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Category: Algebra

Question-98770

Question Number 98770 by bemath last updated on 16/Jun/20 Commented by bemath last updated on 16/Jun/20 $$\mathrm{the}\:\mathrm{choice}\:\mathrm{answer}\: \\ $$$$\left(\mathrm{a}\right)\:\mathrm{12}\:\:\:\:\:\:\left(\mathrm{b}\right)\mathrm{13}\:\:\:\:\:\left(\mathrm{c}\right)\mathrm{14} \\ $$$$\left(\mathrm{d}\right)\mathrm{15}\:\:\:\:\:\left(\mathrm{e}\right)\:\mathrm{16} \\ $$ Commented by…

x-x-x-x-m-x-x-x-m-is-a-real-parameter-

Question Number 98761 by bemath last updated on 16/Jun/20 $$\sqrt{\mathrm{x}+\sqrt{\mathrm{x}}\:}\:−\sqrt{\mathrm{x}−\sqrt{\mathrm{x}}}\:=\:\mathrm{m}\sqrt{\frac{\mathrm{x}}{\mathrm{x}+\sqrt{\mathrm{x}}}} \\ $$$$\mathrm{m}\:\mathrm{is}\:\mathrm{a}\:\mathrm{real}\:\mathrm{parameter} \\ $$ Commented by MJS last updated on 16/Jun/20 $$\mathrm{if}\:{x}\in\mathbb{R} \\ $$$$\Rightarrow\:{x}>\mathrm{0}\wedge{x}−\sqrt{{x}}\geqslant\mathrm{0}\:\Rightarrow\:{x}\geqslant\mathrm{1} \\…

Question-164290

Question Number 164290 by HongKing last updated on 15/Jan/22 Answered by Kamel last updated on 16/Jan/22 $$ \\ $$$$\left.\Omega\left.\left({n}\right)=\int_{\mathrm{0}} ^{\pi} {cot}\left(\frac{{x}}{\mathrm{2}}\right){Arctan}\left(\frac{{nsin}\left({x}\right)}{\mathrm{1}+{ncos}\left({x}\right)}\right){dx};\:{u}=\frac{{n}−\mathrm{1}}{{n}+\mathrm{1}},\:{n}\in\right]−\mathrm{1};\mathrm{1}\right]. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}\int_{\mathrm{0}} ^{+\infty} \frac{\mathrm{1}}{{t}\left(\mathrm{1}+{t}^{\mathrm{2}}…

Solve-the-inequality-x-3-x-1-2-x-x-3-4-6-7x-2-29x-4-gt-0-Z-A-

Question Number 164269 by Zaynal last updated on 15/Jan/22 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{Solve}}\:\boldsymbol{{the}}\:\boldsymbol{{inequality}}; \\ $$$$\:\:\left(\boldsymbol{{x}}+\mathrm{3}\right)\left[\left(\boldsymbol{{x}}−\mathrm{1}\right)^{\mathrm{2}} −\boldsymbol{{x}}\left(\boldsymbol{{x}}+\frac{\mathrm{3}}{\mathrm{4}}\right)\right]+\mathrm{6}+\frac{\mathrm{7}\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{29}\boldsymbol{{x}}}{\mathrm{4}}>\mathrm{0} \\ $$$$\left\{\boldsymbol{{Z}}.\boldsymbol{{A}}\right\} \\ $$ Answered by manxsolar last updated on 15/Jan/22…