ĐK $-1 \le x<0$ hoặc $x \ge 1$
$\sqrt{x+1}=\sqrt{\dfrac{x-1}{x}}+1 \\ \leftrightarrow x+1=\dfrac{x-1}{x}+1+2.\sqrt{\dfrac{x-1}{x}} \\ \rightarrow x^2+x=x-1+x+2x.\sqrt{\dfrac{x-1}{x}} \\ \leftrightarrow x^2-x+1=2.x\sqrt{\dfrac{x-1}{x}}$
Xét $x \ge 1$ có $x^2-x+1=2.\sqrt{x^2-x} \leftrightarrow (\sqrt{x^2-x}-1)^2=0 \leftrightarrow x^2-x-1=0 \leftrightarrow x=\dfrac{1+\sqrt{5}}{2}$
Xét $-1 \le x <0$ có $x^2-x+1=-2.\sqrt{x^2-x}$ (VN)
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