Question Number 201615 by hardmath last updated on 09/Dec/23 $$\mathrm{x},\mathrm{y},\mathrm{z}\:\in\:\mathbb{R} \\ $$$$\mathrm{a},\mathrm{b},\mathrm{c}>\mathrm{0} \\ $$$$\mathrm{prove}\:\mathrm{that}: \\ $$$$\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{a}}\:+\:\frac{\mathrm{y}^{\mathrm{2}} }{\mathrm{b}}\:+\:\frac{\mathrm{z}^{\mathrm{2}} }{\mathrm{c}}\:\geqslant\:\frac{\left(\mathrm{x}\:+\:\mathrm{y}\:+\:\mathrm{z}\right)^{\mathrm{2}} }{\mathrm{a}\:+\:\mathrm{b}\:+\:\mathrm{c}} \\ $$ Answered by AST…
Question Number 201557 by hardmath last updated on 08/Dec/23 $$\mathrm{5}\:\centerdot\:\underset{\:\mathrm{50}} {\underbrace{\mathrm{555}…\mathrm{5}}} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{digits}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{product}. \\ $$ Answered by aleks041103 last updated on 09/Dec/23 $$\mathrm{5}.\mathrm{5}=\mathrm{25}…
Question Number 201547 by Calculusboy last updated on 08/Dec/23 Answered by mr W last updated on 08/Dec/23 $${let}\:{t}=\sqrt{{x}^{\mathrm{2}} +\mathrm{2}}+{x} \\ $$$$\left(\sqrt{{x}^{\mathrm{2}} +\mathrm{2}}−{x}\right){t}=\left(\sqrt{{x}^{\mathrm{2}} +\mathrm{2}}−{x}\:\right)\left(\sqrt{{x}^{\mathrm{2}} +\mathrm{2}}+{x}\right)=\mathrm{2} \\…
Question Number 201477 by York12 last updated on 07/Dec/23 $$\mathrm{how}\:\mathrm{to}\:\mathrm{prove}\:\mathrm{that} \\ $$$$\left(\mathrm{3d}_{\mathrm{3}} +\mathrm{4d}_{\mathrm{2}} +\mathrm{3d}_{\mathrm{1}} \right)^{\mathrm{2}} \leqslant\mathrm{5}\left(\mathrm{d}_{\mathrm{1}} ^{\mathrm{2}} +\mathrm{d}_{\mathrm{2}} ^{\mathrm{2}} +\mathrm{d}_{\mathrm{3}} ^{\mathrm{2}} +\left(\mathrm{d}_{\mathrm{2}} +\mathrm{d}_{\mathrm{1}} \right)^{\mathrm{2}} +\left(\mathrm{d}_{\mathrm{3}}…
Question Number 201464 by hardmath last updated on 06/Dec/23 $$\mathrm{if}\:\:\:\mathrm{f}\left(\mathrm{2}\right)\:=\:\mathrm{3}\:\:\:\mathrm{and}\:\:\:\mathrm{f}\left(\mathrm{4}\right)\:=\:\mathrm{5} \\ $$$$\mathrm{find}\:\:\:\int_{\mathrm{2}} ^{\:\mathrm{4}} \:\mathrm{f}\left(\mathrm{x}\right)\:\centerdot\:\mathrm{f}\:^{'} \left(\mathrm{x}\right)\:\mathrm{dx}\:=\:? \\ $$ Answered by mahdipoor last updated on 06/Dec/23 $$=\left[\frac{\left({f}\left({x}\right)\right)^{\mathrm{2}}…
Question Number 201462 by hardmath last updated on 06/Dec/23 $$\mathrm{Find}: \\ $$$$\int_{\mathrm{0}} ^{\:\boldsymbol{\pi}} \:\sqrt{\mathrm{1}\:-\:\mathrm{4}\:\mathrm{sin}^{\mathrm{2}} \:\frac{\mathrm{x}}{\mathrm{2}}\:\mathrm{cos}^{\mathrm{2}} \:\frac{\mathrm{x}}{\mathrm{2}}}\:\mathrm{dx}\:=\:? \\ $$ Answered by esmaeil last updated on 06/Dec/23…
Question Number 201460 by hardmath last updated on 06/Dec/23 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{smallest}\:\mathrm{positive}\:\mathrm{period}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{function}: \\ $$$$\mathrm{y}\:=\:\mid\:\mathrm{tan}\:\mathrm{2x}\:\mid\:\:+\:\:\mid\:\mathrm{cot}\:\mathrm{2x}\:\mid \\ $$ Answered by Mathspace last updated on 07/Dec/23 $${y}\left({x}\right)=\mid{tan}\left(\mathrm{2}{x}\right)\mid+\frac{\mathrm{1}}{\mid{tan}\left(\mathrm{2}{x}\right)\mid} \\…
Question Number 201430 by hardmath last updated on 06/Dec/23 $$\mathrm{Find}: \\ $$$$\frac{\mathrm{2}}{\mathrm{35}}\:+\:\frac{\mathrm{2}}{\mathrm{63}}\:+\:\frac{\mathrm{2}}{\mathrm{99}}\:+\:\frac{\mathrm{2}}{\mathrm{143}}\:=\:? \\ $$ Answered by Rasheed.Sindhi last updated on 06/Dec/23 $$\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{6}^{\mathrm{2}} −\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{8}^{\mathrm{2}} −\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{10}^{\mathrm{2}} −\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{12}^{\mathrm{2}}…
Question Number 201463 by hardmath last updated on 06/Dec/23 $$\mathrm{a}\:=\:\mathrm{constant}\:\mathrm{number}: \\ $$$$\mathrm{if}\:\:\:\int\mathrm{x}\:\centerdot\:\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{dx}\:=\:\mathrm{x}^{\mathrm{3}} \:-\:\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{4x}\:-\:\frac{\mathrm{a}}{\mathrm{5}} \\ $$$$\mathrm{find}\:\:\:\mathrm{f}\left(\mathrm{2}\right)\:=\:? \\ $$ Answered by mr W last updated on…
Question Number 201441 by dimentri last updated on 06/Dec/23 $${Let}\:{f}\left({x}\right)\:{and}\:{g}\left({x}\right)\:{be}\:{given}\:{by}\: \\ $$$$\:{f}\left({x}\right)=\:\frac{\mathrm{1}}{{x}}\:+\frac{\mathrm{1}}{{x}−\mathrm{2}}\:+\frac{\mathrm{1}}{{x}−\mathrm{4}}\:+\:…\:+\frac{\mathrm{1}}{{x}−\mathrm{2018}} \\ $$$$\:{and}\: \\ $$$$\:\:{g}\left({x}\right)=\frac{\mathrm{1}}{{x}−\mathrm{1}}\:+\frac{\mathrm{1}}{{x}−\mathrm{3}}\:+\frac{\mathrm{1}}{{x}−\mathrm{5}}\:+…+\:\frac{\mathrm{1}}{{x}−\mathrm{2017}}. \\ $$$$\:\:{Prove}\:{that}\:\:\mid\:{f}\left({x}\right)−{g}\left({x}\right)\mid\:>\mathrm{2} \\ $$$$\:\:{for}\:{any}\:{non}−{integer}\:{real}\:{number} \\ $$$$\:\:{x}\:{satisfying}\:\mathrm{0}<{x}<\mathrm{2018}.\: \\ $$ Answered…