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Question-192570




Question Number 192570 by mechanics last updated on 21/May/23
Answered by cortano12 last updated on 21/May/23
 ∣x^2 −4∣<5   (x^2 −9)(x^2 +1)<0   (x+3)(x−3)(x^2 +1)<0    ∴ −3 < x < 3
$$\:\mid\mathrm{x}^{\mathrm{2}} −\mathrm{4}\mid<\mathrm{5} \\ $$$$\:\left(\mathrm{x}^{\mathrm{2}} −\mathrm{9}\right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)<\mathrm{0} \\ $$$$\:\left(\mathrm{x}+\mathrm{3}\right)\left(\mathrm{x}−\mathrm{3}\right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)<\mathrm{0} \\ $$$$\:\:\therefore\:−\mathrm{3}\:<\:\mathrm{x}\:<\:\mathrm{3}\: \\ $$
Answered by Skabetix last updated on 21/May/23
∣x∣+∣x+2∣+∣2−x∣≤8⇒−(8/3)≤x≤(8/3)
$$\mid{x}\mid+\mid{x}+\mathrm{2}\mid+\mid\mathrm{2}−{x}\mid\leqslant\mathrm{8}\Rightarrow−\frac{\mathrm{8}}{\mathrm{3}}\leqslant{x}\leqslant\frac{\mathrm{8}}{\mathrm{3}} \\ $$

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