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## How many real roots does the equation have? ### Solution

$\sqrt{x}=1+\frac{x-1}{1+\sqrt{x}}\;\;\;\;\;\;\;(1)$ Let $$x\geq0$$ and $$\sqrt{x}=t$$, thus $t=1+\frac{t^{2}-1}{1+t}$ $t(1+t)=1+t+t^{2}-1$ $t+t^{2}=1+t+t^{2}-1$ $0=0$ Therefore, the equation $$(1)$$ is an identity if only $$x\geq0$$ $\huge S_{\mathbb{R}}=[0,+\infty[$
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