Question Number 50278 by LYCON TRIX last updated on 15/Dec/18 $$\sqrt{\mathrm{a}+\sqrt{\mathrm{a}−\mathrm{x}}}\:+\:\sqrt{\mathrm{a}−\sqrt{\mathrm{a}+\mathrm{x}}}\:=\:\mathrm{2x} \\ $$$$\mathrm{please}\:\mathrm{i}\:\mathrm{beg}\:\mathrm{u}\:\mathrm{guys}\: \\ $$$$\mathrm{please}\:\mathrm{solve}\:\mathrm{this}\:\mathrm{question} \\ $$ Commented by ajfour last updated on 15/Dec/18 $${a}\:{and}\:{x}\:\in\:\mathbb{R}\:\:{or}\:\mathbb{C}\:?…
Question Number 181318 by mr W last updated on 23/Nov/22 $${if}\:{x}+{y}+{z}=\mathrm{0},\:{find}\:{the}\:{maximum}\:{of} \\ $$$$\frac{\mid{x}+\mathrm{2}{y}+\mathrm{3}{z}\mid}{\:\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} }}. \\ $$ Commented by mr W last updated on…
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Question Number 115773 by bemath last updated on 28/Sep/20 $$\:\frac{\mathrm{1}−{x}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{5}}}\:=\:\frac{\mathrm{3}}{\mathrm{5}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 50226 by Cheyboy last updated on 14/Dec/18 $$\mathrm{Three}\:\mathrm{prizes}\:\mathrm{are}\:\mathrm{awarded}\:\mathrm{each}\:\mathrm{for} \\ $$$$\mathrm{getting}\:\mathrm{more}\:\mathrm{than}\:\mathrm{80\%marks}, \\ $$$$\mathrm{98\%}\:\mathrm{attendance}\:\mathrm{and}\:\mathrm{good} \\ $$$$\mathrm{behaviour}\:\mathrm{in}\:\mathrm{the}\:\mathrm{college}.\mathrm{In}\:\mathrm{how} \\ $$$$\mathrm{many}\:\mathrm{ways}\:\mathrm{the}\:\mathrm{prozes}\:\mathrm{can}\:\mathrm{be} \\ $$$$\mathrm{awarded}\:\mathrm{if}\:\mathrm{15}\:\mathrm{student}\:\mathrm{of}\:\mathrm{the}\:\mathrm{college} \\ $$$$\mathrm{are}\:\mathrm{eligible}\:\mathrm{for}\:\mathrm{the}\:\mathrm{three}\:\mathrm{prizes}? \\ $$ Commented…
Question Number 181279 by mnjuly1970 last updated on 23/Nov/22 Answered by Rasheed.Sindhi last updated on 25/Nov/22 $$\sqrt[{\mathrm{3}}]{{x}+\mathrm{1}}\:+\sqrt[{\mathrm{3}}]{{x}+\mathrm{2}}\:+\sqrt[{\mathrm{3}}]{{x}+\mathrm{3}}\:=\mathrm{0} \\ $$$$\begin{array}{|c|}{\underset{\:\Rightarrow{a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} −\mathrm{3}{abc}=\mathrm{0}} {{a}+{b}+{c}=\mathrm{0}}\:\:\:\:\:\:\:\:}\\\hline\end{array} \\ $$$${x}+\mathrm{1}+{x}+\mathrm{2}+{x}+\mathrm{3}−\mathrm{3}\sqrt[{\mathrm{3}}]{\left({x}+\mathrm{1}\right)}\:\sqrt[{\mathrm{3}}]{{x}+\mathrm{2}}\:\sqrt[{\mathrm{3}}]{{x}+\mathrm{3}}\:=\mathrm{0}…
Question Number 181243 by Agnibhoo98 last updated on 23/Nov/22 $$\mathrm{Solve}\:\mathrm{for}\:{x}\:: \\ $$$$\frac{{x}\:−\:{a}^{\mathrm{2}} }{{b}\:+\:{c}}\:+\:\frac{{x}\:−\:{b}^{\mathrm{2}} }{{c}\:+\:{a}}\:+\:\frac{{x}\:−\:{c}^{\mathrm{2}} }{{a}\:+\:{b}}\:=\:\mathrm{4}\left({a}\:+\:{b}\:+\:{c}\right) \\ $$ Answered by Frix last updated on 23/Nov/22 $${x}=\left({a}+{b}+{c}\right)^{\mathrm{2}}…
Question Number 50163 by Meritguide1234 last updated on 14/Dec/18 Answered by peter frank last updated on 14/Dec/18 $$\mathrm{let}\:\mathrm{x}=\frac{\mathrm{b}−\mathrm{c}}{\mathrm{a}}\:\:\:\: \\ $$$$\:\frac{\mathrm{1}}{\mathrm{x}}=\frac{\mathrm{a}}{\mathrm{b}−\mathrm{c}}\:\: \\ $$$$\mathrm{y}=\frac{\mathrm{c}−\mathrm{a}}{\mathrm{b}} \\ $$$$\frac{\mathrm{1}}{\mathrm{y}}=\frac{\mathrm{b}}{\mathrm{c}−\mathrm{a}} \\…
Question Number 181221 by cortano1 last updated on 23/Nov/22 Answered by mr W last updated on 23/Nov/22 Commented by cortano1 last updated on 23/Nov/22 $$\mathrm{nicesolution}…
Question Number 181196 by depressiveshrek last updated on 22/Nov/22 $${Let}\:{a}_{\mathrm{1}} ,\:{a}_{\mathrm{2}} ,\:{a}_{\mathrm{3}} ,\:…{a}_{\mathrm{2022}} \:{be}\:{numbers} \\ $$$${ranging}\:{from}\:\left(\mathrm{0},\:+\infty\right)\:\backslash\:\left\{\mathrm{1}\right\},\:{for}\:{which} \\ $$$${the}\:{function}\:{f}\::\:\mathbb{R}\rightarrow\mathbb{R}\:{is}\:{defined}\:{as} \\ $$$${f}\left({x}\right)={a}_{\mathrm{1}} ^{{x}} +{a}_{\mathrm{2}} ^{{x}} +{a}_{\mathrm{3}} ^{{x}}…