把下列各式因式分解 (1)ax2 16ay2(2) 2a

2021-03-11 12:00:37 字數 1042 閱讀 2404

1樓:小米米米小米米

(1)ax2-16ay2=a(x+4y)(

duzhix-4y);(dao

專2)-2a3+12a2-18a=-2a(屬a2-6a+9)=-2a (a-3)2;(3)a2-2ab+b2-1=(a-b)2-1=(a-b+1)(a-b-1);(4)a2(x-y)-4b2(x-y)=(x-y)(a2-4b2)=(x-y) (a+2b)(a-2b);(5)(x+2)(x-6)+16=x2-4x+4=(x-2)2.

2樓:科比ai打鐵

(zhi1)ax2-16ay2=a(daox+4y)(內容x-4y);

(2)-2a3+12a2-18a

=-2a(a2-6a+9)

=-2a (a-3)2;

(3)a2-2ab+b2-1

=(a-b)2-1

=(a-b+1)(a-b-1);

(4)a2(x-y)-4b2(x-y)

=(x-y)(a2-4b2)

=(x-y) (a+2b)(a-2b);

(5)(x+2)(x-6)+16=x2-4x+4=(x-2)2.

把下列各式分解因式.(1)9a2-14b2(3)12a2?ab+12b2(3)a2-2ab-4+b2(4)x(x2+1)2-4x3(5)x2-7x-8

3樓:匿名使用者

(1)9a2-1

4b2=(3a+1

2b)(3a-1

2b);

(3)12a

?ab+12b

=12(a2-2ab+b2)=1

2(a-b)2;

(3)a2-2ab-4+b2=(a-b)2-4=(a-b+2)(a-b-2);

(4)x(x2+1)2-4x3=x[(x2+1)2-4x2]=x(x2+1+2x)(x2+1-2x)=x(x+1)2(x-1)2;

(5)x2-7x-8=(x-8)(x+1);

(6)x(x-y)2+y(y-x)=(x-y)[x(x-y)-y]=(x-y)(x2-xy-y).

把下列各式分解因式1mxynyx

1 m x y n baiy x dum n zhix y dao2 2x4 32 2 回x4 16 2 x2 2 42 2 x2 4 答x2 4 2 x2 4 x 2 x 2 3 a3 a a a2 1 a a 1 a 1 4 x y 2 2 x y 1 x y 1 2 5 11 x x 1 x ...

把下列各式分解因式(1)9a2 14b2(3)12a2 a

1 9a2 1 4b2 3a 1 2b 3a 1 2b 3 12a ab 12b 12 a2 2ab b2 1 2 a b 2 3 a2 2ab 4 b2 a b 2 4 a b 2 a b 2 4 x x2 1 2 4x3 x x2 1 2 4x2 x x2 1 2x x2 1 2x x x 1 ...

x 1因式分解, x 1 x 3 1怎麼因式分解

結果為 x1 1 3 i 2,x2 1 3 i 2解題過程如下 x x 1 0 解 x x 1 0 x 3 1 x 1 0 x1 1 3 i 2 x2 1 3 i 2 擴充套件資料一元二次方程的特點 1 能使一元二次方程左右兩邊相等的未知數的值稱為一元二次方程的解。一般情況下,一元二次方程的解也稱為...