Let [t] denote the greatest integer leq t. Th | vedclass

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(D) Given the equation: $[x]^{2}+2[x+2]-7=0$ Using the property $[x+n] = [x]+n$ for any integer $n$,we have $[x+2] = [x]+2$. Substituting this into th

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infinitely many solutions

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(D) Given the equation: $[x]^{2}+2[x+2]-7=0$ Using the property $[x+n] = [x]+n$ for any integer $n$,we have $[x+2] = [x]+2$. Substituting this into the equation: $[x]^{2}+2([x]+2)-7=0$ $[x]^{2}+2[x]+4-7=0$ $[x]^{2}+2[x]-3=0$ Let $y = [x]$,then $y^{2}+2y-3=0$ $(y+3)(y-1)=0$ So,$[x] = 1$ or $[x] = -3$ If $[x] = 1$,then $x \in [1, 2)$ If $[x] = -3$,then $x \in [-3, -2)$ Thus,the solution set is $x \in [-3, -2) \cup [1, 2)$,which contains infinitely many real values.

Let $[t]$ denote the greatest integer $\leq t$. Then the equation in $x$,$[x]^{2}+2[x+2]-7=0$ has

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