Toán Lớp 8: tìm MTC: `2/(9x^2 – 1)`; `(4x)/(1 – 3x)`

TRẢ LỜI

1. MTC:(1-3x)(3x+1)

   2/(9x^2-1)=-2/((3x)^2-1^2)=2/((1-3x)(3x+1))

   (4x)/(1-3x)=(4x(3x+1))/((1-3x)(3x+1))

    

   Trả lời
2. Ta có:

   $9x^2-1=(3x)^2-1=(3x-1)(3x+1)

   $⇒$ $MTC=(3x-1)(3x+1)$

   $\dfrac{2}{9x^2-1}=\dfrac{2}{(3x-1)(3x+1)}$ 

   $\dfrac{4x}{1-3x}=\dfrac{-4x}{3x-1}=\dfrac{-4x(3x+1)}{(3x-1)(3x+1)}$ 

    

   Trả lời