Đáp án:

 

Giải thích các bước giải:

 $A=\dfrac{x+1}{2x+6}+\dfrac{2x+3}{x(x+3)}$

$A=\dfrac{x+1}{2(x+3)}+\dfrac{2x+3}{x(x+3)}$

$A=\dfrac{x(x+1)}{2x(x+3)}+\dfrac{2(2x+3)}{2x(x+3)}$

$A=\dfrac{x^{2}+x+4x+6}{2x(x+3)}$

$A=\dfrac{x^{2}+5x+6}{2x(x+3)}$

$A=\dfrac{x^{2}+2x+3x+6}{2x(x+3)}$

$A=\dfrac{(x^{2}+2x)+(3x+6)}{2x(x+3)}$

$A=\dfrac{x(x+2)+3(x+2)}{2x(x+3)}$

$A=\dfrac{(x+2)(x+3)}{2x(x+3)}$

$A=\dfrac{x+2}{2x}$

(Bài tham khảo)

Câu B thiếu đề bn nhá

Đáp án+Giải thích các bước giải:

 `A=(x+1)/(2x+6)+(2x+3)/[x(x+3)]` (điều kiện xác định: `x \ne 0; -3`)

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

Mẫu thức chung: `2x(x+3)`

`=> A=[x(x+1)]/[2x(x+3)]+[2(2x+3)]/[2x(x+3)]`

`=(x^2+x)/[2x(x+3)]+(4x+6)/[2x(x+3)]`

`=(x^2+x+4x+6)/[2x(x+3)]`

`=(x^2+5x+6)/[2x(x+3)]`

`=(x^2+3x+2x+6)/[2x(x+3)]`

`=[(x^2+3x)+(2x+6)]/[2x(x+3)]`

`=[x(x+3)+2(x+3)]/[2x(x+3)]`

`=[(x+2)(x+3)]/[2x(x+3)]`

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

`B=x/(x^2-4)+1/(x+2)`

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

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

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

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