Question Number 25088 by naziri2013 last updated on 03/Dec/17 $$ \\ $$$$ \\ $$$$ \\ $$$${Q}…\frac{{x}+\mathrm{7}}{{x}+\mathrm{4}}>\mathrm{1},\:\:\:\:\:{x}\in{R} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\…
Question Number 156152 by amin96 last updated on 08/Oct/21 $$\begin{cases}{\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{b}}+\boldsymbol{\mathrm{c}}=\mathrm{1}}\\{\boldsymbol{\mathrm{a}}^{\mathrm{2}} +\boldsymbol{\mathrm{b}}^{\mathrm{2}} +\boldsymbol{\mathrm{c}}^{\mathrm{2}} =\mathrm{2}}\\{\boldsymbol{\mathrm{a}}^{\mathrm{3}} +\boldsymbol{\mathrm{b}}^{\mathrm{3}} +\boldsymbol{\mathrm{c}}^{\mathrm{3}} =\mathrm{3}}\end{cases} \\ $$$$\boldsymbol{\mathrm{a}}^{\mathrm{6}} +\boldsymbol{\mathrm{b}}^{\mathrm{6}} +\boldsymbol{\mathrm{c}}^{\mathrm{6}} =? \\ $$ Commented by…
Question Number 25085 by Tinkutara last updated on 03/Dec/17 $${If}\:{a}_{{n}} −{a}_{{n}−\mathrm{1}} =\mathrm{1}\:{for}\:{every}\:{positive} \\ $$$${integer}\:{greater}\:{than}\:\mathrm{1},\:{then}\:{a}_{\mathrm{1}} +{a}_{\mathrm{2}} +{a}_{\mathrm{3}} \\ $$$$+…{a}_{\mathrm{100}} \:{equals} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{5000}\:.\:{a}_{\mathrm{1}} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{5050}\:.\:{a}_{\mathrm{1}} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{5051}\:.\:{a}_{\mathrm{1}}…
Question Number 25074 by Mr easy last updated on 03/Dec/17 Commented by moxhix last updated on 03/Dec/17 $$\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…