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Question Number 140329 by benjo_mathlover last updated on 06/May/21

$${\mathrm{x}}^{\lfloor \mathrm{x}\rfloor }+{\mathrm{x}}^{\lceil \mathrm{x}\rceil }=\frac{175}{8}$$

Answered by john_santu last updated on 06/May/21

$$0<x<1\to x^0+x^1=\frac{175}{8}$$$$\Rightarrow x=\frac{167}{8}>20$$$$1<x<2\Rightarrow x^1+x^2=\frac{175}{8}$$$$\Rightarrow 8x^2+8x-175=0$$$$\Rightarrow x=\frac{-8\pm \sqrt{64+32.175}}{16}$$$$2<x<3\Rightarrow x^2+x^3=\frac{175}{8}$$$$\Rightarrow 8x^3+8x^2-175=0$$$$\Rightarrow {\left(2x\right)}^3+2{\left(2x\right)}^2-175=0$$$$let2x=y\Rightarrow y^3+2y^2-175=0$$$$(y-5)(y^2+5y+35)=0$$$$\Rightarrow y=5=2x;x=\frac{5}{2}.\left(solution\right)$$

Answered by mr W last updated on 06/May/21

$$letx=n+fwith0⩽f<1$$$${(n+f)}^n+{(n+f)}^{n+1}=\frac{175}{8}$$$$\frac{175}{8}={(n+f)}^n+{(n+f)}^{n+1}⩾n^n+n^{n+1}=(n+1)n^n$$$$\Rightarrow n⩽2$$$$\frac{175}{8}={(n+f)}^n+{(n+f)}^{n+1}⩽{(n+1)}^n+{(n+1)}^{n+1}=(n+2){(n+1)}^n$$$$\Rightarrow n⩾2$$$$\Rightarrow n=2$$$$x^2+x^3=\frac{175}{8}$$$$(x-\frac{5}{2})(x^2+\frac{7x}{2}+\frac{35}{4})=0$$$$\Rightarrow x=\frac{5}{2}$$

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