Tìm x a. sqrt(4x+8) + 4x sqrt( (x+2)/4 ) - sqrt(9x+18) = 0 b. sqrt(x^2-x-6)=sqrt(x-3) c. sqrt( (x-7)/(x-3) )=sqrt(x+3) giúp mình với,mình đang cần gấp

a) $\sqrt{4x + 8}$ + 4$\sqrt{\frac{x + 2}{4}}$ - $\sqrt{9x + 18}$ = 0 Đkxđ: x ≥ -2

⇔ 2$\sqrt{x + 2}$ + 2$\sqrt{x + 2}$ - 3$\sqrt{x + 2}$ = 0

⇔ $\sqrt{x + 2}$ = 0

⇔ x + 2 = 0

⇔ x = -2 (t/m)

Vậy x = -2

b) $\sqrt{x² - x - 6}$ = $\sqrt{x - 3}$ Đkxđ:  x ≥ 3

⇔ x² - x - 6 = x - 3

⇔ x² - 2x - 3 = 0

⇔ x² - 3x + x - 3 = 0

⇔ x.(x - 3) + (x - 3) = 0

⇔ (x - 3)(x + 1) = 0

⇔ \(\left[ \begin{array}{l}x - 3 = 0\\x + 1 = 0\end{array} \right.\) 

⇔ \(\left[ \begin{array}{l}x = 3 (T/m)\\x = -1 (Không t/m)\end{array} \right.\) 

Vậy x = 3

c) $\sqrt{\frac{x - 7}{x - 3}}$ = $\sqrt{x + 3}$ Đkxđ: -3 < x < 3; x ≥ 7

⇔ $\frac{x - 7}{x - 3}$ = x + 3

⇔ x - 7 = (x - 3)(x + 3)

⇔ x - 7 = x² - 9

⇔ x² - x - 2 = 0

⇔ x² - 2x + x - 2 = 0

⇔ x.(x - 2) + (x - 2) = 0

⇔ (x - 2)(x + 1) = 0

⇔ \(\left[ \begin{array}{l}x - 2 = 0\\x + 1 = 0\end{array} \right.\) 

⇔ \(\left[ \begin{array}{l}x = 2 (t/m)\\x = -1 (t/m)\end{array} \right.\) 

Vậy x ∈ {2; -1}