Question Number 165433 by mathlove last updated on 01/Feb/22 Answered by aleks041103 last updated on 01/Feb/22 $$ \\ $$$$\frac{\mathrm{3}}{\mathrm{2}^{{x}} }+\frac{\mathrm{3}}{\mathrm{3}^{{x}} }=\mathrm{2}\frac{\left(\mathrm{3}/\mathrm{2}\right)^{{x}} }{\mathrm{2}^{{x}} }+\mathrm{2}\frac{\left(\mathrm{3}/\mathrm{2}\right)^{−{x}} }{\mathrm{3}^{{x}} }…
Question Number 165435 by mathlove last updated on 01/Feb/22 Answered by Kamel last updated on 01/Feb/22 Commented by mindispower last updated on 02/Feb/22 $${Barak}\:{Alah}\:{fik}\:{Nice}\:{Solution}\:{Sir} \\…
Question Number 34357 by math1967 last updated on 05/May/18 $${If}\:{a}+{b}+{c}=\mathrm{0}\:\:{prove}\:{that} \\ $$$$\left.{i}\right)\left(\frac{{a}}{{b}+{c}}+\:\frac{{b}}{{c}+{a}}\:+\frac{{c}}{{a}+{b}}\right)\left(\frac{{b}+{c}}{{a}}\:+\frac{{c}+{a}}{{b}}\:+\frac{{a}+{b}}{{c}}\right)=\mathrm{9} \\ $$$$\left.{ii}\right)\left(\frac{{a}}{{b}−{c}}\:+\frac{{b}}{{c}−{a}}\:+\frac{{c}}{{a}−{b}}\right)\left(\frac{{b}−{c}}{{a}}\:+\frac{{c}−{a}}{{b}}\:+\frac{{a}−{b}}{{c}}\right)=\mathrm{9} \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 05/May/18 $$\left.{i}\right)\left\{\left(\frac{{a}}{{b}+{c}}+\mathrm{1}−\mathrm{1}\right)+\left(\frac{{b}}{{c}+{a}}+\mathrm{1}−\mathrm{1}\right)+\left(\frac{{c}}{{a}+{b}}+\mathrm{1}−\mathrm{1}\right)\right\}× \\…
Question Number 165419 by mathlove last updated on 01/Feb/22 Answered by TheSupreme last updated on 01/Feb/22 $$\left({a}+\frac{\mathrm{1}}{{a}}\right)={x} \\ $$$${a}^{\mathrm{2}} +\frac{\mathrm{1}}{{a}^{\mathrm{2}} }=\left({a}+\frac{\mathrm{1}}{{a}}\right)^{\mathrm{2}} −\mathrm{2}={x}^{\mathrm{2}} −\mathrm{2} \\ $$$${a}^{\mathrm{3}}…
Question Number 99877 by I want to learn more last updated on 23/Jun/20 Commented by bobhans last updated on 24/Jun/20 $$\left(\mathrm{3}.\mathrm{1}\right)\mathrm{gradient}\:\mathrm{of}\:\mathrm{PQ}\:=\:\mathrm{tan}\:\mathrm{45}^{\mathrm{o}} \:=\:\mathrm{1} \\ $$$$\left(\mathrm{3}.\mathrm{2}\right)\mathrm{MN}\:\mathrm{line}\:\mathrm{parallel}\:\mathrm{to}\:\mathrm{PQ}\:\Rightarrow\mathrm{then}\:\mathrm{equation} \\…
Question Number 99869 by bachamohamed last updated on 23/Jun/20 $$\:\boldsymbol{{tng}}\frac{\boldsymbol{\pi}}{\mathrm{9}}\:\:+\:\mathrm{4}\boldsymbol{{sin}}\frac{\boldsymbol{\pi}}{\mathrm{9}}\:=\sqrt{\mathrm{3}} \\ $$ Answered by smridha last updated on 23/Jun/20 $$\frac{\boldsymbol{{sin}}\left(\frac{\boldsymbol{\pi}}{\mathrm{9}}\right)+\mathrm{2}\boldsymbol{{sin}}\left(\frac{\mathrm{2}\boldsymbol{\pi}}{\mathrm{9}}\right)}{\boldsymbol{{cos}}\left(\frac{\boldsymbol{\pi}}{\mathrm{9}}\right)}=\frac{\boldsymbol{{sin}}\left(\frac{\mathrm{2}\boldsymbol{\pi}}{\mathrm{9}}\right)+\boldsymbol{{cos}}\left(\frac{\boldsymbol{\pi}}{\mathrm{18}}\right)}{\boldsymbol{{cos}}\left(\frac{\boldsymbol{\pi}}{\mathrm{9}}\right)}\:\: \\ $$$$=\frac{\boldsymbol{{sin}}\left(\frac{\mathrm{4}\boldsymbol{\pi}}{\mathrm{9}}\right)+\boldsymbol{{sin}}\left(\frac{\mathrm{2}\boldsymbol{\pi}}{\mathrm{9}}\right)}{\boldsymbol{{cos}}\left(\frac{\boldsymbol{\pi}}{\mathrm{9}}\right)}=\frac{\mathrm{2}.\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}.\boldsymbol{{cos}}\left(\frac{\boldsymbol{\pi}}{\mathrm{9}}\right)}{\boldsymbol{{cos}}\left(\frac{\boldsymbol{\pi}}{\mathrm{9}}\right)}=\sqrt{\mathrm{3}} \\ $$ Answered…
Question Number 165381 by mathlove last updated on 31/Jan/22 Commented by MJS_new last updated on 02/Feb/22 $$\mathrm{we}\:\mathrm{have}\:\mathrm{only}\:\mathrm{one}\:\mathrm{equation}\:\mathrm{for}\:\mathrm{2}\:\mathrm{unknowns} \\ $$$$\Rightarrow\:\mathrm{no}\:\mathrm{unique}\:\mathrm{solution}. \\ $$ Answered by mahdipoor last…
Question Number 99846 by bachamohamed last updated on 23/Jun/20 $$\:\:\:\:\boldsymbol{{li}}\underset{\boldsymbol{{n}}\rightarrow\infty} {\boldsymbol{{m}}}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}!}\right)^{\left(\frac{\mathrm{1}}{\mathrm{2}!}−\frac{\mathrm{1}}{\mathrm{3}!}\right)^{………\left(\frac{\mathrm{1}}{\boldsymbol{{n}}!}−\frac{\mathrm{1}}{\left(\boldsymbol{{n}}+\mathrm{1}\right)!}\right)} } =? \\ $$ Commented by Dwaipayan Shikari last updated on 23/Jun/20 $${Ans}=\mathrm{1} \\…
Question Number 34307 by prof Abdo imad last updated on 03/May/18 $${let}\:{give}\:{A}_{{n}} =\:\:\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\:\:\:\:\frac{\alpha}{{n}}}\\{−\frac{\alpha}{{n}}\:\:\:\:\:\:\mathrm{1}}\end{pmatrix} \\ $$$${calculate}\:{lim}_{{n}\rightarrow+\infty} \:{A}_{{n}} ^{{n}} \:\:\:\:. \\ $$ Commented by math khazana by…
Question Number 34305 by abdo mathsup 649 cc last updated on 03/May/18 $$\left.{let}\right]\:{A}\:=\begin{pmatrix}{\mathrm{1}\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\mathrm{0}}\\{\mathrm{1}\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\mathrm{1}}\end{pmatrix} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{0}\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\mathrm{1}\:\right. \\ $$$$\left.\mathrm{1}\right){find}\:{the}\:{caractetistic}\:{polynom}\:{of}\:{A} \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:{A}^{{n}} \\ $$ Commented by prof Abdo…