Question

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

tìm gt của x 

Answer 1

`x/6+3/[x+2]=2/3`             `ĐK: x \ne -2`

`<=>[x(x+2)+3.6]/[6(x+2)]=[4(x+2)]/[6(x+2)]`

  `=>x^2+2x+18=4x+8`

`<=>x^2-2x+10=0`

`<=>x^2-2x+1+9=0`

`<=>(x-1)^2=-9` (Vô lí vì `(x-1)^2 >= 0` mà `-9 < 0`)

Vậy ptr vô nghiệm

Answer 2

\(\dfrac{x}{6}+\dfrac{3}{x+2}=\dfrac{2}{3}\left(đk:x\ne-2\right)\\ \Leftrightarrow x\left(x+2\right)+18=4\left(x+2\right)\\ \Leftrightarrow x^2+2x+18=4x+8\\ \Leftrightarrow x^2-2x+10=0\\ \Leftrightarrow x^2-2x+1+9=0\\ \Leftrightarrow\left(x-1\right)^2+9=0\left(vô.lý\right)\)