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Question Number 152898 by bobhans last updated on 02/Sep/21

Commented by mathdanisur last updated on 03/Sep/21

$${5\mathrm{x}}^5-{23\mathrm{x}}^4+{39\mathrm{x}}^3-{33\mathrm{x}}^2+24\mathrm{x}=4\Rightarrow =2$$

Commented by MJS_new last updated on 03/Sep/21

$$\mathrm{solution}\mathrm{please}$$

Answered by mindispower last updated on 03/Sep/21

$$(\sqrt{1-x^2}+2+\frac{1+x}{2}+\sqrt{\frac{1+x}{1-x}})=\frac{x^2+1}{x}$$$$(\sqrt{1-x^2}+\sqrt{\frac{1+x}{1-x}})=\frac{2(x^2+1)-5x-x^2}{2x}$$$$(1-x^2+\frac{1+x}{1-x}+2(1+x))={\left(\frac{x^2-5x+2}{2x}\right)}^2$$$$\Rightarrow 4x^2((1-x^2)(1-x)+(1+x)+2(1-x^2))$$$$=(1-x)(x^4+25x^2-10x+4-20x+4x^2)$$$$=4x^2(x^3-3x^2+4)=(1-x)(x^4+29x^2-30x+4)$$$$5x^5-13x^4+29x^3-43x^2+34x-4=0...$$$$$$$$$$$$$$$$$$

Commented by mathdanisur last updated on 03/Sep/21

$${\left(\frac{{\mathrm{x}}^2-5\mathrm{x}+2}{{2\mathrm{x}}^2-4\mathrm{x}}\right)}^2=\frac{1+\mathrm{x}}{1-\mathrm{x}}$$$$(1-\mathrm{x}){({\mathrm{x}}^2-5\mathrm{x}+2)}^2=(1+\mathrm{x}){({2\mathrm{x}}^2-4\mathrm{x})}^2$$$$\mathrm{S}={5\mathrm{x}}^5-{23\mathrm{x}}^4+{39\mathrm{x}}^3-{33\mathrm{x}}^2+24\mathrm{x}=4$$$$\mathrm{S}=\sqrt{4}\Rightarrow \mathrm{S}=2$$

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