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Category: Algebra

Question-170623

Question Number 170623 by cortano1 last updated on 27/May/22 Answered by greougoury555 last updated on 28/May/22 $${d}=\sqrt{{x}^{\mathrm{2}} +\mathrm{81}}\:+\sqrt{\left({x}−\mathrm{5}\right)^{\mathrm{2}} +\mathrm{36}} \\ $$$${d}'=\frac{{x}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{81}}}\:+\frac{{x}−\mathrm{5}}{\:\sqrt{\left({x}−\mathrm{5}\right)^{\mathrm{2}} +\mathrm{36}}}\:=\mathrm{0} \\ $$$$\Rightarrow\frac{{x}}{\:\sqrt{{x}^{\mathrm{2}}…

Question-170568

Question Number 170568 by mathlove last updated on 27/May/22 Answered by Rasheed.Sindhi last updated on 27/May/22 $$\begin{array}{|c|}{\underset{\:\mathrm{f}\left(\mathrm{x}−\mathrm{1}\right)+\mathrm{g}\left(\mathrm{2x}+\mathrm{1}\right)=\mathrm{2x}\:\:\:\:\:\:\:\:\:\:} {\mathrm{f}\left(\mathrm{2x}+\mathrm{2}\right)+\mathrm{2g}\left(\mathrm{4x}+\mathrm{7}\right)=\mathrm{x}−\mathrm{1}}}\\\hline\end{array} \\ $$$$\left(\mathrm{1}\right):\:\:\:\:\mathrm{2x}+\mathrm{2}=\mathrm{y}\Rightarrow\mathrm{x}=\frac{\mathrm{y}−\mathrm{2}}{\mathrm{2}}\:: \\ $$$$\:\:\:\:\mathrm{f}\left(\mathrm{y}\right)+\mathrm{2g}\left(\mathrm{4}\centerdot\frac{\mathrm{y}−\mathrm{2}}{\mathrm{2}}+\mathrm{7}\right)=\frac{\mathrm{y}−\mathrm{2}}{\mathrm{2}}−\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\Rightarrow\mathrm{f}\left(\mathrm{y}\right)+\mathrm{2g}\left(\mathrm{2y}+\mathrm{3}\right)=\frac{\mathrm{y}−\mathrm{4}}{\mathrm{2}}…\left(\mathrm{3}\right) \\…

x-2-y-2-2-1-x-2-y-2-1-1-x-2-2-x-2-y-2-y-2-1-y-2-1-2-1-x-2-x-2-y-2-

Question Number 105023 by bemath last updated on 25/Jul/20 $$\frac{\left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right)^{\mathrm{2}} }{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)\left({y}^{\mathrm{2}} −\mathrm{1}\right)}+\frac{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{\left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right)\left({y}^{\mathrm{2}} −\mathrm{1}\right)}+\frac{\left({y}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)\left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right)}=…

2-3-i-is-a-cubic-root-for-18-3-35i-find-the-other-2-cubic-roots-

Question Number 104973 by malwaan last updated on 25/Jul/20 $$\mathrm{2}\sqrt{\mathrm{3}}\:+\:\boldsymbol{{i}}\:\:\:{is}\:{a}\:{cubic}\:{root}\:{for} \\ $$$$\mathrm{18}\sqrt{\mathrm{3}}\:+\:\mathrm{35}\boldsymbol{{i}}\: \\ $$$$\boldsymbol{{find}}\:\:\boldsymbol{{the}}\:\boldsymbol{{other}}\:\mathrm{2}\:\boldsymbol{{cubic}}\:\boldsymbol{{roots}} \\ $$ Commented by malwaan last updated on 25/Jul/20 شكرا جزيلا سيدي الفاضل واذا كان لديك طريقة او لوحة مفاتيح لكتابة المعادلات والنهايات والتكاملات وغيرها باللغة العربية فاتمنى تساعدني مع فائق احترامي محمد علوان اليمن Answered…