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Category: Algebra

Given-x-1-x-5-then-x-4-1-x-4-x-2-3x-1-

Question Number 135418 by liberty last updated on 13/Mar/21 $${Given}\:{x}+\frac{\mathrm{1}}{{x}}\:=\:\mathrm{5}\:{then}\:\frac{{x}^{\mathrm{4}} +\frac{\mathrm{1}}{{x}^{\mathrm{4}} }}{{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{1}}\:? \\ $$ Answered by SEKRET last updated on 13/Mar/21 $$\:\left(\boldsymbol{\mathrm{x}}+\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}}\right)^{\mathrm{2}} =\mathrm{5}^{\mathrm{2}} \:\:\:\:\:\:\boldsymbol{\mathrm{x}}^{\mathrm{2}}…

Question-135415

Question Number 135415 by JulioCesar last updated on 13/Mar/21 Commented by liberty last updated on 13/Mar/21 $$\left({nx}+\mathrm{1}\right)^{{p}} \:=\:\underset{{i}=\mathrm{0}} {\overset{{p}} {\sum}}\:{C}_{{i}} ^{\:{p}} \:\left({nx}\right)^{{p}−{i}} \: \\ $$$$\:=\:\left({nx}\right)^{{p}}…

Analyze-for-integer-solution-a-b-c-are-fixed-positive-integers-ax-a-by-b-cz-c-

Question Number 4266 by Rasheed Soomro last updated on 06/Jan/16 $$\mathrm{Analyze}\:\mathrm{for}\:\:\mathrm{integer}\:\mathrm{solution}\:: \\ $$$$\mathrm{a},\mathrm{b},\mathrm{c}\:\mathrm{are}\:\mathrm{fixed}\:\mathrm{positive}\:\mathrm{integers}. \\ $$$$\mathrm{ax}^{\mathrm{a}} +\mathrm{by}^{\mathrm{b}} =\mathrm{cz}^{\mathrm{c}} \\ $$ Commented by Yozzii last updated on…

Let-u-ln-3-ln-3-ln-3-ln-2-ln-2-ln-3-ln-2-ln-2-ln-3-ln-3-ln-2-ln-2-ln-3-ln-2-What-is-the-value-of-u-Let-k-ln-x-ln-x-

Question Number 4250 by Yozzii last updated on 06/Jan/16 $${Let}\:{u}=\frac{{ln}\left(\mathrm{3}−\left\{\frac{{ln}\left(\mathrm{3}−\left[\frac{{ln}\left(\mathrm{3}−\left(\ldots\right.\right.}{{ln}\left(\mathrm{2}+\left(\ldots\right.\right.}\right.\right.}{{ln}\left(\mathrm{2}+\left[\frac{{ln}\left(\mathrm{3}−\left(\ldots\right.\right.}{{ln}\left(\mathrm{2}+\left(\ldots\right.\right.}\right.\right.}\right\}\right)}{{ln}\left(\mathrm{2}+\left\{\frac{{ln}\left(\mathrm{3}−\left[\frac{{ln}\left(\mathrm{3}−\left(\ldots\right.\right.}{{ln}\left(\mathrm{2}+\left(\ldots\right.\right.}\right.\right.}{{ln}\left(\mathrm{2}+\left[\frac{{ln}\left(\mathrm{3}−\left(\ldots\right.\right.}{{ln}\left(\mathrm{2}+\left(\ldots\right.\right.}\right.\right.}\right\}\right)}. \\ $$$${What}\:{is}\:{the}\:{value}\:{of}\:{u}?\: \\ $$$$ \\ $$$${Let}\:{k}=\frac{{ln}\left({x}−\left\{\frac{{ln}\left({x}−\left[\frac{{ln}\left({x}−\left(\ldots\right.\right.}{{ln}\left({x}−\mathrm{1}+\left(\ldots\right.\right.}\right.\right.}{{ln}\left({x}−\mathrm{1}+\left[\frac{{ln}\left({x}−\left(\ldots\right.\right.}{{ln}\left({x}−\mathrm{1}+\left(\ldots\right.\right.}\right.\right.}\right\}\right)}{{ln}\left({x}−\mathrm{1}+\left\{\frac{{ln}\left({x}−\left[\frac{{ln}\left({x}−\left(\ldots\right.\right.}{{ln}\left({x}−\mathrm{1}+\left(\ldots\right.\right.}\right.\right.}{{ln}\left({x}−\mathrm{1}+\left[\frac{{ln}\left({x}−\left(\ldots\right.\right.}{{ln}\left({x}−\mathrm{1}+\left(\ldots\right.\right.}\right.\right.}\right\}\right)}. \\ $$$${For}\:{what}\:{values}\:{of}\:{x}\:{does} \\ $$$$\left({i}\right)\:{k}\:{converge}\:\left({ii}\right)\:{k}\:{diverge}? \\ $$ Commented by Yozzii…

Question-135305

Question Number 135305 by JulioCesar last updated on 12/Mar/21 Answered by MJS_new last updated on 12/Mar/21 $$\int\frac{{dx}}{{a}^{{x}} +{b}}= \\ $$$$\:\:\:\:\:\left[{t}={a}^{{x}} +{b}\:\rightarrow\:{dx}=\frac{{dt}}{{a}^{{x}} \mathrm{ln}\:{a}}\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{ln}\:{a}}\int\frac{{dt}}{{t}\left({t}−{b}\right)}=\frac{\mathrm{1}}{{b}\mathrm{ln}\:{a}}\int\left(\frac{\mathrm{1}}{{t}−{b}}−\frac{\mathrm{1}}{{t}}\right){dt}= \\…

n-m-Z-n-n-n-1-n-2-if-n-m-where-we-know-the-value-m-but-not-n-can-we-solve-for-n-

Question Number 4234 by Filup last updated on 04/Jan/16 $${n},\:{m}\in\mathbb{Z} \\ $$$${n}!={n}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)… \\ $$$$ \\ $$$$\mathrm{if}\:{n}!={m}\:\mathrm{where}\:\mathrm{we}\:\mathrm{know}\:\mathrm{the}\:\mathrm{value} \\ $$$${m}\:\mathrm{but}\:\mathrm{not}\:{n},\:\mathrm{can}\:\mathrm{we}\:\mathrm{solve}\:\mathrm{for}\:{n}? \\ $$ Commented by prakash jain last…

2-3-1-2-3-1-2-3-1-2-3-1-

Question Number 135300 by bobhans last updated on 12/Mar/21 $$\left(\sqrt{\mathrm{2}}\:+\sqrt{\mathrm{3}}\:+\mathrm{1}\right)\left(\sqrt{\mathrm{2}}\:−\sqrt{\mathrm{3}}\:+\mathrm{1}\right)\left(\sqrt{\mathrm{2}}\:+\sqrt{\mathrm{3}}−\mathrm{1}\right)\left(\sqrt{\mathrm{2}}−\sqrt{\mathrm{3}}−\mathrm{1}\right)=? \\ $$ Answered by Olaf last updated on 12/Mar/21 $$\left(\sqrt{\mathrm{2}}+\sqrt{\mathrm{3}}+\mathrm{1}\right)\left(\sqrt{\mathrm{2}}−\sqrt{\mathrm{3}}−\mathrm{1}\right)\:=\:\mathrm{2}−\left(\sqrt{\mathrm{3}}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$$=\:−\mathrm{2}−\mathrm{2}\sqrt{\mathrm{3}}\:=\:−\mathrm{2}\left(\mathrm{1}+\sqrt{\mathrm{3}}\right) \\ $$$$\left(\sqrt{\mathrm{2}}+\sqrt{\mathrm{3}}−\mathrm{1}\right)\left(\sqrt{\mathrm{2}}−\sqrt{\mathrm{3}}+\mathrm{1}\right)\:=\:\mathrm{2}−\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)^{\mathrm{2}} \\…