Question Number 172027 by Mikenice last updated on 23/Jun/22 $${solve} \\ $$$$\frac{\mathrm{5}}{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{2}}=\frac{\mathrm{3}}{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{1}}−\frac{\mathrm{4}}{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{8}} \\ $$ Commented by Mikenice last updated on 23/Jun/22 $${thanks}\:{sir}…
Question Number 172020 by Mikenice last updated on 23/Jun/22 $${solve}: \\ $$$$\mathrm{2}^{{x}} =\mathrm{10}{x} \\ $$ Answered by mr W last updated on 23/Jun/22 $${e}^{{x}\mathrm{ln}\:\mathrm{2}} =\mathrm{10}{x}…
Question Number 172021 by Mikenice last updated on 23/Jun/22 $${solve}: \\ $$$$\mathrm{2}^{{x}^{\mathrm{2}} } ={x}^{\mathrm{2}{x}} \\ $$ Answered by puissant last updated on 23/Jun/22 $$\Rightarrow\:\:{x}^{\mathrm{2}} {ln}\mathrm{2}\:=\:\mathrm{2}{xlnx}…
Question Number 172022 by Mikenice last updated on 23/Jun/22 $${find}\:{the}\:{cubic}\:{of} \\ $$$$\sqrt[{\mathrm{3}}]{\mathrm{26}+\mathrm{15}\sqrt{\mathrm{3}}}\:\:\:+\:\sqrt[{\mathrm{3}}]{\mathrm{26}−\mathrm{15}\sqrt{\mathrm{3}}} \\ $$ Answered by Rasheed.Sindhi last updated on 23/Jun/22 $${Let}\:\sqrt[{\mathrm{3}}]{\mathrm{26}+\mathrm{15}\sqrt{\mathrm{3}}}\:\:\:+\:\sqrt[{\mathrm{3}}]{\mathrm{26}−\mathrm{15}\sqrt{\mathrm{3}}}\:={a} \\ $$$$\:{where}\:{a}\in\mathbb{R} \\…
Question Number 172023 by Mikenice last updated on 23/Jun/22 $${if}:\: \\ $$$${bx}^{\mathrm{3}} −\left(\mathrm{3}{b}+\mathrm{2}\right){x}^{\mathrm{2}} −\mathrm{2}\left(\mathrm{5}{b}−\mathrm{3}\right){x}+\mathrm{20}=\mathrm{0} \\ $$$${find}\:{b}\:{in}\:{term}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 172019 by Mikenice last updated on 23/Jun/22 $${solve}: \\ $$$${x}^{\mathrm{3}} +\mathrm{9}{x}^{\mathrm{2}} {y}=\mathrm{10} \\ $$$${y}^{\mathrm{3}} +{xy}^{\mathrm{2}} =\mathrm{2} \\ $$ Terms of Service Privacy Policy…
Question Number 172009 by Mikenice last updated on 23/Jun/22 $${solve}: \\ $$$${if}\:{x}=\mathrm{2017},{y}=\mathrm{2018}\:{and}\:{z}=\mathrm{2019},\: \\ $$$${find}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} −{xy}−{yz}−{zx} \\ $$ Answered by som(math1967) last updated on…
Question Number 172007 by Mikenice last updated on 23/Jun/22 $${solve}: \\ $$$${x}\sqrt{{x}^{\mathrm{2}} +\mathrm{4}}\:−{x}^{\mathrm{2}} =−\mathrm{6} \\ $$ Answered by puissant last updated on 23/Jun/22 $${x}\sqrt{{x}^{\mathrm{2}} +\mathrm{4}}\:=\:{x}^{\mathrm{2}}…
Question Number 172006 by Mikenice last updated on 23/Jun/22 $${solve}: \\ $$$$\sqrt{{x}^{\mathrm{2}} +\mathrm{5}{x}+\mathrm{2}}\:−\sqrt{{x}^{\mathrm{2}} +\mathrm{5}\:}\:=\mathrm{1} \\ $$ Answered by Rasheed.Sindhi last updated on 24/Jun/22 $$\sqrt{{x}^{\mathrm{2}} +\mathrm{5}{x}+\mathrm{2}}\:−\sqrt{{x}^{\mathrm{2}}…
Question Number 171999 by Mikenice last updated on 23/Jun/22 $${resolve}: \\ $$$$\frac{\mathrm{4}\left({x}−\mathrm{4}\right)}{{x}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{3}}\:{into}\:{partial}\:{fraction} \\ $$ Answered by Rasheed.Sindhi last updated on 23/Jun/22 $$\frac{\mathrm{4}\left({x}−\mathrm{4}\right)}{\left({x}−\mathrm{3}\right)\left({x}+\mathrm{1}\right)}=\frac{{a}}{{x}−\mathrm{3}}+\frac{{b}}{{x}+\mathrm{1}} \\ $$$$\mathrm{4}\left({x}−\mathrm{4}\right)={a}\left({x}+\mathrm{1}\right)+{b}\left({x}−\mathrm{3}\right)…