Question Number 78399 by john santu last updated on 17/Jan/20 $${dear}\:{sir}\:{W},\:{Mjs}\: \\ $$$${the}\:{set}\:\left\{\mathrm{1},\mathrm{4},{n}\right\}\:{have}\:{the}\:{condition}\:{that}\: \\ $$$${if}\:{two}\:{different}\:{elements}\:{are} \\ $$$${selected}\:{and}\:\mathrm{2112}\:{is}\:{added}\:{to} \\ $$$${the}\:{result}\:,\:{then}\:{the}\:{result}\: \\ $$$${is}\:{a}\:{perfect}\:{square}\:{if}\:{n}\:{is}\:{a}\: \\ $$$${positif}\:{number}\:.\:{then}\:{the}\:{number}\: \\ $$$${of}\:{possible}\:{values}\:{of}\:{n}\:{is}\:…
Question Number 78390 by mr W last updated on 17/Jan/20 $${solve}\:{for}\:\boldsymbol{{different}}\:{digits}\:{a},{b},{c},{d}\: \\ $$$${such}\:{that}\:\boldsymbol{{abcd}}=\left(\boldsymbol{{ab}}+\boldsymbol{{cd}}\right)^{\mathrm{2}} . \\ $$ Answered by MJS last updated on 17/Jan/20 $$\mathrm{0000} \\…
Question Number 143909 by mathdanisur last updated on 19/Jun/21 Commented by mr W last updated on 19/Jun/21 $${i}\:{remember}\:{i}\:{solved}\:{the}\:{same} \\ $$$${question}\:{some}\:{days}\:{ago}. \\ $$ Commented by amin96…
Question Number 143906 by mathdanisur last updated on 19/Jun/21 Answered by TheHoneyCat last updated on 19/Jun/21 $$\Leftrightarrow{x}^{{x}} −{y}^{{y}} ={e}.\mathrm{ln}\left(\frac{{y}}{{x}}\right)={e}\left(\mathrm{ln}\left({y}\right)−\mathrm{ln}\left({x}\right)\right) \\ $$$$\Leftrightarrow{x}^{{x}} +{e}\mathrm{ln}{x}={y}^{{y}} +{e}\mathrm{ln}{y} \\ $$$$…
Question Number 143899 by liberty last updated on 19/Jun/21 Answered by ajfour last updated on 19/Jun/21 $${let}\:\:{r}^{\mathrm{10}} ={z} \\ $$$${t}={z}^{\mathrm{4}} −{z}^{\mathrm{3}} +{z}^{\mathrm{2}} −{z}+\mathrm{1} \\ $$$${r}^{\mathrm{5}}…
Question Number 78358 by aliesam last updated on 16/Jan/20 $${Q}.\:{solve} \\ $$$$ \\ $$$$\frac{\mathrm{2}^{{sin}^{\mathrm{2}} \left({x}\right)} }{{sin}^{\mathrm{2}} \left({x}\right)\:}\:+\:\frac{\mathrm{3}^{{cos}^{\mathrm{2}} \left({x}\right)} }{{cos}^{\mathrm{2}} \left({x}\right)}\:=\:\mathrm{6} \\ $$ Commented by MJS…
Question Number 78357 by TawaTawa last updated on 16/Jan/20 $$\mathrm{Show}\:\mathrm{that}:\:\:\:\frac{\sqrt{\mathrm{1}\:+\:\mathrm{6x}}\:\:−\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}\:−\:\mathrm{6x}}}}{\:\sqrt{\mathrm{1}\:+\:\mathrm{3x}}\:\:−\:\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}\:−\:\mathrm{3x}}}}\:\:\:\:\:=\:\:\:\mathrm{4}\:+\:\mathrm{6x} \\ $$$$\mathrm{Ignoring}\:\mathrm{higher}\:\mathrm{power}\:\mathrm{of}\:\:\mathrm{x}\:\:\mathrm{in}\:\mathrm{the}\:\mathrm{expansion} \\ $$ Commented by MJS last updated on 16/Jan/20 $$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{understand}\:\mathrm{this}\:\mathrm{question} \\ $$$$\mathrm{the}\:\mathrm{left}\:\mathrm{handed}\:\mathrm{side}\:\mathrm{is}\:\mathrm{defined}\:\mathrm{for} \\…
Question Number 78353 by TawaTawa last updated on 16/Jan/20 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 78351 by loveineq. last updated on 16/Jan/20 $$\mathrm{Let}\:\:{a},{b},{c}\:\in\:\mathrm{R}^{+} \:\:\mathrm{and}\:\:\left({a}+{b}\right)\left({b}+{c}\right)\:=\:\mathrm{1}\:,\:\mathrm{where}\:\:\mathrm{0}\:<{b}\leqslant\:\mathrm{1}\:. \\ $$$$\mathrm{Prove}\:\mathrm{that}\:\:\mid{a}−{b}\mid\mid{b}−{c}\mid\:\geqslant\:\frac{\mid\sqrt{{a}}−\sqrt{{b}}\mid\mid\sqrt{{b}}−\sqrt{{c}}\mid}{\mathrm{2}}\:. \\ $$ Commented by loveineq. last updated on 19/Jan/20 $$\mathrm{More}\:\mathrm{stronger}\:\mathrm{is}\:\:\mid{a}−{b}\mid\mid{b}−{c}\mid\:\geqslant\:\mid\sqrt{{a}}−\sqrt{{b}}\mid\mid\sqrt{{b}}−\sqrt{{c}}\mid\:. \\ $$…
Question Number 78340 by loveineq. last updated on 16/Jan/20 $$\mathrm{Let}\:\:{a},{b},{c}\:>\:\mathrm{0}\:\:\mathrm{and}\:\:{c}^{\mathrm{2}} \:=\:\frac{{ab}+{bc}+{ca}}{\mathrm{3}}\:.\:\mathrm{Prove}\:\mathrm{that} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{{a}^{\mathrm{3}} +{b}^{\mathrm{3}} −\mathrm{2}{c}^{\mathrm{3}} }{{a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} }\:\leqslant\:\mathrm{3}\left(\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{c}^{\mathrm{2}} }{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }\right)…