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Question Number 115908 Answers: 1 Comments: 2
$${what}\:{is}\:{the}\:{cofficient}\:{of}\:{x}^{\mathrm{2}} \\ $$$$\left(\mathrm{1}+{x}\right)\left(\mathrm{1}+\mathrm{2}{x}\right)\left(\mathrm{1}+\mathrm{4}{x}\right)......\left(\mathrm{1}+\mathrm{2}^{{n}} {x}\right) \\ $$
Question Number 115906 Answers: 1 Comments: 0
$${find}\:{all}\:{pairs}\:{of}\:{integers}\:\left({x},{y}\right)\:{such}\:{that} \\ $$$${x}^{\mathrm{3}} +{y}^{\mathrm{3}} =\left({x}+{y}\right)^{\mathrm{2}} \\ $$
Question Number 115897 Answers: 2 Comments: 1
$$\mathrm{Given} \\ $$$${a}_{{n}} \:=\:\sqrt{\mathrm{1}\:+\:\left(\mathrm{1}\:−\:\frac{\mathrm{1}}{{n}}\right)^{\mathrm{2}} }\:+\:\sqrt{\mathrm{1}\:+\:\left(\mathrm{1}\:+\:\frac{\mathrm{1}}{{n}}\right)^{\mathrm{2}} } \\ $$$$\mathrm{The}\:\mathrm{value}\:\mathrm{of}\:\underset{{n}=\mathrm{1}} {\overset{\mathrm{2015}} {\sum}}\left(\frac{\mathrm{4}}{{a}_{{n}} }\right)\:\mathrm{is}\:... \\ $$
Question Number 115858 Answers: 2 Comments: 0
$${What}\:{are}\:{all}\:{ordered}\:{pairs}\:{of}\:{real} \\ $$$${number}\:\left({x},{y}\right)\:{for}\:{which}\: \\ $$$$\mathrm{5}^{{y}−{x}} \:\left({x}+{y}\right)\:=\:\mathrm{1}\:{and}\:\left({x}+{y}\right)^{{x}−{y}} \:=\:\mathrm{5} \\ $$
Question Number 115857 Answers: 1 Comments: 0
$${What}\:{are}\:{all}\:{real}\:{values}\:{of}\:{p}\:{for} \\ $$$${which}\:{the}\:{inequality}\: \\ $$$$−\mathrm{3}<\frac{{x}^{\mathrm{2}} +{px}−\mathrm{2}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}}<\mathrm{2}\:{is}\:{satisfied}\: \\ $$$${by}\:{all}\:{real}\:{values}\:{of}\:{x} \\ $$
Question Number 115871 Answers: 1 Comments: 1
Question Number 115815 Answers: 1 Comments: 0
$$\mathrm{Montrer}\:\mathrm{que}\:\:\forall\left(\mathrm{a},\mathrm{b},\mathrm{c}\right)\in\left(\mathbb{R}_{+} ^{\ast} \right)^{\mathrm{3}} \\ $$$$\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{2}} +\mathrm{bc}}+\frac{\mathrm{1}}{\mathrm{b}^{\mathrm{2}} +\mathrm{ac}}+\frac{\mathrm{1}}{\mathrm{c}^{\mathrm{2}} +\mathrm{ab}}\leqslant\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{ab}}+\frac{\mathrm{1}}{\mathrm{bc}}+\frac{\mathrm{1}}{\mathrm{ac}}\right) \\ $$
Question Number 115773 Answers: 0 Comments: 0
$$\:\frac{\mathrm{1}−{x}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{5}}}\:=\:\frac{\mathrm{3}}{\mathrm{5}} \\ $$
Question Number 115564 Answers: 1 Comments: 0
Question Number 115544 Answers: 1 Comments: 1
$${Given}\:{x}^{\mathrm{2}} +\mathrm{12}\sqrt{{x}}\:=\:\mathrm{5} \\ $$$${then}\:{x}+\mathrm{2}\sqrt{{x}}\:? \\ $$
Question Number 115484 Answers: 1 Comments: 3
$${find}\:{the}\:{value}\:{of}\: \\ $$$$\frac{\mathrm{111}}{\mathrm{1}+\mathrm{1}+\mathrm{1}}\:,\:\frac{\mathrm{222}}{\mathrm{2}+\mathrm{2}+\mathrm{2}}\:,\:\frac{\mathrm{333}}{\mathrm{3}+\mathrm{3}+\mathrm{3}},\:\frac{\mathrm{444}}{\mathrm{4}+\mathrm{4}+\mathrm{4}} \\ $$$$\frac{\mathrm{555}}{\mathrm{5}+\mathrm{5}+\mathrm{5}},\:\frac{\mathrm{666}}{\mathrm{6}+\mathrm{6}+\mathrm{6}}\:,\:\frac{\mathrm{777}}{\mathrm{7}+\mathrm{7}+\mathrm{7}}\:,\:\frac{\mathrm{888}}{\mathrm{8}+\mathrm{8}+\mathrm{8}} \\ $$$$\frac{\mathrm{999}}{\mathrm{9}+\mathrm{9}+\mathrm{9}} \\ $$
Question Number 115436 Answers: 1 Comments: 0
$$\sqrt{\sqrt[{\mathrm{3}}]{{a}}+\sqrt[{\mathrm{3}}]{{b}}}=\frac{\mathrm{1}}{\:\sqrt{{b}−{a}×{s}^{\mathrm{3}} }}\left(\frac{−{s}^{\mathrm{2}} ×\sqrt[{\mathrm{3}}]{{a}^{\mathrm{2}} }}{\mathrm{2}}+{s}\sqrt[{\mathrm{3}}]{{ab}}+\sqrt[{\mathrm{3}}]{{b}^{\mathrm{2}} }\right) \\ $$$${what}\:{is}\:{s}? \\ $$
Question Number 115333 Answers: 1 Comments: 3
Question Number 115325 Answers: 0 Comments: 5
Question Number 115218 Answers: 0 Comments: 0
$$\mathrm{Show}\:\mathrm{that}\:\forall\mathrm{n}\in\mathbb{N},\:\forall\mathrm{u}_{\mathrm{0}} ,\mathrm{u}_{\mathrm{1}} ,...,\mathrm{u}_{\mathrm{n}} ,\mathrm{v}_{\mathrm{0}} ,\mathrm{v}_{\mathrm{1}} ,...\mathrm{v}_{\mathrm{n}} \in\mathbb{C} \\ $$$$\forall\mathrm{k}\leqslant\mathrm{n};\:\mathrm{u}_{\mathrm{k}} =\sum_{\mathrm{i}=\mathrm{0}} ^{\mathrm{k}} \begin{pmatrix}{\mathrm{k}}\\{\mathrm{i}}\end{pmatrix}\mathrm{v}_{\mathrm{i}} \Leftrightarrow\forall\mathrm{k}\leqslant\mathrm{n};\:\mathrm{v}_{\mathrm{k}} =\sum_{\mathrm{i}=\mathrm{0}} ^{\mathrm{k}} \left(−\mathrm{1}\right)^{\mathrm{k}−\mathrm{1}} \begin{pmatrix}{\mathrm{k}}\\{\mathrm{i}}\end{pmatrix}\mathrm{u}_{\mathrm{i}} \\ $$
Question Number 115136 Answers: 1 Comments: 0
Question Number 115135 Answers: 1 Comments: 1
Question Number 115023 Answers: 2 Comments: 0
$${solve}\:\begin{cases}{\mathrm{7}{x}−\mathrm{5}{y}+\mathrm{3}{z}=\mathrm{6}}\\{\mathrm{2}{x}+\mathrm{4}{y}−\mathrm{5}{z}=−\mathrm{5}}\\{\mathrm{9}{x}−\mathrm{8}{y}+\mathrm{2}{z}=−\mathrm{1}}\end{cases}.\:{Find}\:{x}+{y}+{z}\: \\ $$
Question Number 115018 Answers: 2 Comments: 0
$${If}\:\mathrm{9}^{{x}} +\mathrm{9}^{−{x}} \:=\:\mathrm{3}^{\mathrm{2}+{x}} +\mathrm{3}^{\mathrm{2}−{x}} \:−\mathrm{20},\:{then}\: \\ $$$$\mathrm{27}^{{x}} +\mathrm{27}^{−{x}} \:=? \\ $$
Question Number 114956 Answers: 0 Comments: 5
Question Number 114919 Answers: 0 Comments: 0
Question Number 114912 Answers: 0 Comments: 0
Question Number 114917 Answers: 0 Comments: 3
Question Number 114809 Answers: 1 Comments: 1
$$\begin{cases}{\mathrm{1}−\frac{\mathrm{12}}{{y}+\mathrm{3}{x}}=\frac{\mathrm{2}}{\:\sqrt{{x}}}}\\{\mathrm{1}+\frac{\mathrm{12}}{{y}+\mathrm{3}{x}}=\frac{\mathrm{6}}{\:\sqrt{{y}}}}\end{cases} \\ $$
Question Number 114806 Answers: 1 Comments: 0
Question Number 114799 Answers: 1 Comments: 0
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