Question Number 54600 by behi83417@gmail.com last updated on 07/Feb/19 $$\boldsymbol{\mathrm{solve}}\:\boldsymbol{\mathrm{for}}:\:\boldsymbol{\mathrm{x}} \\ $$$$\left.\:\:\:\mathrm{1}\right)\:\sqrt{\mathrm{3}−\boldsymbol{\mathrm{x}}}+\sqrt{\boldsymbol{\mathrm{x}}+\mathrm{1}}>\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\left.\:\:\:\:\mathrm{2}\right)\:\boldsymbol{\mathrm{cos}}^{\mathrm{2}} \boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{cos}}^{\mathrm{2}} \mathrm{2}\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{cos}}^{\mathrm{2}} \mathrm{3}\boldsymbol{\mathrm{x}}=\mathrm{1} \\ $$$$\left.\:\:\:\:\mathrm{3}\right)\sqrt{\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\boldsymbol{\mathrm{p}}}+\mathrm{2}\sqrt{\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\mathrm{1}}=\boldsymbol{\mathrm{x}}\:\:\:\:\:\left[\boldsymbol{\mathrm{p}}\in\boldsymbol{\mathrm{R}}\right] \\ $$ Commented by…
Question Number 54595 by ajfour last updated on 07/Feb/19 $$\:\:\:\:\:{x}+{y}={a} \\ $$$$\:\:\:\:\:{z}+{bx}={c} \\ $$$$\:\:\:\:\:{bz}+{xy}={d} \\ $$$$\:\:\:\:\:{Find}\:{yz}\:{in}\:{terms}\:{of}\:{a},{b},{c},{d}. \\ $$ Commented by mr W last updated on…
Question Number 185665 by mathlove last updated on 25/Jan/23 $${f}\left(\frac{\mathrm{2}{x}−\mathrm{1}}{\mathrm{3}{x}+\mathrm{2}}\right)=\frac{{x}−\mathrm{1}}{{x}−\mathrm{2}}\:\:\:\:\:\:{faind}\:\:\:{f}^{−\mathrm{1}} \left(\mathrm{2}\right)=? \\ $$ Answered by cortano1 last updated on 25/Jan/23 $${f}^{−\mathrm{1}} \left(\frac{{x}−\mathrm{1}}{{x}−\mathrm{2}}\right)=\frac{\mathrm{2}{x}−\mathrm{1}}{\mathrm{3}{x}+\mathrm{2}}\: \\ $$$$\Leftrightarrow\:\frac{{x}−\mathrm{1}}{{x}−\mathrm{2}}\:=\:\mathrm{2}\:;\:{x}−\mathrm{1}=\mathrm{2}{x}−\mathrm{4} \\…
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Question Number 54582 by Tawa1 last updated on 07/Feb/19 $$\mathrm{Solve}\:\mathrm{for}\:\mathrm{x}:\:\:\:\:\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{x}\:+\:\sqrt{\mathrm{x}}\:+\:\mathrm{1}}}\:+\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{x}\:+\:\sqrt{\mathrm{x}}\:−\:\mathrm{2}}}\:\:=\:\sqrt{\mathrm{x}\:+\:\mathrm{1}} \\ $$ Commented by MJS last updated on 07/Feb/19 $$\mathrm{with}\:{t}=\sqrt{{x}}\:\mathrm{and}\:\mathrm{squaring}\:\mathrm{etc}.\:\mathrm{this}\:\mathrm{leads}\:\mathrm{to}\:\mathrm{a} \\ $$$$\mathrm{polynome}\:\mathrm{of}\:\mathrm{degree}\:\mathrm{12} \\ $$$$\mathrm{we}\:\mathrm{can}\:\mathrm{only}\:\mathrm{approximate}. \\…
Question Number 120110 by benjo_mathlover last updated on 29/Oct/20 $$\begin{cases}{\frac{\mathrm{4}{x}^{\mathrm{2}} }{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{1}}\:=\:{y}}\\{\frac{\mathrm{4}{y}^{\mathrm{2}} }{\mathrm{4}{y}^{\mathrm{2}} +\mathrm{1}}\:=\:{z}\:}\\{\frac{\mathrm{4}{z}^{\mathrm{2}} }{\mathrm{4}{z}^{\mathrm{2}} +\mathrm{1}}\:=\:{x}}\end{cases} \\ $$$${where}\:{x},{y},{z}\:\neq\:\mathrm{0}\: \\ $$ Answered by bemath last updated…
Question Number 120108 by behi83417@gmail.com last updated on 29/Oct/20 $$\left(\boldsymbol{\mathrm{x}}^{\mathrm{4}} +\boldsymbol{\mathrm{ax}}^{\mathrm{2}} +\boldsymbol{\mathrm{a}}^{\mathrm{2}} \right)^{\mathrm{2}} +\boldsymbol{\mathrm{x}}^{\mathrm{6}} =\mathrm{1} \\ $$$$\boldsymbol{\mathrm{solve}}\:\boldsymbol{\mathrm{for}}:\:\:\mathrm{x},\mathrm{a}\in\boldsymbol{\mathrm{R}} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 54566 by mr W last updated on 06/Feb/19 $$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\left(\frac{\mathrm{1}+{c}}{\mathrm{1}+\mu^{\frac{\mathrm{1}}{{n}}} }\right)^{{n}} =? \\ $$$${with}\:\mathrm{0}\leqslant{c}\leqslant\mathrm{1},\:\mu>\mathrm{0} \\ $$ Commented by maxmathsup by imad last updated…
Question Number 54562 by Meritguide1234 last updated on 06/Feb/19 Answered by JDamian last updated on 07/Feb/19 $$\sqrt{\mathrm{7}+\mathrm{3}\sqrt{\mathrm{3}}+\mathrm{4}\sqrt{\mathrm{9}\sqrt{\mathrm{3}}−\mathrm{15}}} \\ $$$$\sqrt{\mathrm{7}+\mathrm{3}\sqrt{\mathrm{3}}+\mathrm{4}\sqrt{\mathrm{3}\left(\mathrm{3}\sqrt{\mathrm{3}}−\mathrm{5}\right)}} \\ $$$$\sqrt{\mathrm{7}+\mathrm{3}\sqrt{\mathrm{3}}+\mathrm{2}×\mathrm{2}\sqrt{\mathrm{3}}×\sqrt{\mathrm{3}\sqrt{\mathrm{3}}−\mathrm{5}}} \\ $$$$\sqrt{\mathrm{12}+\mathrm{3}\sqrt{\mathrm{3}}−\mathrm{5}+\mathrm{2}×\mathrm{2}\sqrt{\mathrm{3}}×\sqrt{\mathrm{3}\sqrt{\mathrm{3}}−\mathrm{5}}} \\ $$$$…
Question Number 185624 by mathlove last updated on 24/Jan/23 Answered by MJS_new last updated on 24/Jan/23 $${a}_{\mathrm{1}} =\mathrm{3}^{\mathrm{1}/\mathrm{2}} \\ $$$${a}_{\mathrm{2}} =\left(\mathrm{3}^{\mathrm{1}/\mathrm{2}} \right)^{\mathrm{2}/\mathrm{3}} =\mathrm{3}^{\mathrm{1}/\mathrm{3}} \\ $$$${a}_{\mathrm{3}}…