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Question-153239

Question Number 153239 by rexford last updated on 06/Sep/21 Answered by MJS_new last updated on 06/Sep/21 $${z}={a}+{b}\mathrm{i} \\ $$$${w}=−{b}+{a}\mathrm{i} \\ $$$${z}+{w}\mathrm{i}={a}+{b}\mathrm{i}+\left(−{b}\mathrm{i}−{a}\right)=\mathrm{0} \\ $$$${zw}=\left({a}+{b}\mathrm{i}\right)\left(−{b}+{a}\mathrm{i}\right)=−\mathrm{2}{ab}+\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)\mathrm{i}…

let-a-b-N-a-b-a-b-ab-a-n-a-n-1-a-explicite-a-n-en-fonction-de-a-

Question Number 153214 by pticantor last updated on 05/Sep/21 $$\boldsymbol{{let}}\:\boldsymbol{{a}},\boldsymbol{{b}}\in\mathbb{N}^{\ast} \: \\ $$$$\:\boldsymbol{{a}}\ast\boldsymbol{{b}}=\boldsymbol{{a}}+\boldsymbol{{b}}+\boldsymbol{{ab}} \\ $$$$\boldsymbol{{a}}^{\left(\boldsymbol{{n}}\right)} =\boldsymbol{{a}}^{\left(\boldsymbol{{n}}−\mathrm{1}\right)} \ast\boldsymbol{{a}} \\ $$$$\boldsymbol{{explicite}}\:\boldsymbol{{a}}^{\left(\boldsymbol{{n}}\right)} \:\boldsymbol{{en}}\:\boldsymbol{{fonction}}\:\boldsymbol{{de}}\:\boldsymbol{{a}} \\ $$$$ \\ $$$$ \\…

1-1-ln-x-dx-

Question Number 153164 by alisiao last updated on 05/Sep/21 $$\int_{−\mathrm{1}} ^{\:\mathrm{1}} \boldsymbol{{ln}}\left(\boldsymbol{\Gamma}\left(\boldsymbol{{x}}\right)\right)\:\boldsymbol{{dx}} \\ $$ Answered by Ar Brandon last updated on 05/Sep/21 $$\Omega=\int_{−\mathrm{1}} ^{\mathrm{1}} \mathrm{ln}\left(\Gamma\left({x}\right)\right){dx}=\underset{\Omega_{\mathrm{1}}…

x-y-2-y-2x-6-25-x-y-1-2x-y-5-

Question Number 87614 by mary_ last updated on 05/Apr/20 $$\begin{cases}{\left({x}+{y}\right).\mathrm{2}^{{y}−\mathrm{2}{x}} =\mathrm{6}.\mathrm{25}}\\{\left({x}+{y}\right)^{\frac{\mathrm{1}}{\mathrm{2}{x}−{y}}} =\mathrm{5}}\end{cases} \\ $$ Answered by mahdi last updated on 05/Apr/20 $$\left(\mathrm{x}+\mathrm{y}\right)^{\frac{\mathrm{1}}{\mathrm{2x}−\mathrm{y}}} =\mathrm{5}\Rightarrow\left(\mathrm{x}+\mathrm{y}\right)=\mathrm{5}^{\mathrm{2x}−\mathrm{y}} \\ $$$$\left(\mathrm{x}+\mathrm{y}\right).\mathrm{2}^{\mathrm{y}−\mathrm{2x}}…

Question-153146

Question Number 153146 by SANOGO last updated on 05/Sep/21 Answered by puissant last updated on 05/Sep/21 $${posons}\:{u}\left({n}\right)=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{a}\left({n}\right)\:{et}\:{prenons}\:{f}\:{une} \\ $$$${fonction}\:{de}\:{classe}\:{C}^{\mathrm{1}} ,\:{alors}\:{posons} \\ $$$${a}\left({n}\right)=\mathrm{1}\:{pour}\:{tout}\:{n}\:{et}\:{f}\left({n}\right)=\frac{\mathrm{1}}{{n}} \\…