`\bb a) (2x-1)(3x+2)` `\bb b) (1/5 x^3 – 5x^2+x): (1/10x)`

2 bình luận về “`\bb a) (2x-1)(3x+2)` `\bb b) (1/5 x^3 – 5x^2+x): (1/10x)`”

1. Giải đáp:

   a) (2x-1)(3x+2)

   = 2x(3x+2) – 1(3x+2)

   = 6x^2 + 4x – 3x – 2

   = 6x^2 + (4x-3x) – 2

   = 6x^2 + x – 2

   $\\$

   b) (1/5 x^3 – 5x^2 + x) : (1/10 x)

   = 1/5 x^3 : 1/10 x – 5x^2 : 1/10 x + x : 1/10 x

   = 2x^2 – 50x + 10

    

    

   Trả lời
2. a)

   @ (2x-1)(3x+2)

   = 2x(3x+2)- 1(3x+2)

   = 6x^2+ 4x- 3x- 2

   = 6x^2+ x-2

   b)

   @ (1/5x^3- 5x^2+ x): (1/10x)

   = 1/5x^3: 1/10x- 5x^2: 1/10x+ x: 1/10x

   = 2x^2- 50x+ 10

   ~ $kiddd$ ~

   Trả lời