^2-x+1)
X(y-3)-X3次方(3-y)=X(y-3)+X3次方(y-3)=(y-3)(x+x3)=x(x2+1)(y-3)
原式=x2(x+3)-4(x+3)=(x+3)(x2-4)=(x+3)(x+2)(x-2).
解法1:将常数项8拆成-1+9.原式=x3-9x-1+9=(x3-1)-9x+9=(x-1)(x2+x+1)-9(x-1)=(x-1)(x2+x-8).解法2:将一次项-9x拆成-x-8x.原式=x3
(4x-1)(5x+7)=0x1=1/4x2=-7/53x(x-1)=2-2x3x^2-3x=2-2x3x^2-x-2=0(3x-2)(x+1)=0x=2/3x=-1(2x+3)^2=4(2x+3)(
8252x3-1752x3=(8252-1752)x3=6500x3=19500
x^3+y^3+z^3-3xyz=[(x+y)^3-3x^2y-3xy^2]+z^3-3xyz=[(x+y)^3+z^3]-(3x^2y+3xy^2+3xyz)=(x+y+z)[(x+y)^2-(x+