Question Number 123824 by Snail last updated on 28/Nov/20 $${Show}\:{that}\:{for}\:{all}\:{real}\:{numbers}\:\left({x}/{y}/{z}\right)\:{satisfying}\:\: \\ $$$${x}+{y}+{z}=\mathrm{0}\:{and}\:{xy}\:+{yz}+{zx}=−\mathrm{3}\:\:\:{the}\:{value}\:{of} \\ $$$${expression}\:{x}^{\mathrm{3}} {y}+{y}^{\mathrm{3}} {z}\:+{z}^{\mathrm{3}} {x}\:\:\:{is}\:{a}\:{constant} \\ $$ Commented by Snail last updated on…
Question Number 58248 by tanmay last updated on 20/Apr/19 $${leg}\:{A}_{\mathrm{1}} ,{A}_{\mathrm{2}} ,…{A}_{{n}} \:{and}\:{H}_{\mathrm{1}} ,{H}_{\mathrm{2}} ,…{H}_{{n}} \:{are}\:{n}\:{A}.{M}'{S}\: \\ $$$${and}\:{H}.{M}'{S}\:{respectively}\:{between}\:{a}\:{and}\:{b} \\ $$$${prove}\:{that}\:{A}_{{r}} {H}_{{n}−{r}+\mathrm{1}} ={ab} \\ $$$$\:{n}\geqslant{r}\geqslant\mathrm{1} \\…
Question Number 58246 by tanmay last updated on 20/Apr/19 $${show}\:{that} \\ $$$${P}={x}^{\mathrm{9999}} +{x}^{\mathrm{8888}} +{x}^{\mathrm{7777}} +{x}^{\mathrm{6666}} +{x}^{\mathrm{5555}} +{x}^{\mathrm{4444}} +{x}^{\mathrm{3333}} +{x}^{\mathrm{2222}} +{x}^{\mathrm{1111}} +\mathrm{1} \\ $$$${Q}={x}^{\mathrm{9}} +{x}^{\mathrm{8}} +{x}^{\mathrm{7}}…
Question Number 189302 by mnjuly1970 last updated on 14/Mar/23 $$ \\ $$$$\:\:\:\:{Q}:\:\:\:\:\mathrm{find}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{the}\:\:\mathrm{solutions}\:\:\mathrm{for}\:: \\ $$$$ \\ $$$$\:\:\left(\:{x}_{\:\mathrm{1}} \:+\:{x}_{\:\mathrm{2}} \:\right)^{\:\mathrm{3}} \:+\:{x}_{\:\mathrm{3}} \:+\:{x}_{\:\mathrm{4}} \:+\:{x}_{\:\mathrm{5}} \:=\mathrm{11} \\…
Question Number 123746 by ajfour last updated on 27/Nov/20 Commented by Dwaipayan Shikari last updated on 27/Nov/20 $${What}\:{is}\:{the}\:{question}\:{sir}? \\ $$ Terms of Service Privacy Policy…
Question Number 58210 by salaw2000 last updated on 19/Apr/19 $$\mathrm{find}\:\mathrm{two}\:\mathrm{possible}\:\mathrm{number}\:\mathrm{such}\:\mathrm{that} \\ $$$$\left.\mathrm{1}\right)\:\:\mathrm{xy}=\frac{\mathrm{x}}{\mathrm{y}}=\mathrm{x}−\mathrm{y} \\ $$$$\left.\mathrm{2}\right)\mathrm{xy}=\frac{\mathrm{2x}}{\mathrm{y}}=\mathrm{3}\left(\mathrm{x}−\mathrm{y}\right) \\ $$$$\left.\mathrm{3}\right)\:\:\mathrm{xy}=\frac{\mathrm{x}}{\mathrm{y}}=\mathrm{2}\left(\mathrm{x}−\mathrm{y}\right). \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\…
Question Number 189269 by Rupesh123 last updated on 14/Mar/23 Answered by a.lgnaoui last updated on 15/Mar/23 $${posons}:{E}=\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}+\frac{\mathrm{15}}{{a}+{b}+{c}} \\ $$$$\left({a}+{b}+{c}\right)×{E}=\mathrm{15}+\left({a}+{b}+{c}\right)\left(\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}\right) \\ $$$$=\mathrm{18}+\frac{{a}\left({b}+{c}\right)}{{bc}}+\frac{{b}\left({a}+{c}\right)}{{ac}}+\frac{{c}\left({a}+{b}\right)}{{ab}} \\ $$$$=\mathrm{18}+\frac{\left(\mathrm{1}−\mathrm{2}{abc}−{ab}\right){c}+{ab}\left({a}+{b}\right)}{{abc}} \\ $$$$=\mathrm{18}+\left(\frac{\mathrm{1}−\mathrm{2}{abc}−{ab}}{{ab}}\:\:+\frac{{a}+{b}}{{c}}\right)\:\:\left({i}\right)…
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Question Number 58154 by behi83417@gmail.com last updated on 18/Apr/19 $$\boldsymbol{\mathrm{A}}\left(\mathrm{1},\mathrm{1}+\boldsymbol{\mathrm{i}}\right),\boldsymbol{\mathrm{B}}\left(\sqrt{\mathrm{2}}+\boldsymbol{\mathrm{i}},\mathrm{2}\right),\boldsymbol{\mathrm{C}}\left(\mathrm{1}−\mathrm{3}\boldsymbol{\mathrm{i}},\mathrm{1}−\boldsymbol{\mathrm{i}}\right) \\ $$$$\boldsymbol{\mathrm{are}}\:\boldsymbol{\mathrm{given}}. \\ $$$$\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{angle}}\:\boldsymbol{\mathrm{between}}:\:\:\boldsymbol{\mathrm{AB}}\:\:\boldsymbol{\mathrm{and}}\:\:\boldsymbol{\mathrm{AC}}\:. \\ $$ Answered by MJS last updated on 18/Apr/19 $$\mathrm{if}\:\mathrm{i}=\sqrt{−\mathrm{1}}\:\mathrm{then}\:{A},\:{B},\:{C}\:\in\mathbb{R}^{\mathrm{4}} \:\mathrm{with}\:\mathrm{coordinates}…
Question Number 58156 by hhghg last updated on 18/Apr/19 $$\mathrm{Jaiden}\:\mathrm{buys}\:\mathrm{334}\:\mathrm{cupcakes}.\mathrm{He}\:\mathrm{got}\:\mathrm{14}\:\mathrm{more}\:\mathrm{cupcakes}.\mathrm{How}\:\mathrm{many}\:\mathrm{cupcakes}\:\mathrm{did}\:\mathrm{he}\:\mathrm{got}\:\mathrm{altogether}? \\ $$ Answered by tanmay last updated on 19/Apr/19 $${suppose}\:{you}\:{have}\:\mathrm{3}\:{pen}\:{and}\:{you}\:{get}\:\mathrm{2}\:{more}\:{pen} \\ $$$${then}\:{how}\:{many}\:{do}\:{you}\:{have} \\ $$$$\mathrm{3}+\mathrm{2}=\mathrm{5}\:{pen} \\…