Question Number 187989 by mnjuly1970 last updated on 24/Feb/23 Answered by HeferH last updated on 24/Feb/23 $$\:{b}^{\mathrm{2}} ={ac} \\ $$$$\:{b}^{\mathrm{3}} ={abc} \\ $$$$\:\frac{{b}^{\mathrm{3}} {c}^{\mathrm{3}} +{a}^{\mathrm{3}}…
Question Number 187980 by Mingma last updated on 24/Feb/23 Answered by Rasheed.Sindhi last updated on 24/Feb/23 $$\bullet{x}^{\mathrm{2}} +{bx}+{c}=\mathrm{0}\: \\ $$$$\alpha+\beta=−{b}\:,\:\:\:\alpha\beta={c} \\ $$$$\bullet{x}^{\mathrm{2}} +{bx}−{c}=\mathrm{0} \\ $$$$\gamma+\delta=−{b}\:\:,\:\:\gamma\delta=−{c}…
Question Number 122445 by benjo_mathlover last updated on 17/Nov/20 Answered by TANMAY PANACEA last updated on 17/Nov/20 $${using}\:{Titu}\:{lemma} \\ $$$$\frac{\mathrm{1}^{\mathrm{2}} }{{x}}+\frac{\mathrm{3}^{\mathrm{2}} }{{y}}+\frac{\mathrm{5}^{\mathrm{2}} }{{z}}\geqslant\frac{\left(\mathrm{1}+\mathrm{3}+\mathrm{5}\right)^{\mathrm{2}} }{{x}+{y}+{z}} \\…
Question Number 56904 by pete last updated on 26/Mar/19 $$\mathrm{If}\:\alpha\:\mathrm{and}\:\beta\:\mathrm{are}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\mathrm{3x}^{\mathrm{2}} −\mathrm{x}−\mathrm{3}=\mathrm{0},\:\mathrm{find}\:\mathrm{thevalue}\:\mathrm{of}\:\left(\alpha^{\mathrm{2}} −\beta^{\mathrm{2}} \right) \\ $$$$\mathrm{if}\:\alpha>\beta. \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on…
Question Number 187971 by mnjuly1970 last updated on 24/Feb/23 $$ \\ $$$$\:\:\mathrm{If}\:\:\:,\:\:{x}^{\:\mathrm{5}} \:=\:\mathrm{1}\:\:\:\wedge\:\:{x}\neq\mathrm{1} \\ $$$$ \\ $$$$\:\:\:\:\left(\:\frac{\:\mathrm{1}}{{x}^{\:\mathrm{2}} \:−{x}\:+\mathrm{1}}\:+\:\frac{\mathrm{1}}{{x}^{\:\mathrm{2}} \:+\:{x}\:+\mathrm{1}}\:\right)^{\:\mathrm{10}} =\:? \\ $$$$ \\ $$ Answered…
Question Number 187951 by chhuangakawlni last updated on 24/Feb/23 $${please}\:{help}\:{me}\:{to}\:{solve} \\ $$$$\mathrm{6}{x}^{\mathrm{2}} +{x}−\mathrm{2}=\mathrm{0} \\ $$ Answered by cortano12 last updated on 24/Feb/23 $$\left(\mathrm{x}+\frac{\mathrm{4}}{\mathrm{6}}\right)\left(\mathrm{x}−\frac{\mathrm{3}}{\mathrm{6}}\right)=\mathrm{0} \\ $$$$\:\begin{cases}{\mathrm{x}=−\frac{\mathrm{2}}{\mathrm{3}}}\\{\mathrm{x}=\frac{\mathrm{1}}{\mathrm{2}}}\end{cases}…
Question Number 187916 by mustafazaheen last updated on 23/Feb/23 $${how}\:{is}\:{solution} \\ $$$${y}=\left({cosx}\right)^{\left(\mathrm{3}{x}^{\mathrm{2}} −\mathrm{1}\right)^{{e}^{{x}} } } \\ $$$$\frac{{dy}}{{dx}}=? \\ $$ Answered by mr W last updated…
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Question Number 187905 by Shlock last updated on 23/Feb/23 Commented by Shlock last updated on 23/Feb/23 Let a, b and c be nonnegative real numbers such that a² + b² + c² = 1, prove that Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 122360 by benjo_mathlover last updated on 16/Nov/20 Answered by $@y@m last updated on 16/Nov/20 $$\:\:{Put}\:{t}=\left(\sqrt{\mathrm{5}−\mathrm{2}\sqrt{\mathrm{6}}}\right)^{{x}} \\ $$$$\:\:{t}+\frac{\mathrm{1}}{{t}}=\mathrm{10} \\ $$$${t}^{\mathrm{2}} −\mathrm{10}{t}+\mathrm{1}=\mathrm{0} \\ $$$${t}=\frac{\mathrm{10}\pm\sqrt{\mathrm{100}−\mathrm{4}}}{\mathrm{2}} \\…