PTĐTTNT : a)x^2-xy-6y^2 b)x^3-3x^2-6x+8=0

By Claire

PTĐTTNT :
a)x^2-xy-6y^2
b)x^3-3x^2-6x+8=0

0 bình luận về “PTĐTTNT : a)x^2-xy-6y^2 b)x^3-3x^2-6x+8=0”

  1. $a)x^2-xy-6y^2\\ =x^2-3xy+2xy-6y^2 \\ =x(x-3y)+2y(x-3y) \\ =(x+2y)(x-3y) \\ b)x^3-3x^2-6x+8=0\\ \to x^3-x^2-2x^2+2x-8x+8=0\\ \to x^2(x-1)-2x(x-1)-8(x-1) =0 \\\to (x-1)(x^2-2x-8)=0\\\to(x-1)(x^2+2x-4x-8)=0\\\to(x-1)[x(x+2)-4(x+2)]=0\\\to(x-1)(x+2)(x-4)=0$\(\to\left[ \begin{array}{l}x-1=0\\x+2=0\\x-4=0\end{array} \right.\\\) \(\to\left[ \begin{array}{l}x=1\\x=-2\\x=4\end{array} \right.\)

    Trả lời
  2. Đáp án:

    Giải thích các bước giải:

    a)3x^2-12x+12

    =3x^2-6x-6x+12

    =3x(x-2)-6(x-2)

    =(3x-6)(x-2)

    b)x^2-xy-6y^2

    =x^2-3xy+2xy-6y^2

    =x(x-3y)+2y(x-3y)

    =(x+2y)(x-3y)

    c)x^2+7x+7y-y^2

    =x^2+xy-xy-y^2+7x+7y

    =x(x+y)-y(x+y)+7(x+y)

    =(x-y+7)(x+y)

    d)x^3-3x^2-6x+8=0

    =x^3-3x^2-6x+8

    =x^3-x^2-2x^2+2x-8x+8

    = x(x-1)-2x(x-1)-8(x-1)

    = (x-1)(x-2x-8)

    Trả lời

Viết một bình luận