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Question Number 120355 by TITA last updated on 30/Oct/20 Commented by TITA last updated on 30/Oct/20 $${please}\:{help} \\ $$ Commented by TITA last updated on…
Question Number 54808 by turbo msup by abdo last updated on 11/Feb/19 $${let}\:{p}\left({x}\right)=\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{x}^{\mathrm{4}} \right)…\left(\mathrm{1}+{x}^{\mathrm{2}^{{n}} } \right) \\ $$$${with}\:{n}\:{integr}\:{natural} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{a}\:{simple}\:{form}\:{of}\:{p}\left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{find}\:{roots}\:{of}\:{p}\left({x}\right){and}\:{decompose} \\ $$$${p}\left({x}\right)\:{inside}\:{C}\left[{x}\right]…
Question Number 185876 by MATHEMATICSAM last updated on 29/Jan/23 $${x}\:=\:\frac{\mathrm{2}{ab}}{{a}+{b}}\:\mathrm{then}\:\mathrm{prove}\:\mathrm{that}\: \\ $$$$\frac{{x}\:+\:{a}}{{x}\:−\:{a}}\:+\:\frac{{x}\:+\:{b}}{{x}\:−\:{b}}\:=\:\mathrm{2} \\ $$ Answered by Rasheed.Sindhi last updated on 29/Jan/23 $${x}\:=\:\frac{\mathrm{2}{ab}}{{a}+{b}}\:\mathrm{then}\:\mathrm{prove}\:\mathrm{that}\: \\ $$$$\frac{{x}\:+\:{a}}{{x}\:−\:{a}}\:+\:\frac{{x}\:+\:{b}}{{x}\:−\:{b}}\:=\:\mathrm{2} \\…
Question Number 120325 by Ar Brandon last updated on 30/Oct/20 $$\mathrm{Let}\:{f}:\mathbb{R}\rightarrow\mathbb{R}\:\mathrm{be}\:\mathrm{a}\:\mathrm{function}\:\mathrm{satisfying}\:\mathrm{the} \\ $$$$\mathrm{functional}\:\mathrm{relation} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left({f}\left(\mathrm{x}\right)\right)^{\mathrm{y}} +\left({f}\left(\mathrm{y}\right)\right)^{\mathrm{x}} =\mathrm{2}{f}\left(\mathrm{xy}\right) \\ $$$$\mathrm{for}\:\mathrm{all}\:\mathrm{x},\:\mathrm{y}\:\in\mathbb{R}\:\mathrm{and}\:\mathrm{it}\:\mathrm{is}\:\mathrm{given}\:\mathrm{that}\:{f}\left(\mathrm{1}\right)=\mathrm{1}/\mathrm{2}.\:\mathrm{Answer} \\ $$$$\mathrm{the}\:\mathrm{following}\:\mathrm{questions}. \\ $$$$\left(\boldsymbol{\mathrm{i}}\right)\:\:\:\:{f}\left(\mathrm{x}+\mathrm{y}\right)= \\ $$$$\:\:\:\:\:\:\:\:\:\left(\mathrm{A}\right)\:{f}\left(\mathrm{x}\right)+{f}\left(\mathrm{y}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{B}\right)\:\:{f}\left(\mathrm{x}\right){f}\left(\mathrm{y}\right)…
Question Number 120324 by mathocean1 last updated on 30/Oct/20 $$\mathrm{we}\:\mathrm{are}\:\mathrm{in}\:\mathbb{C}. \\ $$$$\left(\mathrm{E}\right):\:\mathrm{z}^{\mathrm{3}} +\left(\mathrm{4}−\mathrm{5i}\right)\mathrm{z}^{\mathrm{2}} +\left(\mathrm{8}−\mathrm{20i}\right)\mathrm{z}−\mathrm{40i}=\mathrm{0} \\ $$$$\left.\mathrm{1}\right)\:\mathrm{Show}\:\mathrm{that}\:\left(\mathrm{E}\right)\:\mathrm{has}\:\mathrm{one}\:\mathrm{imaginary}\:\mathrm{pure}\:\mathrm{root} \\ $$$$\left.\mathrm{2}\right)\:\mathrm{solve}\:\left(\mathrm{E}\right) \\ $$ Answered by Olaf last updated…
Question Number 54787 by behi83417@gmail.com last updated on 10/Feb/19 $$\:{if}:\:\:\frac{{a}}{\:\sqrt{{b}}+\sqrt{{c}}}=\frac{{b}}{\:\sqrt{{c}}+\sqrt{{d}}}=\frac{{c}}{\:\sqrt{{d}}+\sqrt{{a}}}=\frac{{d}}{\:\sqrt{{a}}+\sqrt{{b}}} \\ $$$$\Rightarrow\frac{{a}}{{d}}=? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 54786 by behi83417@gmail.com last updated on 10/Feb/19 $${solve}\:{for}:{a},{b},{c},{d}\in\boldsymbol{{R}} \\ $$$$\:\:\:\:\:\:{a}^{\mathrm{2}} ={b}+\sqrt{{c}} \\ $$$$\:\:\:\:\:\:{b}^{\mathrm{2}} ={c}+\sqrt{{d}} \\ $$$$\:\:\:\:\:\:{c}^{\mathrm{2}} ={d}+\sqrt{{a}} \\ $$$$\:\:\:\:\:\:{d}^{\mathrm{2}} ={a}+\sqrt{{b}} \\ $$ Commented…
Question Number 54775 by MJS last updated on 10/Feb/19 $$\mathrm{found}\:\mathrm{something}\:\mathrm{interesting},\:\mathrm{it}\:\mathrm{was}\:\mathrm{published} \\ $$$$\mathrm{by}\:\mathrm{Tschirnhaus}\:\mathrm{in}\:\mathrm{1683} \\ $$$$\mathrm{we}\:\mathrm{can}\:\mathrm{reduce} \\ $$$${x}^{\mathrm{3}} +{ax}^{\mathrm{2}} +{bx}+{c}=\mathrm{0} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{to} \\ $$$${y}^{\mathrm{3}} +{py}+{q}=\mathrm{0} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{and}\:\mathrm{further}\:\mathrm{to}…
Question Number 185840 by Shrinava last updated on 28/Jan/23 Commented by mr W last updated on 29/Jan/23 $${here}\:{someone}\:{likes}\:{the}\:{question}\: \\ $$$${without}\:{to}\:{know}\:{what}\:{the}\:{question}\:{is}. \\ $$ Terms of Service…