ĐK: $x∈\Bbb R$
$\dfrac{7}{x+\sqrt x+1}=4\\↔\dfrac{7}{x+\sqrt x+1}=\dfrac{4(x+\sqrt x+1)}{x+\sqrt x+1}\\→7=4x+4\sqrt x+4\\↔-4x-4\sqrt x+3=0\\↔4x+4\sqrt x-3=0\\↔4x+6\sqrt x-2\sqrt x-3=0\\↔(4x+6\sqrt x)-(2\sqrt x+3)=0\\↔2\sqrt x(2\sqrt x+3)-(2\sqrt x+3)=0\\↔(2\sqrt x-1)(2\sqrt x+3)=0$
Nhận thấy: $\sqrt x\ge 0$
$→2\sqrt x\ge 0\\↔2\sqrt x+3\ge 3>0$
mà $(2\sqrt x-1)(2\sqrt x+3)=0$
$→2\sqrt x-1=0\\↔2\sqrt x=1\\↔\sqrt x=\dfrac{1}{2}\\↔x=\dfrac{1}{4}(TM)$
Vậy $S=\{\dfrac{1}{4}\}$
