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} \\ $$$${CD}=\left[{ba}\right]={a}+\mathrm{10}{b} \\ $$$${with}\:\mathrm{1}\leqslant{a},{b}\leqslant\mathrm{9} \\ $$$$…
Question Number 127785 by Engr_Jidda last updated on 02/Jan/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 62234 by aliesam last updated on 18/Jun/19 Answered by tanmay last updated on 18/Jun/19 $$\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} \\ $$$$−{a}\leqslant{x},{y}\leqslant{a} \\ $$$${x}={y}…
Question Number 62211 by Tony Lin last updated on 18/Jun/19 $$\frac{{x}}{\:\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }+\mathrm{3}}{Max}=\frac{\mathrm{5}}{\mathrm{3}}? \\ $$ Commented by MJS last updated on 17/Jun/19 $$\mathrm{what}\:\mathrm{does}\:\mathrm{this}\:\mathrm{mean}? \\ $$ Commented…
Question Number 127742 by arash sharifi last updated on 01/Jan/21 $$\int{xdx} \\ $$ Answered by Olaf last updated on 01/Jan/21 $$\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} +\mathrm{C}_{\mathrm{1}} \\ $$$$\mathrm{are}\:\mathrm{you}\:\mathrm{serious}\:?\: \\…
Question Number 127743 by arash sharifi last updated on 01/Jan/21 $${dx}+{ydy}={x}^{\mathrm{2}} {ydy} \\ $$ Commented by mr W last updated on 01/Jan/21 $${you}\:{have}\:{asked}\:{the}\:{same}\:{question}\: \\ $$$${before}\:{and}\:{it}\:{was}\:{answered}!…
Question Number 62184 by aliesam last updated on 17/Jun/19 $$\sqrt[{\mathrm{3}}]{\mathrm{2}{x}−\mathrm{1}}\:+\:\sqrt{\mathrm{3}{x}+\mathrm{1}}\:=\:\mathrm{3}\sqrt[{\mathrm{4}}]{{x}} \\ $$ Commented by MJS last updated on 17/Jun/19 $${x}=\mathrm{0}\:\vee\:{x}=\mathrm{1} \\ $$$$\mathrm{trying}\:\mathrm{2}{x}−\mathrm{1}={n}^{\mathrm{3}} \:\Rightarrow\:{x}=\frac{{n}^{\mathrm{3}} +\mathrm{1}}{\mathrm{2}} \\…
Question Number 127716 by liberty last updated on 01/Jan/21 $$\:\mathrm{Let}\:\mathrm{p}\:\mathrm{and}\:\mathrm{q}\:\mathrm{be}\:\mathrm{two}\:\mathrm{positive}\:\mathrm{real}\:\mathrm{number} \\ $$$$\mathrm{such}\:\mathrm{that}\:\begin{cases}{\mathrm{p}\sqrt{\mathrm{p}}\:+\mathrm{q}\sqrt{\mathrm{q}}\:=\:\mathrm{32}}\\{\mathrm{p}\sqrt{\mathrm{q}}\:+\:\mathrm{q}\sqrt{\mathrm{p}}\:=\:\mathrm{31}}\end{cases} \\ $$$$\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\frac{\mathrm{5}\left(\mathrm{p}+\mathrm{q}\right)?}{\mathrm{7}} \\ $$ Answered by mindispower last updated on 01/Jan/21 $$\left(\sqrt{{p}}+\sqrt{{q}}\right)^{\mathrm{3}} ={p}\sqrt{{p}}+{q}\sqrt{{q}}+\mathrm{3}\left({p}\sqrt{{q}}+{p}\sqrt{{q}}\right)=\mathrm{32}+\mathrm{3}.\mathrm{31}=\mathrm{125}…