giúp e bài này phần 2 với ạ

`1.`

        `(x^2-4)(4x-1)=(x-2)(16x^2-1)`

`⇔(x^2-4)(4x-1)-(x-2)(16x^2-1)=0`

`⇔(x-2)(x+2)(4x-1)-(x-2)(4x-1)(4x+1)=0`

`⇔(x-2)(4x-1)[(x+2)-(4x+1)]=0`

`⇔(x-2)(4x-1)(x+2-4x-1)=0`

`⇔(x-2)(4x-1)(1-3x)=0`

`⇔[(x-2=0),(4x-1=0),(1-3x=0):}`

`⇔[(x=2),(x=1/4),(x=1/3):}`

Vậy `S={2;1/4;1/3}`

`2.`

`M=( 1/(\sqrt{x}-1) + (\sqrt{x})/(x-1) ):( (\sqrt{x})/(\sqrt{x}-1) -1 )`

    `=( (\sqrt{x}+1)+ \sqrt{x} )/( (\sqrt{x}-1)(\sqrt{x}+1) ) : (\sqrt{x} - (\sqrt{x}-1) )/(\sqrt{x}-1)`

    `= (2\sqrt{x}+1)/( (\sqrt{x}-1)(\sqrt{x}+1) ) : 1/(\sqrt{x}-1)`

    `= (2\sqrt{x}+1)/( (\sqrt{x}-1)(\sqrt{x}+1) ) . (\sqrt{x}-1)`

    `=(2\sqrt{x}+1)/(\sqrt{x}+1)`

Vậy `M=(2\sqrt{x}+1)/(\sqrt{x}+1)` với `x≥0, x≠1`

`#` `ccUh`