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

Commented by mathdanisur last updated on 03/Sep/21

5x^5 −23x^4 +39x^3 −33x^2 +24x=4 ⇒ =2

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

Commented by MJS_new last updated on 03/Sep/21

solution please

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

Answered by mindispower last updated on 03/Sep/21

((√(1−x^2 ))+2+((1+x)/2)+(√((1+x)/(1−x))))=((x^2 +1)/x) ((√(1−x^2 ))+(√((1+x)/(1−x))))=((2(x^2 +1)−5x−x^2 )/(2x)) (1−x^2 +((1+x)/(1−x))+2(1+x))=(((x^2 −5x+2)/(2x)))^2 ⇒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...

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

Commented by mathdanisur last updated on 03/Sep/21

(((x^2 -5x+2)/(2x^2 -4x)))^2 = ((1+x)/(1-x)) (1-x)(x^2 -5x+2)^2 =(1+x)(2x^2 -4x)^2 S=5x^5 -23x^4 +39x^3 -33x^2 +24x=4 S=(√4) ⇒ S=2

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

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