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3x 2 5xy 2y 2 x 9y 4

3x^2+5xy-2y^2+x+9y-4=3x^2+(5y+1)x-(2y^2-9y+4)=3x^2+(5y+1)x-(y-4)(2y-1)=(3x-y+4)(x+2y-1)十字交叉

答案如下:

原式=(3x-y)(x+2y)+x+9y-4 =(3x-y)(x+2y)-(3x-y)+4x+8y-4 =(3x-y)(x+2y-1)+4(x+2y-1) =(3x-y+4)(x+2y-1)

(ax+by+c)(mx+ny+p)类型的。先配二次项,(3x-y)(x+2y),再配x,(3x+4)(x-1)。合到一块检查y,答案是(3x-y+4)(x+2y-1)

12ab-6(a^2+b^2)=-6(a^2-2ab+b^2)=-6(a-b)^2 3x^2+5xy-2y^2+x+9y-4=(3x-y)(x+2y)+x+9y-4=(3x-y+4)(x+2y-1)

(1)3x2+5xy-2y2+x+9y-4=(3x-y)(x+2y)+x+9y-4=(3x-y)(x+2y)-(3x-y)+4x+8y-4=(3x-y)(x+2y-1)+4(x+2y-1)=(3x-y+4)(x+2y-1);(2)a3+1=(a+1)(a2-a+1)(3)4x4-13x2+9=(4x2-9)(x-1)=(2x+3)(2x-3)(x+1)(x-1);...

用十字叉乘因式分解

原式=(x3+3x2y-5xy2+9y3)+(-2y3+2xy2+x2y-2x3)-(4x2y-x3-3xy2+7y3)=x3+3x2y-5xy2+9y3-2y3+2xy2+x2y-2x3-4x2y+x3+3xy2-7y3=(1-2+1)x3+(3+1-4)x2y+(-5+2+3)xy2+(9-2-7)y3=0∴无论x,y取何值,原式的值均为常数0.

y/x=3/4 所以y=(3/4)x 所以原式=[3x²-5x*(3/4)x+2*(9/16)x²]/[2x²+3x*(3/4)x-5*(9/16)x²] =[(3/8)x²]/[(23/16)x²] =6/23

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