Question

Tính
1) \(\dfrac{x^3-8}{5x+20}\cdot\dfrac{x^2+4x}{x^2+2x+4}\)

2) \(\dfrac{x+1}{x^2-2x-8}\cdot\dfrac{4-x}{x^2+x}\)

3)  \(\dfrac{3x^2-x}{x^2-1}\cdot\dfrac{1-x^4}{\left(1-3x\right)^3}\)

 

mn giúp mik với ạ ;-;

Answer 1

`1)[x^3-8]/[5x+20].[x^2+4x]/[x^2+2x+4]`

`=[(x-2)(x^2+2x+4)]/[5(x+4)].[x(x+4)]/[x^2+2x+4]`

`=[x(x-2)]/5=[x^2-2x]/5`

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`2)[x+1]/[x^2-2x-8].[4-x]/[x^2+x]`

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

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

_____________________________________________

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

`=[x(3x-1)]/[x^2-1].[-(x^2-1)(x^2+1)]/[-(3x-1)(9x^2+3x+1)]`

`=[x(x^2+1)]/[9x^2+3x+1]=[x^3+x]/[9x^2+3x+1]`