Question

`(x-9)/48+(x+6)/12=(x-3)/4`

Answer 1

Ta có: \(\dfrac{x-9}{48}+\dfrac{x+6}{12}=\dfrac{x-3}{4}\)

\(\Leftrightarrow\dfrac{x-9}{48}+\dfrac{4\left(x+6\right)}{48}=\dfrac{12\left(x-3\right)}{48}\)

\(\Leftrightarrow x-9+4x+24=12x-36\)

\(\Leftrightarrow5x+15-12x+36=0\)

\(\Leftrightarrow-7x+51=0\)

\(\Leftrightarrow-7x=-51\)

hay \(x=\dfrac{51}{7}\)

Vậy: \(x=\dfrac{51}{7}\)

Answer 2

\(\dfrac{x-9}{48}+\dfrac{x+6}{12}=\dfrac{x-3}{4}\)

\(\Leftrightarrow\dfrac{x-9}{48}+\dfrac{4x+24}{48}=\dfrac{12x-36}{48}\)

\(\Leftrightarrow7x=51\Leftrightarrow x=\dfrac{51}{7}\)

Answer 3

`(x-9)/48+(x+6)/12=(x-3)/4`

`=>x-9+4(x+6)=12(x-3)`

`=>x-9+4x+24=12x-36`

`=>5x+15=12x-36`

`=>7x=51`

`=>x=51/7`

Vậy `x=51/7`