Question Number 142556 by iloveisrael last updated on 02/Jun/21 $$\begin{cases}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{2}{yz}+\mathrm{2}−{z}^{\mathrm{2}} }\\{{z}=\mathrm{4022}−{x}−{y}}\end{cases} \\ $$$$\:{x},{y},{z}\in\mathbb{Z}^{+} \:.\:{then}\:{z}= \\ $$ Answered by iloveisrael last updated on 02/Jun/21…
Question Number 77024 by Sathvik last updated on 02/Jan/20 $$\mathrm{The}\:\mathrm{possible}\:\mathrm{value}\:\mathrm{of}\:\mathrm{p}\:\mathrm{for}\:\mathrm{which}\: \\ $$$$\mathrm{graph}\:\mathrm{of}\:\mathrm{the}\:\mathrm{function}\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{2p}^{\mathrm{2}} − \\ $$$$\mathrm{3ptan}\:\mathrm{x}+\mathrm{tan}\:^{\mathrm{2}} \mathrm{x}+\mathrm{1}\:\mathrm{does}\:\mathrm{not}\:\mathrm{lie}\:\mathrm{below}\: \\ $$$$\mathrm{x}-\mathrm{axis}\:\mathrm{for}\:\mathrm{all}\:\mathrm{x}\in\left(\frac{−\Pi}{\mathrm{2}},\frac{\Pi}{\mathrm{2}}\right)\:\mathrm{is} \\ $$$$\left(\mathrm{a}\right)\mathrm{0}\:\:\:\:\:\:\left(\mathrm{b}\right)\mathrm{4}\:\:\:\:\:\:\:\:\left(\mathrm{c}\right)\mathrm{3}\:\:\:\:\:\:\:\:\left(\mathrm{d}\right)\mathrm{8} \\ $$ Answered by MJS…
Question Number 142548 by ivoo last updated on 02/Jun/21 Commented by ivoo last updated on 02/Jun/21 How many integer solutions? Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 77014 by aliesam last updated on 02/Jan/20 Commented by mr W last updated on 02/Jan/20 $${look}\:{and}\:{think}: \\ $$$${x}={y}=\pm\mathrm{1} \\ $$ Commented by mr…
Question Number 11476 by tawa last updated on 26/Mar/17 Answered by sandy_suhendra last updated on 27/Mar/17 $$\frac{\mathrm{1}}{\left(\mathrm{a}−\mathrm{b}\right)\left(\mathrm{b}−\mathrm{c}\right)}\:+\:\frac{\mathrm{1}}{\left(\mathrm{b}−\mathrm{c}\right)\left(\mathrm{c}−\mathrm{a}\right)}\:+\:\frac{\mathrm{1}}{\left(\mathrm{c}−\mathrm{a}\right)\left(\mathrm{a}−\mathrm{b}\right)}\:\: \\ $$$$=\frac{\mathrm{c}−\mathrm{a}+\mathrm{a}−\mathrm{b}+\mathrm{b}−\mathrm{c}}{\left(\mathrm{a}−\mathrm{b}\right)\left(\mathrm{b}−\mathrm{c}\right)\left(\mathrm{c}−\mathrm{a}\right)}\:=\:\mathrm{0} \\ $$$$ \\ $$$$\frac{\mathrm{1}}{\left(\mathrm{a}−\mathrm{b}\right)^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{\left(\mathrm{b}−\mathrm{c}\right)^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{\left(\mathrm{c}−\mathrm{a}\right)^{\mathrm{2}}…
Question Number 11477 by @ANTARES_VY last updated on 26/Mar/17 $$\frac{\boldsymbol{\mathrm{a}}^{\mathrm{8}} +\boldsymbol{\mathrm{a}}^{\mathrm{4}} +\mathrm{1}}{\boldsymbol{\mathrm{a}}^{\mathrm{6}} +\mathrm{1}}. \\ $$$$\boldsymbol{\mathrm{solves}}…. \\ $$ Answered by sm3l2996 last updated on 26/Mar/17 $$\mathrm{a}^{\mathrm{8}}…
Question Number 77009 by Master last updated on 02/Jan/20 Commented by Master last updated on 02/Jan/20 $$\mathrm{lose}\:\mathrm{x}\:\mathrm{from}\:\mathrm{the}\:\mathrm{system}\left(\mathrm{find}\:\mathrm{the}\:\mathrm{connection}\:\mathrm{between}\:\mathrm{a}\:\mathrm{and}\:\mathrm{c}\right) \\ $$ Commented by MJS last updated on…
Question Number 142514 by mnjuly1970 last updated on 01/Jun/21 $$\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:…..\:{number}\:\:{theory}….. \\ $$$$\:\:\:\:\:\:\:{Solve}\:{in}\:\mathbb{Z}\:: \\ $$$$\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}+\frac{\mathrm{1}}{{xy}}\:=\frac{\mathrm{1}}{\mathrm{4}}\:….? \\ $$$$\:\:\:\:\:……… \\ $$ Answered by ArielVyny last updated…
Question Number 142499 by ajfour last updated on 01/Jun/21 Commented by ajfour last updated on 01/Jun/21 $${The}\:{red}\:{circle}\:{has}\:{radius}\:{c}. \\ $$$${The}\:{blue}\:{pair}\:{of}\:{lines}\:{are} \\ $$$${mutually}\:{perpendicular}. \\ $$$${Locate}\:{A}.\:\:\:\:\left({c}<\frac{\mathrm{2}}{\mathrm{3}\sqrt{\mathrm{3}}}\right) \\ $$…
Question Number 11423 by tawa last updated on 25/Mar/17 $$\mathrm{please}\:\mathrm{solve}.\: \\ $$$$\mathrm{ax}^{\mathrm{4}} \:+\:\mathrm{bx}^{\mathrm{3}} \:+\:\mathrm{cx}^{\mathrm{2}} \:+\:\mathrm{dx}\:+\:\mathrm{e}\:=\:\mathrm{0} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com