a, $\frac{\sqrt {x-2}}{\sqrt {x+2}}$ =$\frac 12$
⇔$2{\sqrt {x-2}}$ =${\sqrt {x+2}}$
⇔ $\left \{ {4x-8\geq0 \atop x+2=4(x-2)} \right.$
⇔ $\left \{ {4x\geq8 \atop x+2=4x-8} \right.$
⇔ $\left \{ {x\geq2 \atop x-4x=-8-2} \right.$
⇔ $\left \{ {x\geq2 \atop -3x=-10} \right.$
⇔ $\left \{ {x\geq2 \atop x=\frac{10}3 (N)} \right.$
S={$\frac{10}3$}
b, 2 $\frac{\sqrt x+1}{\sqrt x+1}$ =$\frac 32$
⇔2*1 =$\frac 32$ (!)
⇔S=∅
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