Question Number 202638 by depressiveshrek last updated on 31/Dec/23 $$\sqrt{{x}−\frac{\mathrm{1}}{{x}}}−\sqrt{\mathrm{1}−\frac{\mathrm{1}}{{x}}}=\mathrm{1}−\frac{\mathrm{1}}{{x}} \\ $$ Commented by AST last updated on 31/Dec/23 $${x}=\mathrm{1}.{Suppose}\:{x}\neq\mathrm{1},{let}\:{a}=\sqrt{{x}−\frac{\mathrm{1}}{{x}}};{b}=\sqrt{\mathrm{1}−\frac{\mathrm{1}}{{x}}} \\ $$$${a}−{b}=\mathrm{1}−\frac{\mathrm{1}}{{x}};\left({a}+{b}\right)^{\mathrm{2}} =\mathrm{1}+{x} \\ $$$${a}^{\mathrm{2}}…
Question Number 202616 by hardmath last updated on 30/Dec/23 $$\mathrm{If}\:\:\:\:\:\frac{\mathrm{a}}{\mathrm{a}\:+\:\mathrm{b}}\:−\:\frac{\mathrm{1}\:−\:\mathrm{a}}{\mathrm{a}\:−\:\mathrm{b}}\:=\:\mathrm{x} \\ $$$$\mathrm{Find}\:\:\:\:\:\frac{\mathrm{1}\:−\:\mathrm{b}}{\mathrm{a}\:−\:\mathrm{b}}\:+\:\frac{\mathrm{b}}{\mathrm{a}\:+\:\mathrm{b}}\:=\:? \\ $$ Answered by MATHEMATICSAM last updated on 30/Dec/23 $$\frac{{a}}{{a}\:+\:{b}}\:−\:\frac{\mathrm{1}\:−\:{a}}{{a}\:−\:{b}}\:=\:{x} \\ $$$$\Rightarrow\:\frac{{a}}{{a}\:+\:{b}}\:−\:\mathrm{1}\:−\:\frac{\mathrm{1}\:−\:{a}}{{a}\:−\:\:{b}}\:−\:\mathrm{1}\:=\:{x}\:−\:\mathrm{2} \\…
Question Number 202604 by MATHEMATICSAM last updated on 30/Dec/23 $$\mathrm{If}\:{x}\:=\:\frac{\sqrt{{a}\:+\:\mathrm{1}}\:+\:\sqrt{{a}\:−\:\mathrm{1}}}{\:\sqrt{{a}\:+\:\mathrm{1}}\:−\:\sqrt{{a}\:−\:\mathrm{1}}}\:\mathrm{and}\: \\ $$$${y}\:=\:\frac{\sqrt{{a}\:+\:\mathrm{1}}\:−\:\sqrt{{a}\:−\:\mathrm{1}}}{\:\sqrt{{a}\:+\:\mathrm{1}}\:+\:\sqrt{{a}\:−\:\mathrm{1}}}\:\mathrm{then}\:\mathrm{show}\:\mathrm{that} \\ $$$$\frac{{x}^{\mathrm{2}} \:−\:{xy}\:+\:{y}^{\mathrm{2}} }{{x}^{\mathrm{2}} \:+\:{xy}\:+\:{y}^{\mathrm{2}} }\:=\:\frac{\mathrm{4}{a}^{\mathrm{2}} \:−\:\mathrm{3}}{\mathrm{4}{a}^{\mathrm{2}} \:−\:\mathrm{1}}\:. \\ $$ Answered by Rasheed.Sindhi…
Question Number 202605 by hardmath last updated on 30/Dec/23 Commented by mr W last updated on 30/Dec/23 $${what}\:{do}\:{you}\:{mean}\:{when}\:{you}\:{write} \\ $$$$\underset{={b}} {{max}}\:\underset{={a}} {{min}}\:\left(…\right)\:? \\ $$ Commented…
Question Number 202633 by hardmath last updated on 30/Dec/23 Answered by cortano12 last updated on 31/Dec/23 $$\:\mathrm{5}\clubsuit\mathrm{6}\:=\:\mathrm{5}−\mathrm{6}=−\mathrm{1} \\ $$$$\:\left(\mathrm{5}\clubsuit\mathrm{6}\right)\clubsuit\mathrm{9}\:=\:−\mathrm{1}\clubsuit\mathrm{9}\:=\:−\mathrm{1}−\mathrm{9}=−\mathrm{10} \\ $$ Commented by hardmath last…
Question Number 202580 by hardmath last updated on 29/Dec/23 $$\mathrm{If} \\ $$$$\underset{\:=\:\boldsymbol{\mathrm{b}}} {\mathrm{max}}\:\underset{\:=\:\boldsymbol{\mathrm{a}}} {\mathrm{min}}\:\left(\mathrm{a}^{\mathrm{2}} \:-\:\mathrm{2ab}\:-\:\mathrm{b}^{\mathrm{2}} \:-\:\mathrm{2ax}\:+\:\mathrm{10bx}\right)\:=\:\mathrm{7} \\ $$$$ \\ $$$$\mathrm{Find}:\:\:\:\boldsymbol{\mathrm{x}}\:=\:? \\ $$ Terms of Service…
Question Number 202551 by aba last updated on 29/Dec/23 $$\underset{\mathrm{i}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\left(\mathrm{x}+\mathrm{y}\right)_{\mathrm{i}} =\left(\mathrm{x}+\mathrm{y}\right)_{\mathrm{1}} +\left(\mathrm{x}+\mathrm{y}\right)_{\mathrm{2}} +…\left(\mathrm{x}+\mathrm{y}\right)_{\mathrm{n}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{x}_{\mathrm{1}} +\mathrm{y}_{\mathrm{1}} +\mathrm{x}_{\mathrm{2}} +\mathrm{y}_{\mathrm{2}} +…+\mathrm{x}_{\mathrm{n}} +\mathrm{y}_{\mathrm{n}} \: \\ $$$$\mathrm{please}\:\mathrm{it}'\mathrm{s}\:\mathrm{correct}\:?…
Question Number 202540 by MATHEMATICSAM last updated on 29/Dec/23 $$\mathrm{If}\:\frac{{b}}{{a}\:+\:{b}}\:=\:\frac{\mathrm{3}{a}\:−\:{b}\:−\:{c}}{\mathrm{2}{b}\:+\:{c}\:−\:{a}}\:=\:\frac{\mathrm{3}{c}\:−\:{a}}{\mathrm{2}{a}\:−\:{b}\:+\:\mathrm{3}{c}}\: \\ $$$$\left[{a}\:+\:{b}\:+\:{c}\:\neq\:\mathrm{0}\right]\:\mathrm{then}\:\mathrm{show}\:\mathrm{that}\:{a}\:=\:{b}\:=\:{c}. \\ $$ Answered by som(math1967) last updated on 29/Dec/23 $$\:\frac{{b}}{{a}+{b}}=\frac{\mathrm{3}{a}−{b}−{c}}{\mathrm{2}{b}+{c}−{a}}=\frac{\mathrm{3}{c}−{a}}{\mathrm{2}{a}−{b}+\mathrm{3}{c}} \\ $$$$\frac{\mathrm{3}{b}}{\mathrm{3}{a}+\mathrm{3}{b}}=\frac{\mathrm{3}{a}−{b}−{c}}{\mathrm{2}{b}+{c}−{a}}=\frac{\mathrm{3}{c}−{a}}{\mathrm{2}{a}−{b}+\mathrm{3}{c}} \\…
Question Number 202532 by justenspi last updated on 29/Dec/23 $${please}\:{help} \\ $$ Commented by justenspi last updated on 29/Dec/23 Terms of Service Privacy Policy Contact:…
Question Number 202535 by Rasheed.Sindhi last updated on 29/Dec/23 $$\boldsymbol{\mathrm{Solve}}\:\boldsymbol{\mathrm{for}}\:\boldsymbol{{x}} \\ $$$$\sqrt[{\mathrm{3}}]{{x}^{\mathrm{3}} +\mathrm{3}{x}−\mathrm{3}}\:+\sqrt[{\mathrm{3}}]{−{x}^{\mathrm{3}} −\mathrm{3}{x}+\mathrm{5}}\:=\mathrm{2} \\ $$$$\left(\mathrm{An}\:\mathrm{alteration}\:\mathrm{of}\:\mathrm{Q}#\mathrm{202500}\right) \\ $$ Answered by mr W last updated on…