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Question Number 188704 by mathlove last updated on 05/Mar/23

Answered by a.lgnaoui last updated on 05/Mar/23

posons   Z=  avec:A=x(√x) −x  =x((√x)  −1)        B= (((x^3 )^(1/4)  −1)/( (x)^(1/4)  −1))   −(√x)       C=((((x^3  ))^(1/4)  +1)/( (x)^(1/4)  +1))−(√(x ))   B=((  (√x) −1 )/( (x)^(1/4)  −1))  C=((1−(√x))/( (x)^(1/4)  +1))  B×C=1−(√x)   ⇒ (A/(B×C))=((x((√x) −1))/( (√x) −1))      Z^3 =(((x(√x) −1)/(1−(√x))))^3 =    [((x(√x) −x)/(((((x^3 )^(1/4)  −1)/( (x)^(1/4)  −1))−(√x) )((((x^3 )^(1/4)  +1)/( (x)^(1/4)  +1))−(√x) )))]^3 =−x^3 .

$${posons}\:\:\:{Z}=\:\:{avec}:{A}={x}\sqrt{{x}}\:−{x}\:\:={x}\left(\sqrt{{x}}\:\:−\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:{B}=\:\frac{\sqrt[{\mathrm{4}}]{{x}^{\mathrm{3}} }\:−\mathrm{1}}{\:\sqrt[{\mathrm{4}}]{{x}}\:−\mathrm{1}}\:\:\:−\sqrt{{x}}\:\:\:\:\:\:\:{C}=\frac{\sqrt[{\mathrm{4}}]{{x}^{\mathrm{3}} \:}\:+\mathrm{1}}{\:\sqrt[{\mathrm{4}}]{{x}}\:+\mathrm{1}}−\sqrt{{x}\:}\: \\ $$$${B}=\frac{\:\:\sqrt{{x}}\:−\mathrm{1}\:}{\:\sqrt[{\mathrm{4}}]{{x}}\:−\mathrm{1}}\:\:{C}=\frac{\mathrm{1}−\sqrt{{x}}}{\:\sqrt[{\mathrm{4}}]{{x}}\:+\mathrm{1}}\:\:{B}×{C}=\mathrm{1}−\sqrt{{x}}\: \\ $$$$\Rightarrow\:\frac{{A}}{{B}×{C}}=\frac{{x}\left(\sqrt{{x}}\:−\mathrm{1}\right)}{\:\sqrt{{x}}\:−\mathrm{1}}\:\:\:\: \\ $$$${Z}^{\mathrm{3}} =\left(\frac{\boldsymbol{{x}}\sqrt{\boldsymbol{{x}}}\:−\mathrm{1}}{\mathrm{1}−\sqrt{\boldsymbol{{x}}}}\right)^{\mathrm{3}} = \\ $$$$ \\ $$$$\left[\frac{\boldsymbol{{x}}\sqrt{\boldsymbol{{x}}}\:−\boldsymbol{{x}}}{\left(\frac{\sqrt[{\mathrm{4}}]{\boldsymbol{{x}}^{\mathrm{3}} }\:−\mathrm{1}}{\:\sqrt[{\mathrm{4}}]{\boldsymbol{{x}}}\:−\mathrm{1}}−\sqrt{\boldsymbol{{x}}}\:\right)\left(\frac{\sqrt[{\mathrm{4}}]{\boldsymbol{{x}}^{\mathrm{3}} }\:+\mathrm{1}}{\:\sqrt[{\mathrm{4}}]{\boldsymbol{{x}}}\:+\mathrm{1}}−\sqrt{\boldsymbol{{x}}}\:\right)}\right]^{\mathrm{3}} =−\boldsymbol{{x}}^{\mathrm{3}} . \\ $$

Commented by mathlove last updated on 06/Mar/23

thanks

$${thanks} \\ $$

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