Question Number 191393 by Mingma last updated on 23/Apr/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 191395 by Mingma last updated on 23/Apr/23 Answered by Rasheed.Sindhi last updated on 23/Apr/23 $${xy}+{yz}+{zx}=\mathrm{0};\:\:\frac{{y}+{z}}{{x}}+\frac{{x}+{z}}{{y}}+\frac{{x}+{y}}{{z}}=? \\ $$$$\frac{{y}+{z}}{{x}}+\frac{{x}+{z}}{{y}}+\frac{{x}+{y}}{{z}} \\ $$$$=\frac{{y}+{z}}{{x}}+\mathrm{1}+\frac{{x}+{z}}{{y}}+\mathrm{1}+\frac{{x}+{y}}{{z}}+\mathrm{1}−\mathrm{3} \\ $$$$=\frac{{x}+{y}+{z}}{{x}}+\frac{{x}+{y}+{z}}{{y}}+\frac{{x}+{y}+{z}}{{z}}−\mathrm{3} \\ $$$$=\left({x}+{y}+{z}\right)\left(\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}+\frac{\mathrm{1}}{{z}}\right)−\mathrm{3}…
Question Number 191394 by Mingma last updated on 23/Apr/23 Answered by Rasheed.Sindhi last updated on 23/Apr/23 $$\sqrt{{a}}\:+\frac{\mathrm{2022}}{\:\sqrt{{b}}\:}=\mathrm{2023} \\ $$$$\mathrm{2023}\sqrt{{b}}\:−\sqrt{{a}}\:\sqrt{{b}}\:=\mathrm{2022} \\ $$$$\sqrt{{b}}\:\left(\mathrm{2023}−\sqrt{{a}}\:\right)=\mathrm{2}×\mathrm{3}×\mathrm{337} \\ $$$$\begin{cases}{\sqrt{{b}}\:=\mathrm{2}\:\wedge\:\mathrm{2023}−\sqrt{{a}}\:=\mathrm{3}×\mathrm{337}}\\{\sqrt{{b}}\:=\mathrm{3}\:\wedge\:\mathrm{2023}−\sqrt{{a}}\:=\mathrm{2}×\mathrm{337}}\\{\sqrt{{b}}=\mathrm{337}\:\wedge\:\mathrm{2023}−\sqrt{{a}}\:=\mathrm{2}×\mathrm{3}\:}\\{\sqrt{{b}}\:=\mathrm{2}×\mathrm{3}\:\wedge\:\mathrm{2023}−\sqrt{{a}}\:=\mathrm{337}}\\{\sqrt{{b}}\:=\mathrm{2}×\mathrm{337}\:\wedge\:\mathrm{2023}−\sqrt{{a}}\:=\mathrm{3}}\\{\sqrt{{b}}\:=\mathrm{3}×\mathrm{337}\:\wedge\:\mathrm{2023}−\sqrt{{a}}\:=\mathrm{2}}\\{\sqrt{{b}}\:=\mathrm{2}×\mathrm{3}×\mathrm{337}\:\wedge\:\mathrm{2023}−\sqrt{{a}}\:=\mathrm{1}}\end{cases}\: \\ $$$$\begin{cases}{{b}\:=\mathrm{4}\:\wedge\:{a}\:=\left(\mathrm{2023}−\mathrm{3}×\mathrm{337}\right)^{\mathrm{2}}…
Question Number 191389 by MATHEMATICSAM last updated on 23/Apr/23 $$\mathrm{Factorize}: \\ $$$${x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }+\frac{\mathrm{11}}{\mathrm{8}} \\ $$ Answered by Frix last updated on 23/Apr/23 $${x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}}…
Question Number 191343 by mnjuly1970 last updated on 23/Apr/23 $$ \\ $$$$\:\:\:\:\:\:\mathrm{solve}\:\:{in}\:\:\mathbb{R} \\ $$$$ \\ $$$$\:\:\:\:\:\:\lfloor\:\frac{\mathrm{1}}{{x}}\:\rfloor\:+\:\lfloor\:\frac{\mathrm{2}}{{x}}\:\rfloor\:+\:\lfloor\:\frac{\mathrm{3}}{{x}}\:\rfloor\:=\:\mathrm{1} \\ $$$$ \\ $$ Answered by mr W last…
Question Number 60255 by ANTARES VY last updated on 19/May/19 $$\boldsymbol{\mathrm{b}}=\sqrt[{\mathrm{3}}]{\frac{\boldsymbol{\mathrm{kT}}}{\boldsymbol{\mathrm{P}}}}.\:\boldsymbol{\mathrm{distance}}\:\:\boldsymbol{\mathrm{molekular}} \\ $$$$\boldsymbol{\mathrm{prove}}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 125785 by Mathgreat last updated on 13/Dec/20 $$\oplus+\oplus+\oplus=\mathrm{33} \\ $$$$\blacktriangle+\blacktriangle+\mathrm{9}\blacktrinagledown=\mathrm{25} \\ $$$$\mathrm{9}\blacktrinagledown+\blacktriangle+\mathrm{2}\bigstar=\mathrm{19} \\ $$$$\mathrm{2}\oplus+\mathrm{8}\blacktrinagledown\:\centerdot\:\bigstar=\:? \\ $$$$\:\: \\ $$ Answered by Olaf last updated…
Question Number 191304 by Mingma last updated on 22/Apr/23 Answered by cortano12 last updated on 22/Apr/23 $$\:\mathrm{x}=\mathrm{t}^{\mathrm{2}} −\mathrm{2}\: \\ $$$$\:\left(\mathrm{t}^{\mathrm{2}} −\mathrm{2}\right)^{\mathrm{3}} −\mathrm{3}\left(\mathrm{t}^{\mathrm{2}} −\mathrm{2}\right)−\mathrm{t}=\mathrm{0} \\ $$$$\:\left(\mathrm{t}−\mathrm{2}\right)\left(\mathrm{2t}+\mathrm{1}−\sqrt{\mathrm{5}}\right)\left(\mathrm{2t}+\mathrm{1}+\sqrt{\mathrm{5}}\right)\left(\frac{\mathrm{1}}{\mathrm{4}}\mathrm{t}^{\mathrm{3}}…
Question Number 191300 by Mingma last updated on 22/Apr/23 Answered by Tinku Tara last updated on 22/Apr/23 $$\left({a}^{\mathrm{2}} −\mathrm{3}\right)\left({a}^{\mathrm{2}} −\mathrm{2}\right)\left({a}^{\mathrm{2}} −\mathrm{1}\right)\left({a}^{\mathrm{2}} \right)\left({a}^{\mathrm{2}} +\mathrm{1}\right)\left({a}^{\mathrm{2}} +\mathrm{2}\right)\left({a}^{\mathrm{2}} +\mathrm{3}\right)=\mathrm{8}!…
Question Number 191301 by Mingma last updated on 22/Apr/23 Answered by a.lgnaoui last updated on 23/Apr/23 $$\mathrm{a};\mathrm{b};\mathrm{c}>\mathrm{0}\:\:\:\Rightarrow\left(\mathrm{a}+\mathrm{b}+\mathrm{c}>\mathrm{0}\:\:\:\mathrm{et}\:\mathrm{abc}>\mathrm{0}\right) \\ $$$$\mathrm{poxons}\:\:\mathrm{x}=\mathrm{abc}\:\:\:\:\:\mathrm{y}=\mathrm{a}+\mathrm{b}+\mathrm{c}\:\:\:\mathrm{xy}=\mathrm{1} \\ $$$$\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{b}}=\boldsymbol{\mathrm{y}}−\boldsymbol{\mathrm{c}}\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{c}}=\boldsymbol{\mathrm{y}}−\boldsymbol{\mathrm{b}} \\ $$$$\:\:\left(\boldsymbol{\mathrm{y}}−\boldsymbol{\mathrm{c}}\right)\left(\boldsymbol{\mathrm{y}}−\boldsymbol{\mathrm{b}}\right)\:=\boldsymbol{\mathrm{y}}^{\mathrm{2}} −\left(\boldsymbol{\mathrm{b}}+\boldsymbol{\mathrm{c}}\right)\boldsymbol{\mathrm{y}}+\boldsymbol{\mathrm{bc}} \\…