ĐK: -2 ≤ x ≤ 6
$\sqrt[]{x+2}$+ $\sqrt[]{6-x}$=4
<=> ($\sqrt[]{x+2}$+ $\sqrt[]{6-x}$)² = 16
<=> x+2+ 6-x+2$\sqrt[]{(x+2)(6-x)}$= 16
<=> 2$\sqrt[]{(x+2)(6-x)}$= 8
<=> $\sqrt[]{(x+2)(6-x)}$ =4
<=> (x+2)(6-x)= 16
<=> x²- 4x+ 8=0
<=> x²- 4x+ 4+ 4=0
<=> (x-2)²= -4
=> pt vô nghiệm (vì (x-2)²≥ 0)