Question Number 43707 by Tawa1 last updated on 14/Sep/18 $$\mathrm{Simplify}:\:\:\: \\ $$$$\left(\mathrm{x}\:+\:\mathrm{y}\:+\:\mathrm{z}\right)\left(\mathrm{x}^{−\mathrm{1}} \:+\:\mathrm{y}^{−\mathrm{1}} \:+\:\mathrm{z}^{−\mathrm{1}} \right)\:=\:\left(\mathrm{x}^{−\mathrm{1}} \:\mathrm{y}^{−\mathrm{1}} \:\mathrm{z}^{−\mathrm{1}} \right)\left(\mathrm{x}\:+\:\mathrm{y}\right)\left(\mathrm{y}\:+\:\mathrm{z}\right)\left(\mathrm{z}\:+\:\mathrm{x}\right) \\ $$ Commented by math1967 last updated…
Question Number 43706 by Tawa1 last updated on 14/Sep/18 $$\mathrm{If}\:\:\mathrm{pqr}\:=\:\mathrm{1} \\ $$$$\mathrm{Hence}\:\mathrm{evaluate}:\:\:\frac{\mathrm{1}}{\mathrm{1}\:+\:\mathrm{e}\:+\:\mathrm{f}^{−\mathrm{1}} }\:\:+\:\:\frac{\mathrm{1}}{\mathrm{1}\:+\:\mathrm{f}\:+\:\mathrm{g}^{−\mathrm{1}} }\:\:+\:\:\frac{\mathrm{1}}{\mathrm{1}\:+\:\mathrm{g}\:+\:\mathrm{e}^{−\mathrm{1}} } \\ $$ Commented by Joel578 last updated on 14/Sep/18 $${p},{q},{r}\:\mathrm{and}\:{e},{f},{g}\:?…
Question Number 43702 by rajesh123456 last updated on 14/Sep/18 $${simplify}\:\:\:\left[\frac{\mathrm{12}^{\mathrm{1}/\mathrm{5}} }{\mathrm{27}^{\mathrm{1}/\mathrm{5}} }\right]^{\mathrm{5}/\mathrm{2}} \\ $$ Answered by Joel578 last updated on 14/Sep/18 $$\left(\frac{\mathrm{12}^{\mathrm{1}/\mathrm{5}} }{\mathrm{27}^{\mathrm{1}/\mathrm{5}} }\right)^{\mathrm{5}/\mathrm{2}} \:=\:\frac{\sqrt{\mathrm{12}}}{\:\sqrt{\mathrm{27}}}\:=\:\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{3}\sqrt{\mathrm{3}}}\:=\:\frac{\mathrm{2}}{\mathrm{3}}…
Question Number 43705 by Tawa1 last updated on 14/Sep/18 $$\mathrm{Prove}\:\mathrm{that}\:\mathrm{to}\:\mathrm{each}\:\mathrm{quadratic}\:\mathrm{factor}\:\mathrm{in}\:\mathrm{the}\:\mathrm{denominator}\:\mathrm{of}\:\mathrm{the}\:\mathrm{form}\: \\ $$$$\mathrm{ax}^{\mathrm{2}} \:+\:\mathrm{bx}\:+\:\mathrm{c}\:\:\:\mathrm{which}\:\mathrm{does}\:\mathrm{not}\:\mathrm{have}\:\mathrm{linear}\:\mathrm{factors},\:\mathrm{there}\:\mathrm{corresponds}\:\mathrm{to} \\ $$$$\mathrm{a}\:\mathrm{partial}\:\mathrm{fraction}\:\mathrm{of}\:\mathrm{the}\:\mathrm{form}\:\:\:\:\frac{\mathrm{Ax}\:+\:\mathrm{B}}{\mathrm{ax}^{\mathrm{2}} \:+\:\mathrm{bx}\:+\:\mathrm{c}}\:\:\:\mathrm{where}\:\:\mathrm{A}\:\mathrm{and}\:\mathrm{B}\:\mathrm{are}\:\mathrm{constant}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 109237 by bobhans last updated on 22/Aug/20 $$\:\:\frac{\flat{o}\flat{hans}}{\angle\angle\angle\angle\angle} \\ $$$$\begin{cases}{{x}^{\mathrm{3}} +{x}^{\mathrm{2}} {y}\:=\:\mathrm{9}}\\{{y}^{\mathrm{3}} +{y}^{\mathrm{2}} {x}\:=\:\mathrm{25}}\end{cases}.\:{find}\:{x}\:{and}\:{y}. \\ $$ Answered by bemath last updated on 22/Aug/20…
Question Number 109222 by bemath last updated on 22/Aug/20 $$\:\:\:\:\frac{\ldots\flat{em}\mathcal{ATH}\ldots}{\cong\cong\cong\cong\cong\cong} \\ $$$$\sqrt{\mathrm{5}+\sqrt{\mathrm{10}}}\:=\:{x}.\left(\sqrt{\mathrm{5}+\sqrt{\mathrm{15}}\:}\:+\sqrt{\mathrm{5}−\sqrt{\mathrm{15}}}\:\right) \\ $$$${x}\:=? \\ $$ Answered by bobhans last updated on 22/Aug/20 $$\:\:\:\frac{\_\flat{o}\flat\mathcal{HAN}\varsigma\_}{\triangleright\triangleleft} \\…
Question Number 174754 by mnjuly1970 last updated on 10/Aug/22 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 174755 by mnjuly1970 last updated on 10/Aug/22 Commented by kaivan.ahmadi last updated on 11/Aug/22 $${I}\subseteq\sqrt{{I}},{J}\subseteq\sqrt{{J}}\Rightarrow{I}+{J}\subseteq\sqrt{{I}}+\sqrt{{J}} \\ $$$$\Rightarrow\sqrt{{I}+{J}}\subseteq\sqrt{\sqrt{{I}}+\sqrt{{J}}} \\ $$$${on}\:{the}\:{other}\:{hand} \\ $$$$\sqrt{{I}}+\sqrt{{J}}\subseteq\sqrt{{I}+{J}}\Rightarrow\sqrt{\sqrt{{I}}+\sqrt{{J}}}\subseteq \\ $$$$\sqrt{\sqrt{{I}+{J}}}=\sqrt{{I}+{J}}…
Question Number 43665 by peter frank last updated on 13/Sep/18 $$\left.\mathrm{1}\right)\:{if}\:\:{s}_{{n}\:\:} \:=\alpha^{{n}} +\beta^{{n}} +\lambda^{{n}\:} \:{where}\:\alpha,\beta,\lambda \\ $$$${are}\:{the}\:{root}\:{of}\:{ax}^{\mathrm{3}} +{bx}^{\mathrm{2}} +{cx}+{d}=\mathrm{0} \\ $$$$\:{then}\:\:{show}\:{that}\:{s}_{\mathrm{4}\:} =\frac{\mathrm{4}{abd}+\mathrm{4}{b}^{\mathrm{2}} {c}−\mathrm{2}{c}}{{a}^{\mathrm{3}} } \\…
Question Number 174735 by Shrinava last updated on 09/Aug/22 Terms of Service Privacy Policy Contact: info@tinkutara.com