Question

rut gon bieu thuc :

\(A=\left(\dfrac{6x+1}{x^2-6x}+\dfrac{6x-1}{x^2+6x}\right).\dfrac{x^2-36}{x^2+1}\)

Answer 1

Với x khác 0 ; x khác 6 ; -6 

\(A=\left(\dfrac{\left(6x+1\right)\left(x+6\right)+\left(6x-1\right)\left(x-6\right)}{x\left(x+6\right)\left(x-6\right)}\right):\dfrac{x^2+1}{x^2-36}\)

\(=\dfrac{6x^2+37x+6+6x^2-37x+6}{x\left(x+6\right)\left(x-6\right)}:\dfrac{x^2+1}{x^2-36}=\dfrac{12x^2+12}{x\left(x+6\right)\left(x-6\right)}:\dfrac{x^2+1}{x^2-36}=\dfrac{12}{x}\)

 

Answer 2

Với `x \ne 0,x \ne +-6` có:

`A=([6x+1]/[x^2-6x]+[6x-1]/[x^2+6x]).[x^2-36]/[x^2+1]`

`A=[(6x+1)(x+6)+(6x-1)(x-6)]/[x(x+6)(x-6)].[(x-6)(x+6)]/[x^2+1]`

`A=[6x^2+36x+x+6+6x^2-36x-x+6]/[x(x^2+1)]`

`A=[12x^2+12]/[x(x^2+1)]`

A=[12(x^2+1)]/[x(x^2+1)]=12/x`

Answer 3

Với `x \ne 0,x \ne +-6` có:

`A=([6x+1]/[x^2-6x]+[6x-1]/[x^2+6x]).[x^2-36]/[x^2+1]`

`A=[(6x+1)(x+6)+(6x-1)(x-6)]/[x(x+6)(x-6)].[(x-6)(x+6)]/[x^2+1]`

`A=[6x^2+36x+x+6+6x^2-36x-x+6]/[x(x^2+1)]`

`A=[12x^2+12]/[x(x^2+1)]`

`A=[12(x^2+1)]/[x(x^2+1)]=12/x`