Question Number 202383 by MATHEMATICSAM last updated on 25/Dec/23 $$\mathrm{Show}\:\mathrm{that}\:\frac{{a}\sqrt{{b}}\:−\:{b}\sqrt{{a}}}{{a}\sqrt{{b}}\:+\:{b}\sqrt{{a}}}\:=\:\frac{\mathrm{1}}{{a}\:−\:{b}}\left({a}\:+\:{b}\:−\:\mathrm{2}\sqrt{{ab}}\right). \\ $$ Answered by AST last updated on 25/Dec/23 $$=\frac{\left({a}\sqrt{{b}}−{b}\sqrt{{a}}\right)^{\mathrm{2}} }{{ab}\left({a}−{b}\right)}\:=\frac{{a}^{\mathrm{2}} {b}+{b}^{\mathrm{2}} {a}−\mathrm{2}{ab}\sqrt{{ab}}}{{ab}\left({a}−{b}\right)}=\frac{{a}+{b}−\mathrm{2}\sqrt{{ab}}}{{a}−{b}} \\ $$…
Question Number 202300 by hardmath last updated on 24/Dec/23 $$ \\ $$Parvin leaves home to go to school. After walking 3/8 of the way, he…
Question Number 202303 by hardmath last updated on 24/Dec/23 $$\mathrm{If}\:\:\:\mathrm{a}_{\mathrm{1}} \:=\:\mathrm{1}\:\:\:\mathrm{and}\:\:\:\mathrm{a}_{\mathrm{1}} \:\centerdot\:\mathrm{a}_{\mathrm{2}} \:\centerdot\:…\:\centerdot\:\mathrm{a}_{\boldsymbol{\mathrm{n}}} \:=\:\mathrm{n}^{\mathrm{2}} \\ $$$$\mathrm{Find}:\:\:\:\mathrm{a}_{\mathrm{2}} \:+\:\mathrm{a}_{\mathrm{13}} \:=\:? \\ $$ Answered by MATHEMATICSAM last updated…
Question Number 202302 by hardmath last updated on 24/Dec/23 $$\mathrm{If}\:\:\:\frac{\mathrm{3cosx}\:+\:\mathrm{2sinx}}{\mathrm{cosx}\:−\:\mathrm{sinx}}\:=\:\mathrm{4}\:\:\:\mathrm{find}\:\:\:\mathrm{ctgx}\:=\:? \\ $$ Answered by cortano12 last updated on 24/Dec/23 $$\:\:\Leftrightarrow\:\:\underbrace{\mathcal{X}} \\ $$ Answered by MM42…
Question Number 202296 by hardmath last updated on 24/Dec/23 $$\mathrm{Find}: \\ $$$$\mathrm{log}_{\mathrm{2}} \:\left(\mathrm{log}_{\mathrm{3}} \:\left(\mathrm{log}_{\mathrm{4}} \:\mathrm{64}\right)\right)\:=\:? \\ $$ Answered by MATHEMATICSAM last updated on 24/Dec/23 $$\mathrm{log}_{\mathrm{2}}…
Question Number 202297 by hardmath last updated on 24/Dec/23 $$\mathrm{If}\:\:\:\mathrm{a}^{\mathrm{2}} \:+\:\mathrm{a}\:=\:\mathrm{5} \\ $$$$\mathrm{Find}\:\:\:\frac{\mathrm{a}^{−\boldsymbol{\mathrm{b}}} \:+\:\mathrm{a}^{\mathrm{1}−\boldsymbol{\mathrm{b}}} \:+\:\mathrm{a}^{\mathrm{2}−\boldsymbol{\mathrm{b}}} }{\mathrm{a}^{−\boldsymbol{\mathrm{b}}} }\:=\:? \\ $$ Answered by MATHEMATICSAM last updated on…
Question Number 202298 by hardmath last updated on 24/Dec/23 $$\mathrm{x}\:\in\:\mathbb{R} \\ $$$$\mathrm{Find}:\:\:\:\boldsymbol{\mathrm{max}}\left(\frac{\mathrm{5}}{\mathrm{x}^{\mathrm{2}} \:−\:\mathrm{6x}\:+\:\mathrm{11}}\right)\:=\:? \\ $$ Answered by Rasheed.Sindhi last updated on 24/Dec/23 $$\boldsymbol{\mathrm{max}}\left(\frac{\mathrm{5}}{\mathrm{x}^{\mathrm{2}} \:−\:\mathrm{6x}\:+\:\mathrm{11}}\right)=\boldsymbol{\mathrm{max}}\left(\frac{\mathrm{5}}{\left({x}−\mathrm{3}\right)^{\mathrm{2}} +\mathrm{2}}\right)…
Question Number 202290 by MATHEMATICSAM last updated on 24/Dec/23 $$\mathrm{If}\:\frac{\mathrm{3}{a}\:−\:{b}}{{x}\:+\:{y}}\:=\:\frac{\mathrm{3}{b}\:−\:{c}}{{y}\:+\:{z}}\:=\:\frac{\mathrm{3}{c}\:−\:{a}}{{z}\:+\:{x}}\:\mathrm{then}\:\mathrm{show} \\ $$$$\mathrm{that}\:\frac{{a}\:+\:{b}\:+\:{c}}{{x}\:+\:{y}\:+\:{z}}\:=\:\frac{{a}^{\mathrm{2}\:} \:+\:{b}^{\mathrm{2}} \:+\:{c}^{\mathrm{2}} }{{ax}\:+\:{by}\:+\:{cz}}\:. \\ $$ Answered by Rasheed.Sindhi last updated on 24/Dec/23 $$\mathrm{If}\:\frac{\mathrm{3}{a}\:−\:{b}}{{x}\:+\:{y}}\:=\:\frac{\mathrm{3}{b}\:−\:{c}}{{y}\:+\:{z}}\:=\:\frac{\mathrm{3}{c}\:−\:{a}}{{z}\:+\:{x}}\:\mathrm{then}\:\mathrm{show}…
Question Number 202348 by York12 last updated on 24/Dec/23 $$ \\ $$$$\mathrm{Let}\:{a},{b},{c}\:\:\in\mathbb{R}^{+} \:,\:{a}+{b}+{c}=\mathrm{3}\:\mathrm{prove}\:\mathrm{the}\:\mathrm{following}\:\mathrm{inequality} \\ $$$$\frac{\left(\mathrm{2}{a}−\mathrm{3}\right)^{\mathrm{2}} }{{b}}+\frac{\left(\mathrm{2}{b}−\mathrm{3}\right)^{\mathrm{2}} }{{c}}+\frac{\left(\mathrm{2}{c}−\mathrm{3}\right)^{\mathrm{2}} }{{a}}\geqslant\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{{a}+{b}}+\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }{{b}+{c}}+\frac{{c}^{\mathrm{2}} +{a}^{\mathrm{2}} }{{c}+{a}} \\…
Question Number 202340 by hardmath last updated on 24/Dec/23 Answered by MATHEMATICSAM last updated on 25/Dec/23 $$\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}\:+\:\sqrt{\mathrm{2}}}\:=\:\sqrt{\mathrm{3}}\:−\:\sqrt{\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{4}}\:+\:\sqrt{\mathrm{3}}}\:=\:\sqrt{\mathrm{4}}\:−\:\sqrt{\mathrm{3}} \\ $$$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}\:+\:\sqrt{\mathrm{4}}}\:=\:\sqrt{\mathrm{5}}\:−\:\sqrt{\mathrm{4}} \\ $$$$. \\ $$$$.…