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Question Number 62334 by smartsmith459@gmail.com last updated on 19/Jun/19 $${if}\:\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} =\:\left(\alpha+\beta\right)^{\mathrm{2}} −\mathrm{2}\alpha\beta\:{evaluate}\left(\alpha−\beta\right) \\ $$ Answered by Kunal12588 last updated on 20/Jun/19 $$\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} =\left(\alpha+\beta\right)^{\mathrm{2}}…
Question Number 62281 by behi83417@gmail.com last updated on 19/Jun/19 $$\begin{cases}{\boldsymbol{\mathrm{x}}^{\mathrm{3}} +\boldsymbol{\mathrm{y}}^{\mathrm{3}} =\mathrm{3}\boldsymbol{\mathrm{xy}}}\\{\boldsymbol{\mathrm{x}}^{\mathrm{4}} +\boldsymbol{\mathrm{y}}^{\mathrm{4}} =\mathrm{4}\boldsymbol{\mathrm{xy}}}\end{cases}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[\boldsymbol{\mathrm{x}},\boldsymbol{\mathrm{y}}\neq\mathrm{0}\right] \\ $$ Answered by mr W last updated on 19/Jun/19 $${let}\:{u}={x}+{y},\:{v}={xy}…
Question Number 62276 by behi83417@gmail.com last updated on 18/Jun/19 $$\begin{cases}{\frac{\sqrt{\boldsymbol{\mathrm{x}}}}{\boldsymbol{\mathrm{a}}}+\frac{\sqrt{\boldsymbol{\mathrm{y}}}}{\boldsymbol{\mathrm{b}}}=\mathrm{1}}\\{\frac{\sqrt{\boldsymbol{\mathrm{a}}}}{\boldsymbol{\mathrm{x}}}+\frac{\sqrt{\boldsymbol{\mathrm{b}}}}{\boldsymbol{\mathrm{y}}}=\mathrm{1}}\end{cases}\:\:\:\boldsymbol{\mathrm{a}},\boldsymbol{\mathrm{b}}\in\mathrm{R}^{+} \\ $$ Answered by mr W last updated on 19/Jun/19 $${let}\:{X}=\sqrt{{x}},\:{Y}=\sqrt{{y}},\:{A}=\sqrt{{a}},\:{B}=\sqrt{{b}} \\ $$$$\frac{{X}}{{A}^{\mathrm{2}} }+\frac{{Y}}{{B}^{\mathrm{2}} }=\mathrm{1}…
Question Number 62275 by behi83417@gmail.com last updated on 18/Jun/19 $$\begin{cases}{\boldsymbol{\mathrm{a}}\sqrt{\boldsymbol{\mathrm{x}}}+\boldsymbol{\mathrm{b}}\sqrt{\boldsymbol{\mathrm{y}}}=\mathrm{2}\sqrt{\boldsymbol{\mathrm{ab}}}}\\{\boldsymbol{\mathrm{x}}\sqrt{\boldsymbol{\mathrm{a}}}+\boldsymbol{\mathrm{y}}\sqrt{\boldsymbol{\mathrm{b}}}=\mathrm{2}\sqrt{\boldsymbol{\mathrm{ab}}}}\end{cases}\:\:\:\:\:\:\boldsymbol{\mathrm{a}},\boldsymbol{\mathrm{b}}\in\mathrm{R}^{+} \\ $$ Answered by mr W last updated on 19/Jun/19 $${let}\:{X}=\sqrt{{x}},{Y}=\sqrt{{y}},{A}=\sqrt{{a}},{B}=\sqrt{{b}} \\ $$$${A}^{\mathrm{2}} {X}+{B}^{\mathrm{2}} {Y}=\mathrm{2}{AB}…
Question Number 127788 by peter frank last updated on 02/Jan/21 Answered by mr W last updated on 02/Jan/21 $${AB}=\left[{ab}\right]=\mathrm{10}{a}+{b} \\ $$$${CD}=\left[{ba}\right]={a}+\mathrm{10}{b} \\ $$$${with}\:\mathrm{1}\leqslant{a},{b}\leqslant\mathrm{9} \\ $$$$…
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Question Number 62234 by aliesam last updated on 18/Jun/19 Answered by tanmay last updated on 18/Jun/19 $$\mathrm{9}\left({x}+{y}\right)=\left({x}+{y}\right)\left({x}^{\mathrm{2}} −{xy}+{y}^{\mathrm{2}} \right) \\ $$$$\left({x}+{y}\right)\left(\mathrm{9}−{x}^{\mathrm{2}} +{xy}−{y}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$${eithdr}\:{x}+{y}=\mathrm{0}\:\:\:{or}\:{x}^{\mathrm{2}}…
Question Number 62228 by behi83417@gmail.com last updated on 17/Jun/19 $$\begin{cases}{\sqrt{\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{x}}}+\sqrt{\boldsymbol{\mathrm{a}}−\boldsymbol{\mathrm{y}}}=\mathrm{2}\boldsymbol{\mathrm{a}}}\\{\sqrt{\boldsymbol{\mathrm{a}}−\boldsymbol{\mathrm{x}}}+\sqrt{\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{y}}}=\mathrm{2}\boldsymbol{\mathrm{a}}}\end{cases}\:\:\:\:\:\boldsymbol{\mathrm{a}}\in\boldsymbol{\mathrm{R}}. \\ $$ Commented by mr W last updated on 18/Jun/19 $${a}>\mathrm{0} \\ $$$$−{a}\leqslant{x},{y}\leqslant{a} \\ $$$${x}={y}…