Đáp án:
A=($\frac{x+\sqrt[]{x}}{x\sqrt[]{x}+x+\sqrt[]{x}+1}$+$\frac{1}{x+1}$): $\frac{\sqrt[]{x}-1}{x+1}$
=($\frac{\sqrt[]{x(\sqrt[]{x}+1)}}{(x+1)(\sqrt[]{x}+1)}$+ $\frac{1}{x+1}$ ):$\frac{\sqrt[]{x}-1}{x+1}$
=($\frac{\sqrt[]{x}}{x+1}$+ $\frac{1}{x+1}$): $\frac{\sqrt[]{x}-1}{x+1}$
=$\frac{\sqrt[]{x}+1}{\sqrt[]{x}-1}$
Để A có nghĩ ⇔ $\frac{\sqrt[]{x}+1}{\sqrt[]{x}-1}$ $\geq$ 0
⇔$\sqrt[]{x}$ +1$\geq$ 0
$\sqrt[]{x}$ -1 $\geq$ 0
x$\geq$ 0
⇔x$\neq$ ±1,x$\geq$ 0
b) Ta có : $\frac{\sqrt[]{x}+1}{\sqrt[]{x}-1}$ =1+$\frac{2}{\sqrt[]{x}-1}$
Để A ∈ Z ⇔ $\frac{2}{\sqrt[]{x}-1}$ ∈ Z
⇒ $\sqrt[]{x}$ -1∈Ư(2) = {±1,±2}
Ta có bảng gtri:
$\sqrt[]{x}$ -1 -1 1 -2 2
$\sqrt[]{x}$ 0 2 -1 3
x 0 4 9
ĐCĐK TM TM TM
Vậy ..............
