Question Number 79497 by john santu last updated on 25/Jan/20
$$\frac{\left(\mathrm{4}{x}−\mid{x}−\mathrm{6}\mid\right)\left(\mathrm{log}_{\frac{\mathrm{1}}{\mathrm{3}}} \left({x}+\mathrm{4}\right)+\mathrm{1}\right)}{\mathrm{2}^{{x}^{\mathrm{2}} −\mathrm{2}^{\mid{x}\mid} } }\geqslant\mathrm{0} \\ $$
Answered by john santu last updated on 25/Jan/20
![(((4x−∣x−6∣)((1/3)−1)(x+4−3))/((2−1)(x^2 −∣x∣)))≥0 ÷ (−(2/3)) (((4x−∣x−6∣)(x+1))/(∣x∣(∣x∣−1)))≤0 (((4x−∣x−6∣)(x+1))/(∣x∣(x−1)(x+1)))≤0 ((4x−∣x−6∣)/(x−1))≤0 , x≠−1, x≠0 , x>−4 ((x−1.2)/(x−1)) ≤0 ⇒ x∈(1, (6/5)]](Q79498.png)
$$\frac{\left(\mathrm{4}{x}−\mid{x}−\mathrm{6}\mid\right)\left(\frac{\mathrm{1}}{\mathrm{3}}−\mathrm{1}\right)\left({x}+\mathrm{4}−\mathrm{3}\right)}{\left(\mathrm{2}−\mathrm{1}\right)\left({x}^{\mathrm{2}} −\mid{x}\mid\right)}\geqslant\mathrm{0}\:\:\boldsymbol{\div}\:\left(−\frac{\mathrm{2}}{\mathrm{3}}\right)\:\: \\ $$$$\frac{\left(\mathrm{4}{x}−\mid{x}−\mathrm{6}\mid\right)\left({x}+\mathrm{1}\right)}{\mid{x}\mid\left(\mid{x}\mid−\mathrm{1}\right)}\leqslant\mathrm{0} \\ $$$$\frac{\left(\mathrm{4}{x}−\mid{x}−\mathrm{6}\mid\right)\left({x}+\mathrm{1}\right)}{\mid{x}\mid\left({x}−\mathrm{1}\right)\left({x}+\mathrm{1}\right)}\leqslant\mathrm{0} \\ $$$$\frac{\mathrm{4}{x}−\mid{x}−\mathrm{6}\mid}{{x}−\mathrm{1}}\leqslant\mathrm{0}\:,\:{x}\neq−\mathrm{1},\:{x}\neq\mathrm{0}\:,\:{x}>−\mathrm{4} \\ $$$$\frac{{x}−\mathrm{1}.\mathrm{2}}{{x}−\mathrm{1}}\:\leqslant\mathrm{0}\:\Rightarrow\:{x}\in\left(\mathrm{1},\:\frac{\mathrm{6}}{\mathrm{5}}\right] \\ $$
Commented by jagoll last updated on 26/Jan/20
$$\mathrm{mr}.\:\mathrm{how}\:\mathrm{to}\:\mathrm{get}\:\mathrm{4x}−\mid\mathrm{x}−\mathrm{6}\mid\:=\:\mathrm{x}−\mathrm{1}.\mathrm{2}? \\ $$