Question Number 25114 by Mr easy last updated on 04/Dec/17 Commented by moxhix last updated on 04/Dec/17 $${x}=\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+{x}}}}>\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+\mathrm{0}}} \\ $$$$\therefore{x}>\sqrt{\mathrm{2}} \\ $$$${put}\:\phi:\:\:\phi=\sqrt{\mathrm{1}+\phi}\:\:\left(\phi>\mathrm{1}\right) \\ $$$$\phi^{\mathrm{2}} −\phi−\mathrm{1}=\mathrm{0}…
Question Number 25112 by Mr easy last updated on 04/Dec/17 Commented by prakash jain last updated on 04/Dec/17 $$\left({a}+{b}\right)^{\mathrm{2}{n}} =\underset{{i}=\mathrm{0}} {\overset{\mathrm{2}{n}} {\sum}}\:^{\mathrm{2}{n}} {C}_{{i}} {a}^{{i}} {b}^{\mathrm{2}{n}−{i}}…
Question Number 156180 by SANOGO last updated on 08/Oct/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 156176 by MathSh last updated on 08/Oct/21 Answered by ghimisi last updated on 09/Oct/21 $$\frac{{a}^{\mathrm{2}} }{{x}^{\mathrm{2}} }+\frac{\mathrm{1}}{{y}^{\mathrm{2}} }+\frac{{c}^{\mathrm{2}} }{{z}^{\mathrm{2}} }=\mathrm{36} \\ $$$$\left.\mathrm{36}\centerdot\mathrm{4}=\left(\frac{{a}^{\mathrm{2}} }{{x}^{\mathrm{2}}…
Question Number 156172 by MathSh last updated on 08/Oct/21 Commented by Rasheed.Sindhi last updated on 09/Oct/21 $$\mathrm{Infinitely}\:\mathrm{many}\:\mathrm{sets}. \\ $$ Commented by MathSh last updated on…
Question Number 156170 by MathSh last updated on 08/Oct/21 Answered by mindispower last updated on 09/Oct/21 $${Hello}\:{sir}\:{withe}\:{resodue}\left[{theorm}\:{is}\:{diractly}\:\right. \\ $$$$\frac{{p}\left({x}\right)}{{q}\left({x}\right)}=\:{deg}\:{q}>{deg}\left({p}\right)+\mathrm{1}\:{in}\:{this}\:{case} \\ $$$$ \\ $$$$ \\ $$…
Question Number 156164 by MathSh last updated on 08/Oct/21 Answered by ghimisi last updated on 09/Oct/21 $$\mid{x}\mid=\frac{\mathrm{1}}{\mathrm{2}}\mid{x}+{y}+{x}−{y}\mid\leqslant\frac{\mathrm{1}}{\mathrm{2}}\mid{x}+{y}\mid+\frac{\mathrm{1}}{\mathrm{2}}\mid{y}−{x}\mid\:\:\left(\mathrm{1}\right) \\ $$$$\mid{y}\mid=\frac{\mathrm{1}}{\mathrm{2}}\mid{y}+{x}+{y}−{x}\mid\leqslant\frac{\mathrm{1}}{\mathrm{2}}\mid{x}+{y}\mid+\frac{\mathrm{1}}{\mathrm{2}}\mid{y}−{x}\mid\mid\:\:\left(\mathrm{2}\right) \\ $$$$\mid\mathrm{3}{x}+\mathrm{2}{y}\mid=\mid\mathrm{4}{x}+\mathrm{3}{y}−\left({x}+{y}\right)\mid\leqslant\mid\mathrm{4}{x}+\mathrm{3}{y}\mid+\mid{x}+{y}\mid\:\:\left(\mathrm{3}\right) \\ $$$$\left(\mathrm{1}\right)+\left(\mathrm{2}\right)+\left(\mathrm{3}\right)\Rightarrow \\ $$$$\mid{x}\mid+\mid{y}\mid+\mid\mathrm{3}{x}+\mathrm{2}{y}\mid\leqslant\mid\mathrm{4}{x}+\mathrm{3}{y}\mid+\mathrm{2}\mid{x}+{y}\mid+\mid{y}−{x}\mid…
Question Number 25088 by naziri2013 last updated on 03/Dec/17 $$ \\ $$$$ \\ $$$$ \\ $$$${Q}…\frac{{x}+\mathrm{7}}{{x}+\mathrm{4}}>\mathrm{1},\:\:\:\:\:{x}\in{R} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\…
Question Number 156152 by amin96 last updated on 08/Oct/21 $$\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}}…