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Question-145829




Question Number 145829 by bramlexs22 last updated on 08/Jul/21
Answered by EDWIN88 last updated on 08/Jul/21
(1)(√(x+((15)/4))) ≠ 1 ; x≠ −((11)/4)    and x >−((15)/4)  (2)(x^2 −(8/3)x)>0⇒x(x−(8/3))>0  ⇒x<0 ∪ x>(8/3)  (3) let log _(√(x+((15)/4))) (x^2 −(8/3)x)=t  ⇒t+(4/t) ≤ 4 ; ((t^2 −4t+4)/t) ≤ 0  ⇒ (((t−2)^2 )/t) ≤ 0 we get t <0  log _(√(x+((15)/4))) (x^2 −(8/3)x)< 0  ⇒x^2 −(8/3)x−1< 0  ⇒3x^2 −8x−3 <0  ⇒(3x+1)(x−3)<0  ⇒−(1/3)<x<3  solution set we get from   (1)∩(2)∩(3)  ⇒ (8/3)<x<3 ∪−(1/3)< x<0
$$\left(\mathrm{1}\right)\sqrt{{x}+\frac{\mathrm{15}}{\mathrm{4}}}\:\neq\:\mathrm{1}\:;\:{x}\neq\:−\frac{\mathrm{11}}{\mathrm{4}} \\ $$$$\:\:{and}\:{x}\:>−\frac{\mathrm{15}}{\mathrm{4}} \\ $$$$\left(\mathrm{2}\right)\left({x}^{\mathrm{2}} −\frac{\mathrm{8}}{\mathrm{3}}{x}\right)>\mathrm{0}\Rightarrow{x}\left({x}−\frac{\mathrm{8}}{\mathrm{3}}\right)>\mathrm{0} \\ $$$$\Rightarrow{x}<\mathrm{0}\:\cup\:{x}>\frac{\mathrm{8}}{\mathrm{3}} \\ $$$$\left(\mathrm{3}\right)\:{let}\:\mathrm{log}\:_{\sqrt{{x}+\frac{\mathrm{15}}{\mathrm{4}}}} \left({x}^{\mathrm{2}} −\frac{\mathrm{8}}{\mathrm{3}}{x}\right)={t} \\ $$$$\Rightarrow{t}+\frac{\mathrm{4}}{{t}}\:\leqslant\:\mathrm{4}\:;\:\frac{{t}^{\mathrm{2}} −\mathrm{4}{t}+\mathrm{4}}{{t}}\:\leqslant\:\mathrm{0} \\ $$$$\Rightarrow\:\frac{\left({t}−\mathrm{2}\right)^{\mathrm{2}} }{{t}}\:\leqslant\:\mathrm{0}\:{we}\:{get}\:{t}\:<\mathrm{0} \\ $$$$\mathrm{log}\:_{\sqrt{{x}+\frac{\mathrm{15}}{\mathrm{4}}}} \left({x}^{\mathrm{2}} −\frac{\mathrm{8}}{\mathrm{3}}{x}\right)<\:\mathrm{0} \\ $$$$\Rightarrow{x}^{\mathrm{2}} −\frac{\mathrm{8}}{\mathrm{3}}{x}−\mathrm{1}<\:\mathrm{0} \\ $$$$\Rightarrow\mathrm{3}{x}^{\mathrm{2}} −\mathrm{8}{x}−\mathrm{3}\:<\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{3}{x}+\mathrm{1}\right)\left({x}−\mathrm{3}\right)<\mathrm{0} \\ $$$$\Rightarrow−\frac{\mathrm{1}}{\mathrm{3}}<{x}<\mathrm{3} \\ $$$${solution}\:{set}\:{we}\:{get}\:{from}\: \\ $$$$\left(\mathrm{1}\right)\cap\left(\mathrm{2}\right)\cap\left(\mathrm{3}\right) \\ $$$$\Rightarrow\:\frac{\mathrm{8}}{\mathrm{3}}<{x}<\mathrm{3}\:\cup−\frac{\mathrm{1}}{\mathrm{3}}<\:{x}<\mathrm{0}\: \\ $$

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