Question Number 25074 by Mr easy last updated on 03/Dec/17 Commented by moxhix last updated on 03/Dec/17 $$\left({x}−{a}\right)\left({x}−{b}\right)\left({x}−{c}\right)\left({x}−{d}\right)=\mathrm{0} \\ $$$${x}^{\mathrm{4}} −\left({a}+{b}+{c}+{d}\right){x}^{\mathrm{3}} +\left({ab}+{bc}+{cd}+{da}\right){x}^{\mathrm{2}} −\left({abc}+{bcd}+{cda}+{dab}\right){x}+{abcd}=\mathrm{0} \\ $$$$\begin{cases}{{a}+{b}+{c}+{d}=\mathrm{0}}\\{{ab}+{bc}+{cd}+{da}=\mathrm{0}}\\{{abc}+{bcd}+{cda}+{dab}={k}}\\{{abcd}=−\mathrm{15}}\end{cases}…
Question Number 25066 by Mr easy last updated on 02/Dec/17 $${let}\:{a},{b},{c},{x},{y}\:{and}\:{z}\:{be}\:{complex}\:{number} \\ $$$${such}\:{that}\:{a}=\frac{{b}+{c}}{{x}−\mathrm{2}}\:,{b}=\frac{{c}+{a}}{{y}−\mathrm{2}}\:\:\:\:{c}=\frac{{a}+{b}}{{z}−\mathrm{2}}. \\ $$$${xy}\:+{yz}\:+{zx}=\mathrm{1000}\:{and}\:{x}+{y}+{z}=\mathrm{2016} \\ $$$${find}\:{the}\:{value}\:{of}\:{xyz}. \\ $$ Answered by ajfour last updated on…
Question Number 25054 by Sattwik Chakraborty last updated on 02/Dec/17 $${If}\:{x}:{y}=\mathrm{5}:\mathrm{2},\:{then}\:{find}\:{the}\:{value}\:{of}\:\left(\mathrm{8}{x}+\mathrm{9}{y}\right)/\left(\mathrm{8}{x}+\mathrm{27}\right) \\ $$ Answered by $@ty@m last updated on 02/Dec/17 $$\frac{\mathrm{8}{x}+\mathrm{9}{y}}{\mathrm{8}{x}+\mathrm{27}{y}} \\ $$$${Divide}\:{Nr}\:\&{Dr}\:{by}\:{y} \\ $$$$\frac{\mathrm{8}\frac{{x}}{{y}}+\mathrm{9}}{\mathrm{8}\frac{{x}}{{y}}+\mathrm{27}}…
Question Number 25053 by Sattwik Chakraborty last updated on 02/Dec/17 $${If}\:\mathrm{2}{A}=\mathrm{3}{B}=\mathrm{4}{C}\:{find}\:{the}\:{value}\:{of}\:{A}:{B}:{C} \\ $$ Answered by $@ty@m last updated on 02/Dec/17 $$\mathrm{2}{A}=\mathrm{3}{B} \\ $$$$\Rightarrow{A}:{B}=\mathrm{3}:\mathrm{2}\:\:{or}\:{A}:{B}=\mathrm{6}:\mathrm{4} \\ $$$$\:\:{Similarly}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{B}:{C}=\mathrm{4}:\mathrm{3}…
Question Number 25046 by Tinkutara last updated on 02/Dec/17 $${Show}\:{that} \\ $$$$\left({a}\right)\:{N}=\frac{\mathrm{10}^{\mathrm{143}} −\mathrm{1}}{\mathrm{9}}\:{is}\:{composite},\:{and} \\ $$$$\left({b}\right)\:{N}\:{has}\:{two}\:{factors}\:{each}\:{of}\:{which}\:{is} \\ $$$${a}\:{series}\:{of}\:{a}\:{G}.{P}. \\ $$ Answered by Rasheed.Sindhi last updated on…
Question Number 25049 by Tinkutara last updated on 02/Dec/17 $$\mathrm{If}\:{I}\:=\:\underset{{k}=\mathrm{1}} {\overset{\mathrm{98}} {\sum}}\underset{{k}} {\overset{{k}+\mathrm{1}} {\int}}\frac{{k}\:+\:\mathrm{1}}{{x}\left({x}\:+\:\mathrm{1}\right)}{dx},\:\mathrm{then} \\ $$$$\left(\mathrm{1}\right)\:{I}\:>\:\frac{\mathrm{49}}{\mathrm{50}} \\ $$$$\left(\mathrm{2}\right)\:{I}\:<\:\frac{\mathrm{49}}{\mathrm{50}} \\ $$$$\left(\mathrm{3}\right)\:{I}\:<\:\mathrm{log}_{{e}} \mathrm{99} \\ $$$$\left(\mathrm{4}\right)\:{I}\:>\:\mathrm{log}_{{e}} \mathrm{99} \\…
Question Number 156119 by Ghaniy last updated on 08/Oct/21 $$\mathrm{solve}\:: \\ $$$$\:\frac{\mathrm{1}+\mathrm{2x}}{\mathrm{1}+\sqrt{\mathrm{1}+\mathrm{2x}}}+\frac{\mathrm{1}−\mathrm{2x}}{\mathrm{1}−\sqrt{\mathrm{1}−\mathrm{2x}}}=\mathrm{1} \\ $$$$ \\ $$ Commented by immortel last updated on 08/Oct/21 Commented by…
Question Number 156107 by cortano last updated on 08/Oct/21 $$\:\:\:\:\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}}\right)^{\mathrm{x}+\mathrm{1}} =\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2019}}\right)^{\mathrm{2019}} \\ $$ Commented by mr W last updated on 08/Oct/21 $$\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)^{{x}+\mathrm{1}} \\ $$$$=\left(\frac{{x}+\mathrm{1}}{{x}}\right)^{{x}+\mathrm{1}} \\…
Question Number 156103 by MathSh last updated on 07/Oct/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 25025 by tawa tawa last updated on 01/Dec/17 $$\mathrm{If}\:\:\:\:\mathrm{a}^{\mathrm{4}} \:+\:\mathrm{b}^{\mathrm{4}} \:+\:\mathrm{c}^{\mathrm{4}} \:+\:\mathrm{d}^{\mathrm{4}} \:=\:\mathrm{16},\:\:\mathrm{prove}\:\mathrm{that}:\:\:\mathrm{a}^{\mathrm{5}} \:+\:\mathrm{b}^{\mathrm{5}} \:+\:\mathrm{c}^{\mathrm{5}} \:+\:\mathrm{d}^{\mathrm{5}} \:\leqslant\:\mathrm{32} \\ $$$$\mathrm{for}\:\:\mathrm{a},\:\mathrm{b},\:\mathrm{c},\:\mathrm{d}\:\in\:\mathbb{R} \\ $$ Answered by…