Bài `2.`

`a. (x+y)/[2.(x-y)] - (x-y)/[2.(x+y)] + (2y^2)/(x^2 - y^2)`

`= [(x+y).(x+y) - (x-y).(x-y) + 4y^2]/[2.(x-y).(x+y)]`

`= (x^2 + 2xy + y^2 - x^2 + 2xy - y^2 + 4y^2)/[2.(x-y).(x+y)]`

`= (4y^2 + 4xy)/[2.(x-y).(x+y)]`

`= [4y.(y + x)]/[2.(x-y).(x+y)]`

`= (2y)/(x-y)`

`b. (x+5)/(2x-4) . (4-2x)/(x+2)`

`= (x+5)/[2.(x-2)] . [-2.(x - 2)]/(x+2)`

`= -(x+5)/(x+2)`

`= (-x - 5)/(x+2)`

`c. 8/(x^2 + 2x - 3) + 2/(x+3) + 1/(x-1)`

`= [8 + 2.(x-1) + 1.(x+3)]/[(x-1).(x+3)]`

`= (8 + 2x - 2 + x + 3)/[(x-1).(x+3)]`

`= (3x + 9)/[(x-1).(x+3)]`

`= [3.(x + 3)]/[(x-1).(x+3)]`

`= 3/(x-1)`

`d. (1-4x^2)/(x^2 + 4x) . (3x)/(2-4x)`

`= [(1 - 2x).(1+2x)]/[x.(x+4)] . (3x)/[2.(1-2x)]`

`= (1+2x)/(x+4) . 3/2`

`= (3 + 6x)/(2x + 8)`

Đáp án:

 a) $\frac{x + y}{2(x-y)}$-$\frac{x-y}{2(x+y)}$+$\frac{2y^2}{x^2-y^2}$ 

= $\frac{x+y}{2(x-y)}$-$\frac{x-y}{2(x+y)}$+$\frac{2y^2}{(x-y)(x+y)}$

= (x+y)^2-(x-y)^2+4y^2

 ____________________________

       2(x-y)(x+y)

=  2y.2x+4y^2

__________________

    2(x-y)  (x+y)

= 4xy + 4y^2

 ________________

    2(x-y).(x+y)

= 4y (x+y)

______________

   2(x-y)(x+y)

= $\frac{2y}{x-y}$ 

b) $\frac{x+5}{2x-4}$ . $\frac{4-2x}{x+2}$ 

= $\frac{x + 5}{2x-4}$. -$\frac{(2x-4)}{x+2}$

= -$\frac{x-5}{x+2}$ 

c) Mình lười ghi đề

= $\frac{8+ 2(x-1)+x+3}{x^2+2x-3}$ 

= $\frac{8+2x-2+x+3}{x^2+2x-3}$ 

= $\frac{3x+9}{x^2+2x-3}$ 

= $\frac{3(x+3)}{(x+3)(x-1)}$ 

= $\frac{3}{x-1}$ 

d) Mình lười ghi

= $\frac{(1-2x) (1+2x).3x}{x.(x+4).2(1-2x)}$ 

= $\frac{3(1+2x)}{2(x+4)}$ 

mong vote

#huyentran