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Question Number 78335 by john santu last updated on 16/Jan/20

find the solution of  (x/((x−2)^3 +(x−3)^3 −1)) ≥ 0

$${find}\:{the}\:{solution}\:{of} \\ $$$$\frac{{x}}{\left({x}−\mathrm{2}\right)^{\mathrm{3}} +\left({x}−\mathrm{3}\right)^{\mathrm{3}} −\mathrm{1}}\:\geqslant\:\mathrm{0} \\ $$

Answered by MJS last updated on 16/Jan/20

⇒ x>0∧((x−2)^3 +(x−3)^3 −1)>0 ∨        x=0∧((x−2)^3 +(x−3)^3 −1)≠0 ∨        x<0∧((x−2)^3 +(x−3)^3 −1)<0  (x−2)^3 +(x−3)^3 −1=(x−3)(2x^2 −9x+12)  ⇒  { ((((x−2)^3 +(x−3)^3 −1)<0; x<3)),((((x−2)^3 +(x−3)^3 −1)>0; x>3)) :}  ⇒  x≤0∨x>3

$$\Rightarrow\:{x}>\mathrm{0}\wedge\left(\left({x}−\mathrm{2}\right)^{\mathrm{3}} +\left({x}−\mathrm{3}\right)^{\mathrm{3}} −\mathrm{1}\right)>\mathrm{0}\:\vee \\ $$$$\:\:\:\:\:\:{x}=\mathrm{0}\wedge\left(\left({x}−\mathrm{2}\right)^{\mathrm{3}} +\left({x}−\mathrm{3}\right)^{\mathrm{3}} −\mathrm{1}\right)\neq\mathrm{0}\:\vee \\ $$$$\:\:\:\:\:\:{x}<\mathrm{0}\wedge\left(\left({x}−\mathrm{2}\right)^{\mathrm{3}} +\left({x}−\mathrm{3}\right)^{\mathrm{3}} −\mathrm{1}\right)<\mathrm{0} \\ $$$$\left({x}−\mathrm{2}\right)^{\mathrm{3}} +\left({x}−\mathrm{3}\right)^{\mathrm{3}} −\mathrm{1}=\left({x}−\mathrm{3}\right)\left(\mathrm{2}{x}^{\mathrm{2}} −\mathrm{9}{x}+\mathrm{12}\right) \\ $$$$\Rightarrow\:\begin{cases}{\left(\left({x}−\mathrm{2}\right)^{\mathrm{3}} +\left({x}−\mathrm{3}\right)^{\mathrm{3}} −\mathrm{1}\right)<\mathrm{0};\:{x}<\mathrm{3}}\\{\left(\left({x}−\mathrm{2}\right)^{\mathrm{3}} +\left({x}−\mathrm{3}\right)^{\mathrm{3}} −\mathrm{1}\right)>\mathrm{0};\:{x}>\mathrm{3}}\end{cases} \\ $$$$\Rightarrow \\ $$$${x}\leqslant\mathrm{0}\vee{x}>\mathrm{3} \\ $$

Commented by john santu last updated on 16/Jan/20

thanks sir

$${thanks}\:{sir} \\ $$

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