Question Number 191839 by MATHEMATICSAM last updated on 01/May/23 $$\mathrm{2}^{{a}} \:=\:\mathrm{3}^{{b}} \:=\:\mathrm{36}^{{c}} \:\mathrm{then}\:\mathrm{prove}\:\mathrm{that} \\ $$$${ab}\:=\:\mathrm{2}{c}\left({a}\:+\:{b}\right). \\ $$ Answered by AST last updated on 01/May/23 $${a}={blog}_{\mathrm{2}}…
Question Number 191833 by Mingma last updated on 01/May/23 Answered by AST last updated on 01/May/23 $${a}={log}_{\mathrm{2}} \mathrm{900};{b}={log}_{\mathrm{3}} \mathrm{900};{c}={log}_{\mathrm{5}} \mathrm{900} \\ $$$$\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}={log}_{\mathrm{900}} \mathrm{2}+{log}_{\mathrm{900}} \mathrm{3}+{log}_{\mathrm{900}} \mathrm{5}…
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Question Number 60745 by Tawa1 last updated on 25/May/19 Commented by Prithwish sen last updated on 25/May/19 $$\frac{\left[\left(\mathrm{n}+\mathrm{1}\right)!\right]^{\mathrm{n}} }{\left(\mathrm{n}!\right)^{\mathrm{n}+\mathrm{1}} }\:=\left[\frac{\left(\mathrm{n}+\mathrm{1}\right)!}{\mathrm{n}!}\right]^{\mathrm{n}} .\frac{\mathrm{1}}{\mathrm{n}!}\:=\frac{\left(\mathrm{n}+\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{n}!} \\ $$$$=\frac{\left(\mathrm{n}+\mathrm{1}\right)}{\mathrm{1}}.\frac{\left(\mathrm{n}+\mathrm{1}\right)}{\mathrm{2}}…………\frac{\left(\mathrm{n}+\mathrm{1}\right)}{\mathrm{n}}\:>\mathrm{1} \\…
Question Number 60734 by Tawa1 last updated on 25/May/19 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{product}\:\mathrm{of}\:\mathrm{the}\:\mathrm{real}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\:\:\:\:\:\:\:\:\:\left(\mathrm{x}\:+\:\mathrm{2}\:+\:\sqrt{\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{4x}\:+\:\mathrm{3}}\right)^{\mathrm{5}} \:−\:\:\mathrm{32}\left(\mathrm{x}\:+\:\mathrm{2}\:−\:\sqrt{\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{4x}\:+\:\mathrm{3}}\right)^{\mathrm{5}} \:\:=\:\:\mathrm{31} \\ $$ Commented by Prithwish sen last updated on…
Question Number 191798 by Abdullahrussell last updated on 30/Apr/23 Commented by Frix last updated on 30/Apr/23 $$\mathrm{Qu}\:\mathrm{191753};\:\mathrm{just}\:\mathrm{add}\:−\mathrm{3}+\mathrm{24}=\mathrm{21} \\ $$ Commented by BaliramKumar last updated on…
Question Number 60723 by alphaprime last updated on 25/May/19 $${solve}\:{for}\:{x}\: \\ $$$$\sqrt{{a}−\sqrt{{a}+{x}}}\:+\:\sqrt{{a}+\sqrt{{a}−{x}}}\:=\:\mathrm{2}{x} \\ $$ Commented by MJS last updated on 25/May/19 $$\mathrm{what}'\mathrm{s}\:\mathrm{the}\:\mathrm{source}\:\mathrm{of}\:\mathrm{this}\:\mathrm{problem}? \\ $$ Commented…
Question Number 191790 by mathlove last updated on 30/Apr/23 $${prove}\:{that} \\ $$$$\frac{\mathrm{2}{x}−\mathrm{4}}{\mathrm{2}\centerdot\mathrm{3}\centerdot\mathrm{4}}+\frac{\mathrm{3}{x}−\mathrm{5}}{\mathrm{3}\centerdot\mathrm{4}\centerdot\mathrm{5}}+\frac{\mathrm{4}{x}−\mathrm{6}}{\mathrm{4}\centerdot\mathrm{5}\centerdot\mathrm{6}}+…..+\frac{\mathrm{100}{x}−\mathrm{102}}{\mathrm{100}\centerdot\mathrm{101}\centerdot\mathrm{102}}=\frac{\mathrm{103}}{\mathrm{102}} \\ $$$$ \\ $$ Commented by BaliramKumar last updated on 01/May/23 $$\mathrm{if}\:\mathrm{x}=\mathrm{1}\:\mathrm{then}\:\mathrm{all}\:\mathrm{term}\:\left(−\mathrm{ve}\right)\:\mathrm{but}\:\frac{\mathrm{103}}{\mathrm{102}}\:\mathrm{is}\:\left(+\mathrm{ve}\right)\: \\…
Question Number 60705 by behi83417@gmail.com last updated on 24/May/19 Answered by MJS last updated on 25/May/19 $$\mathrm{the}\:\mathrm{only}\:\mathrm{solution}\:\mathrm{is}\:{x}={y}=\mathrm{1} \\ $$ Commented by behi83417@gmail.com last updated on…
Question Number 191775 by Mingma last updated on 30/Apr/23 Answered by AST last updated on 30/Apr/23 $$\overset{\_\_\_\_\_\_\_\_\_\_\_\_} {{ABCDEF}}=\mathrm{10000}\overset{\_\_\_\_\_\_\_} {{ABCD}}+\overset{\_\_\_} {{EF}} \\ $$$$\mathrm{17}\left[\mathrm{10000}\overset{\_\_\_} {{AB}}+\overset{\_\_\_\_\_\_\_} {{CDEF}}\right]=\mathrm{12}\left[\mathrm{100}\overset{\_\_\_\_\_\_\_} {{CDEF}}+\overset{\_\_\_}…