Question Number 227249 by Spillover last updated on 10/Jan/26 Answered by breniam last updated on 10/Jan/26 $$\frac{{x}−{y}\left({x}\right)+\mathrm{1}}{{x}+{y}\left({x}\right)−\mathrm{1}}=−\mathrm{1}+\frac{\mathrm{2}{x}+\mathrm{1}}{{x}+{y}\left({x}\right)−\mathrm{1}} \\ $$$${z}\left({x}\right)={x}+{y}\left({x}\right) \\ $$$${y}\left({x}\right)={z}\left({x}\right)−{x} \\ $$$${y}'\left({x}\right)={z}'\left({x}\right)−\mathrm{1} \\ $$$${z}'\left({x}\right)−\mathrm{1}=−\mathrm{1}+\frac{\mathrm{2}{x}+\mathrm{1}}{{z}\left({x}\right)−\mathrm{1}}…
Question Number 227250 by Spillover last updated on 10/Jan/26 Answered by breniam last updated on 10/Jan/26 $${z}\left({x}\right)={x}+{y}\left({x}\right) \\ $$$${y}\left({x}\right)={z}\left({x}\right)−{x} \\ $$$${y}'\left({x}\right)={z}'\left({x}\right)−\mathrm{1} \\ $$$${z}\left({x}\right)\left({z}'\left({x}\right)−\mathrm{1}\right)=\mathrm{2}{x}−{z}\left({x}\right)+\mathrm{2} \\ $$$${z}'\left({x}\right){z}\left({x}\right)=\mathrm{2}\left({x}+\mathrm{1}\right)…
Question Number 227236 by AlanMuhamad last updated on 09/Jan/26 $$\mid{x}−\mathrm{1}\mid\geqslant\mathrm{6} \\ $$$${x}−\mathrm{1}\geqslant\mathrm{6}\:\:\:\:\:{or}\:\:\:\:\:{x}−\mathrm{1}\leqslant−\mathrm{6} \\ $$$${x}\geqslant\mathrm{7}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}\leqslant−\mathrm{5} \\ $$$${s}_{\mathrm{1}} =\left[\mathrm{7},+\infty\right)\:\:\:\:\:\:\:\:{s}_{\mathrm{2}} =\left(−\infty,−\mathrm{5}\right] \\ $$$${s}=\left(−\infty,−\mathrm{5}\right]\:{U}\:\left[\mathrm{7},+\infty\right) \\ $$$${R}/\left(−\mathrm{5}\:,\:\mathrm{7}\:\right) \\ $$ Terms…
Question Number 227238 by as_ last updated on 09/Jan/26 $$ \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\underline{\boldsymbol{\mathrm{Essaie}}\:\boldsymbol{\mathrm{de}}\:\boldsymbol{\mathrm{corriger}}\:\boldsymbol{\mathrm{sur}}\:\boldsymbol{\mathrm{la}}\:\boldsymbol{\mathrm{trigos}}}\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\underline{\boldsymbol{\mathrm{Exercice}}}\:\mathrm{2}\:: \\ $$$$\:\:\:\mathrm{1}-\underline{\mathrm{Montrons}\:\mathrm{que}}\::\:\forall\mathrm{x}\in\mathbb{R},\:\mathrm{cos}^{\mathrm{6}} \mathrm{x}+\mathrm{sin}^{\mathrm{6}} \mathrm{x}=\frac{\mathrm{1}}{\mathrm{8}}\left(\mathrm{5}+\mathrm{3cos4x}\right)\:\:\:\:\: \\ $$$$\:\:\mathrm{Soit}\:\mathrm{x}\in\mathbb{R},\:\:\mathrm{On}\:\mathrm{a}\::\: \\…
Question Number 227232 by fantastic2 last updated on 07/Jan/26 $$\left(\mathrm{64}−\left(\mathrm{8}\pi+\mathrm{4}\pi\right)\right)−\left(\mathrm{8}\pi−\left(\mathrm{4}\left(\mathrm{tan}^{−\mathrm{1}} \mathrm{2}+\mathrm{4tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}}−\mathrm{2}\right)+\mathrm{8}\left(\pi−\mathrm{2}\right)\right)\right)−\left(\mathrm{64}−\left(\mathrm{16}\pi+\mathrm{8}\pi−\mathrm{16}\left(\mathrm{tan}^{−\mathrm{1}} \mathrm{2}+\mathrm{4tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}}−\mathrm{2}\right)\right)\right)−\left(\mathrm{3}.\mathrm{2}+\mathrm{2}+\left(\mathrm{32tan}^{−\mathrm{1}} \frac{\mathrm{4}.\mathrm{8}}{\mathrm{6}.\mathrm{4}}−\mathrm{32sin}\:\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{4}.\mathrm{8}}{\mathrm{6}.\mathrm{4}}\right)−\pi\right)+\mathrm{4}\left(\mathrm{tan}^{−\mathrm{1}} \mathrm{2}+\mathrm{4tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}}−\mathrm{2}\right) \\ $$ Terms of Service Privacy Policy…
Question Number 227220 by Spillover last updated on 06/Jan/26 $${Solve}\: \\ $$$$\:\:\:\:\:\:\:\frac{{dy}}{{dx}}=\frac{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }{{xy}} \\ $$$$\:\:\:\:\mathrm{6}/\mathrm{1}/\mathrm{2026} \\ $$ Answered by breniam last updated on 09/Jan/26…
Question Number 227221 by Spillover last updated on 06/Jan/26 $${Solve}\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\left({x}^{\mathrm{2}} +{xy}\right)\frac{{dy}}{{dx}}={xy}−{y}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\mathrm{6}/\mathrm{1}/\mathrm{2026} \\ $$ Answered by som(math1967) last updated on 07/Jan/26 $$\:\frac{{dy}}{{dx}}=\frac{{xy}−{y}^{\mathrm{2}}…
Question Number 227222 by Spillover last updated on 06/Jan/26 $$\:\:\:\:{Solve} \\ $$$$\:\:\:\:\:\:\:{x}\frac{{dy}}{{dx}}+{y}={x}^{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\mathrm{6}/\mathrm{1}/\mathrm{2026} \\ $$$$ \\ $$ Answered by som(math1967) last updated on 07/Jan/26…
Question Number 227200 by Tawa11 last updated on 06/Jan/26 Commented by Tawa11 last updated on 07/Jan/26 $$\mathrm{Thanks}\:\mathrm{sir}. \\ $$$$\mathrm{I}\:\mathrm{appreciate}. \\ $$ Commented by fantastic2 last…
Question Number 227219 by Spillover last updated on 06/Jan/26 $${Solve}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{{dy}}{{dx}}=\frac{{y}^{\mathrm{2}} +{xy}^{\mathrm{2}} }{{x}^{\mathrm{2}} {y}−{x}^{\mathrm{2}} } \\ $$$$\:\:\:\:\mathrm{6}/\mathrm{1}/\mathrm{2026} \\ $$$$ \\ $$ Answered by peace2…