a) $\sqrt{x+3}$ + $\sqrt{9x+27}$ - $\sqrt{4x+12}$ = `6` ( đk : `x` ≥ `-3` )
⇔ $\sqrt{x+3}$ + 3$\sqrt{x+3}$ - 2$\sqrt{x+3}$ = `6`
⇔ 2$\sqrt{x+3}$ = `6`
⇔ $\sqrt{x+3} = `3`
⇔ `x`+`3` = `9` ⇒ `x` = `6` ⊂ đkxđ
Vậy : `x` = `6`
b) $\sqrt{9-30x+25x²}$ = `2`
⇔ $\sqrt{(3-5x)²}$ = `2`
⇔ l `3` - `5x` l = `2`
⇔ \(\left[ \begin{array}{l}3-5x=2\\3-5x=-2\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=\frac{1}{5} \\x=1\end{array} \right.\)
Vậy : x = $\frac{1}{5}$ ; x = `1`