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Question Number 206104 Answers: 2 Comments: 0
$$\:\:\:\:\downharpoonleft\underline{\:} \\ $$
Question Number 206096 Answers: 1 Comments: 0
$$\int\frac{\mathrm{1}}{{x}^{\mathrm{3}} \sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}\:.{dx} \\ $$
Question Number 206095 Answers: 2 Comments: 0
Question Number 206093 Answers: 2 Comments: 0
Question Number 206082 Answers: 2 Comments: 0
$$\mathrm{If}\:\:\:\mathrm{cos}\boldsymbol{\alpha}\:=\:\mathrm{sin}\boldsymbol{\alpha}\:+\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$$$\mathrm{Find}\:\:\:\mathrm{sin2}\boldsymbol{\alpha}\:=\:? \\ $$
Question Number 206079 Answers: 3 Comments: 0
Question Number 206078 Answers: 0 Comments: 1
Question Number 206074 Answers: 0 Comments: 1
$$\left({x}_{\mathrm{1}} \:−\:{x}_{\mathrm{2}} \:+\:{x}_{\mathrm{3}} \right)^{\mathrm{2}} \:=\:{x}_{\mathrm{2}} \left({x}_{\mathrm{4}} \:+\:{x}_{\mathrm{5}} \:−\:{x}_{\mathrm{2}} \right) \\ $$$$\left({x}_{\mathrm{2}} \:−\:{x}_{\mathrm{3}} \:+\:{x}_{\mathrm{4}} \right)^{\mathrm{2}} \:=\:{x}_{\mathrm{3}} \left({x}_{\mathrm{5}} \:+\:{x}_{\mathrm{1}} \:−\:{x}_{\mathrm{3}} \right) \\ $$$$\left({x}_{\mathrm{3}} \:−\:{x}_{\mathrm{4}} \:+\:{x}_{\mathrm{5}} \right)^{\mathrm{2}} \:=\:{x}_{\mathrm{4}} \left({x}_{\mathrm{1}} \:+\:{x}_{\mathrm{2}} \:−\:{x}_{\mathrm{4}} \right) \\ $$$$\left({x}_{\mathrm{4}} \:−\:{x}_{\mathrm{5}} \:+\:{x}_{\mathrm{1}} \right)^{\mathrm{2}} \:=\:{x}_{\mathrm{5}} \left({x}_{\mathrm{2}} \:+\:{x}_{\mathrm{3}} \:−\:{x}_{\mathrm{5}} \right) \\ $$$$\left({x}_{\mathrm{5}} \:−\:{x}_{\mathrm{1}} \:+\:{x}_{\mathrm{2}} \right)^{\mathrm{2}} \:=\:{x}_{\mathrm{1}} \left({x}_{\mathrm{3}} \:+\:{x}_{\mathrm{4}} \:−\:{x}_{\mathrm{1}} \right) \\ $$$$\mathrm{Find}\:\frac{\mathrm{2}{x}_{\mathrm{1}} \:+\:{x}_{\mathrm{2}} \:+\:{x}_{\mathrm{3}} }{\mathrm{3}{x}_{\mathrm{4}} \:−\:{x}_{\mathrm{5}} }\:. \\ $$
Question Number 206072 Answers: 0 Comments: 0
$$\int_{\mathrm{0}} ^{\pi} {arctan}\left(\frac{{ln}\left({sin}\left({x}\right)\right)}{{x}}\right){dx}=...? \\ $$
Question Number 206069 Answers: 1 Comments: 0
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{−{x}^{\mathrm{3}} +{x}}{\mathrm{sin}\:{x}} \\ $$
Question Number 206066 Answers: 2 Comments: 0
$$\mathrm{Find}: \\ $$$$\frac{\mathrm{3}}{\mathrm{4}}\:\centerdot\:\frac{\mathrm{8}}{\mathrm{9}}\:\centerdot\:\frac{\mathrm{15}}{\mathrm{16}}\:\centerdot\:...\:\centerdot\:\frac{\mathrm{120}}{\mathrm{121}}\:=\:? \\ $$
Question Number 206064 Answers: 3 Comments: 1
Question Number 206063 Answers: 2 Comments: 0
$$\mathrm{If}\:\left({x}\:+\:\sqrt{\mathrm{1}\:+\:{x}^{\mathrm{2}} }\right)\left({y}\:+\:\sqrt{\mathrm{1}\:+\:{y}^{\mathrm{2}} }\right)\:=\:\mathrm{1}\:\mathrm{then} \\ $$$$\mathrm{find}\:\left({x}\:+\:{y}\right)^{\mathrm{2}} . \\ $$
Question Number 206060 Answers: 1 Comments: 0
Question Number 206053 Answers: 1 Comments: 1
Question Number 206048 Answers: 1 Comments: 0
$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{2}^{\mathrm{sin}^{\mathrm{2}} \theta} \:+\:\mathrm{2}^{\mathrm{cos}^{\mathrm{2}} \theta} \:\geqslant\:\mathrm{2}\sqrt{\mathrm{2}}. \\ $$
Question Number 206047 Answers: 1 Comments: 0
Question Number 206045 Answers: 2 Comments: 1
Question Number 206037 Answers: 2 Comments: 0
$${is}\:{it}\:{a}\:{polynomial}? \\ $$$${x}^{\mathrm{3}} −\mathrm{2}{x}^{\mathrm{2}} +\sqrt{{x}^{\mathrm{2}} }+\mathrm{10} \\ $$
Question Number 206036 Answers: 1 Comments: 0
$${is}\:{it}\:{a}\:{polynomial}? \\ $$$$\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}{x}−\frac{\mathrm{2}}{{x}^{−\mathrm{2}} } \\ $$
Question Number 206038 Answers: 2 Comments: 0
$$\sqrt{\mathrm{1}\:+\:\mathrm{2023}\sqrt{\mathrm{1}\:+\:\mathrm{2024}\sqrt{\mathrm{1}+\:\mathrm{2025}\sqrt{\mathrm{1}\:+\:\mathrm{2026}\sqrt{\mathrm{1}\:+\:..............\infty}}}}}\:=\:?\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$
Question Number 206025 Answers: 2 Comments: 0
$$\:\:\:\:\:\mathrm{2}^{\mathrm{2024}} \:=\:{x}\:\left({mod}\:\mathrm{10}\right)\: \\ $$
Question Number 206024 Answers: 2 Comments: 0
$${proove} \\ $$$${e}^{{i}\pi} +\mathrm{1}=\mathrm{0} \\ $$
Question Number 206020 Answers: 1 Comments: 0
Question Number 206007 Answers: 1 Comments: 1
$$\left({E},<,>\:\right):\:\:\:{prouve} \\ $$$$<{x},{y}>=\frac{\mathrm{1}}{\mathrm{4}}\underset{{k}={o}} {\overset{\mathrm{3}} {\sum}}{i}^{{k}} \mid\mid{x}\:+\:{i}^{{k}} {y}\mid\mid^{\mathrm{2}} \\ $$
Question Number 206003 Answers: 2 Comments: 0
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