Question Number 181375 by Tawa11 last updated on 24/Nov/22 $$\mathrm{If}\:\:\mathrm{a}\:=\:\mathrm{1},\:\mathrm{b}\:=\:\mathrm{2},\:\mathrm{c}\:=\:\mathrm{3},\:…\:\mathrm{z}\:\:=\:\:\mathrm{26} \\ $$$$\mathrm{then}\:\:\:\:\left(\mathrm{t}\:−\:\mathrm{a}\right)\left(\mathrm{t}\:−\:\mathrm{b}\right)\left(\mathrm{t}\:−\:\mathrm{c}\right)\:…\:\left(\mathrm{t}\:−\:\mathrm{y}\right)\left(\mathrm{t}\:−\:\mathrm{z}\right)\:=\:\:\:?? \\ $$ Commented by mr W last updated on 24/Nov/22 $$\:…\:\:{is}\:\left({t}−{d}\right)…\left({t}−{s}\right)\left({t}−{t}\right)\left({t}−{u}\right)…\left({t}−{x}\right) \\ $$…
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Question Number 115815 by Ar Brandon last updated on 28/Sep/20 $$\mathrm{Montrer}\:\mathrm{que}\:\:\forall\left(\mathrm{a},\mathrm{b},\mathrm{c}\right)\in\left(\mathbb{R}_{+} ^{\ast} \right)^{\mathrm{3}} \\ $$$$\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{2}} +\mathrm{bc}}+\frac{\mathrm{1}}{\mathrm{b}^{\mathrm{2}} +\mathrm{ac}}+\frac{\mathrm{1}}{\mathrm{c}^{\mathrm{2}} +\mathrm{ab}}\leqslant\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{ab}}+\frac{\mathrm{1}}{\mathrm{bc}}+\frac{\mathrm{1}}{\mathrm{ac}}\right) \\ $$ Answered by 1549442205PVT last updated…
Question Number 50279 by Saorey last updated on 15/Dec/18 $$\begin{cases}{\mathrm{x}+\mathrm{y}=\mathrm{6}}\\{\mathrm{y}+\mathrm{z}=\mathrm{10}}\end{cases}\:\:\left(\mathrm{x},\mathrm{y},\mathrm{z}>\mathrm{0}\right) \\ $$ Answered by mr W last updated on 15/Dec/18 $${if}\:{x},{y},{z}\:{are}\:{integers},{then} \\ $$$$\left({x},{y},{z}\right)=\left(\mathrm{1},\mathrm{5},\mathrm{5}\right)/\left(\mathrm{2},\mathrm{4},\mathrm{6}\right)/\left(\mathrm{3},\mathrm{3},\mathrm{7}\right)/\left(\mathrm{4},\mathrm{2},\mathrm{8}\right)/\left(\mathrm{5},\mathrm{1},\mathrm{9}\right) \\ $$$${if}\:{x},{y},{z}\:{are}\:{real},\:{then}…
Question Number 50278 by LYCON TRIX last updated on 15/Dec/18 $$\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}…