Question Number 185982 by Shrinava last updated on 30/Jan/23 Answered by Mathspace last updated on 30/Jan/23 $${we}\:{developp}\:{first}\:\left({x}+\mathrm{1}\right)^{\mathrm{2}{n}} \\ $$$$=\sum_{{k}=\mathrm{0}} ^{\mathrm{2}{n}} {C}_{\mathrm{2}{n}} ^{{k}} {x}^{{k}} \:\:{but} \\…
Question Number 185973 by normans last updated on 30/Jan/23 $$\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\left[\boldsymbol{{prove}}\:\boldsymbol{{that}};\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}\:+\:\mathrm{2}\:=\:\mathrm{3} \\ $$$$ \\ $$ Commented by MJS_new last updated on 30/Jan/23…
Question Number 185972 by Michaelfaraday last updated on 30/Jan/23 Answered by Rasheed.Sindhi last updated on 30/Jan/23 $${x}=\sqrt[{\mathrm{3}}]{{p}}\:+\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{p}}\:}\:,\:{y}=\sqrt{{p}}\:+\frac{\mathrm{1}}{\:\sqrt{{p}}\:} \\ $$$${Show}\:{that}\:\:\:{y}^{\mathrm{2}} −\mathrm{2}={x}\left({x}^{\mathrm{2}} −\mathrm{3}\right) \\ $$$${x}^{\mathrm{3}} =\left(\sqrt[{\mathrm{3}}]{{p}}\:+\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{p}}\:}\right)^{\mathrm{3}} ={p}+\frac{\mathrm{1}}{{p}}+\mathrm{3}\left(\sqrt[{\mathrm{3}}]{{p}}\:+\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{p}}\:}\right)…
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Question Number 54857 by Tawa1 last updated on 13/Feb/19 Answered by tanmay.chaudhury50@gmail.com last updated on 13/Feb/19 $$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }+\lambda{y}=\mathrm{0} \\ $$$${y}={e}^{{mx}} \\ $$$${m}^{\mathrm{2}} {e}^{{mx}} +\lambda{e}^{{mx}}…
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Question Number 120355 by TITA last updated on 30/Oct/20 Commented by TITA last updated on 30/Oct/20 $${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)…