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Question Number 105605 Answers: 2 Comments: 0
$${prove}\:{by}\:{mathematical}\:{induction}\: \\ $$$$\mathrm{2}^{\mathrm{3}} +\mathrm{4}^{\mathrm{3}} +\mathrm{6}^{\mathrm{3}} +\mathrm{8}^{\mathrm{3}} +...+\left(\mathrm{2}{n}\right)^{\mathrm{3}} =\:\mathrm{2}{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)^{\mathrm{2}} \\ $$
Question Number 105536 Answers: 1 Comments: 0
Question Number 105534 Answers: 1 Comments: 2
Question Number 105456 Answers: 2 Comments: 0
$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{curve}\:\mathrm{y}=\mathrm{x}^{\mathrm{4}} +\mathrm{3x}^{\mathrm{2}} +\mathrm{2x} \\ $$$$\mathrm{does}\:\mathrm{not}\:\mathrm{meet}\:\mathrm{the}\:\mathrm{straight}\:\mathrm{line} \\ $$$$\mathrm{y}=\mathrm{2x}−\mathrm{1}\:\mathrm{and}\:\mathrm{find}\:\mathrm{the}\:\mathrm{distace}\:\mathrm{between} \\ $$$$\mathrm{their}\:\mathrm{nearest}\:\mathrm{points}.\left(\mathrm{Answer}\:\frac{\mathrm{1}}{\sqrt{\mathrm{5}}}\right) \\ $$
Question Number 105448 Answers: 3 Comments: 2
$$\mathcal{G}{iven}\:\begin{cases}{\frac{\mathrm{4}{x}^{\mathrm{2}} }{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{1}}\:=\:{y}}\\{\frac{\mathrm{4}{y}^{\mathrm{2}} }{\mathrm{4}{y}^{\mathrm{2}} +\mathrm{1}}\:=\:{z}}\\{\frac{\mathrm{4}{z}^{\mathrm{2}} }{\mathrm{4}{z}^{\mathrm{2}} +\mathrm{1}}\:=\:{x}}\end{cases}\:\:\:.\:{find} \\ $$$${x}+{y}+{z}\:?\: \\ $$
Question Number 105414 Answers: 0 Comments: 2
Question Number 105413 Answers: 0 Comments: 0
Question Number 105406 Answers: 0 Comments: 0
Question Number 105340 Answers: 1 Comments: 0
$$\mathrm{1}\backslash\mathrm{Show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{polynomial},\:\mathrm{P}\left(\mathrm{x}\right)=\mathrm{x}^{\mathrm{n}} +\mathrm{ax}+\mathrm{b},\:\mathrm{n}\geqslant\mathrm{2} \\ $$$$\left(\mathrm{a},\mathrm{b}\right)\in\mathbb{R}^{\mathrm{2}} \:\mathrm{has}\:\mathrm{at}\:\mathrm{most}\:\mathrm{3}\:\mathrm{real}\:\mathrm{distinct}\:\mathrm{roots}. \\ $$$$\mathrm{2}\backslash\mathrm{Show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{x}^{\mathrm{4}} +\mathrm{4ax}+\mathrm{b}=\mathrm{0}\:\mathrm{has}\:\mathrm{no} \\ $$$$\mathrm{more}\:\mathrm{than}\:\mathrm{2}\:\mathrm{real}\:\mathrm{distinct}\:\mathrm{roots}. \\ $$
Question Number 105322 Answers: 0 Comments: 1
Question Number 105320 Answers: 0 Comments: 0
$$\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{C}} \\ $$$${kx}^{\mathrm{2}} +{ky}^{\mathrm{2}} +{z}^{\mathrm{2}} \geqslant{C} \\ $$$${x},{y},{z},{k}\in\mathbb{R}\:\:{such}\:{that} \\ $$$${xy}+{yz}+{zx}=\mathrm{1} \\ $$
Question Number 105234 Answers: 0 Comments: 0
$${f}:\mathbb{R}\rightarrow\mathbb{R}\:\:{x},{y}\in\mathbb{R} \\ $$$${f}\left({f}\left({x}\right){f}\left({y}\right)\right)+{f}\left({x}+{y}\right)={f}\left({xy}\right) \\ $$$${f}\left({x}\right)=? \\ $$
Question Number 105222 Answers: 0 Comments: 0
Question Number 105189 Answers: 0 Comments: 2
Question Number 105158 Answers: 4 Comments: 0
$$\begin{cases}{{xy}+{x}+{y}\:=\:\mathrm{20}}\\{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \:=\:\mathrm{40}}\end{cases}{find}\:{x}\:{and}\:{y} \\ $$
Question Number 105124 Answers: 1 Comments: 0
$${x}^{\mathrm{4}} +{ax}^{\mathrm{3}} +{bx}^{\mathrm{2}} +{cx}+{d}=\mathrm{0} \\ $$$${Find}\:{x}. \\ $$
Question Number 105120 Answers: 1 Comments: 0
$${f}\left(\mathrm{2}{f}\left({x}\right)+{f}\left({y}\right)\right)=\mathrm{2}{x}+{y}\:\:\:{f}:\mathbb{R}\rightarrow\mathbb{R} \\ $$$${f}\left({x}\right)=?\:\:\:\:{x},{y}\in\mathbb{R} \\ $$$$ \\ $$
Question Number 105083 Answers: 0 Comments: 0
Question Number 105082 Answers: 0 Comments: 0
Question Number 105023 Answers: 1 Comments: 0
$$\frac{\left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right)^{\mathrm{2}} }{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)\left({y}^{\mathrm{2}} −\mathrm{1}\right)}+\frac{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{\left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right)\left({y}^{\mathrm{2}} −\mathrm{1}\right)}+\frac{\left({y}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)\left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right)}= \\ $$$$ \\ $$
Question Number 104973 Answers: 1 Comments: 1
$$\mathrm{2}\sqrt{\mathrm{3}}\:+\:\boldsymbol{{i}}\:\:\:{is}\:{a}\:{cubic}\:{root}\:{for} \\ $$$$\mathrm{18}\sqrt{\mathrm{3}}\:+\:\mathrm{35}\boldsymbol{{i}}\: \\ $$$$\boldsymbol{{find}}\:\:\boldsymbol{{the}}\:\boldsymbol{{other}}\:\mathrm{2}\:\boldsymbol{{cubic}}\:\boldsymbol{{roots}} \\ $$
Question Number 104859 Answers: 2 Comments: 0
Question Number 104832 Answers: 1 Comments: 0
$$\begin{cases}{{x}+{y}+\frac{{x}^{\mathrm{2}} }{{y}^{\mathrm{2}} }\:=\:\mathrm{7}}\\{\frac{\left({x}−{y}\right){x}^{\mathrm{2}} }{{y}^{\mathrm{2}} }\:=\:\mathrm{12}\:}\end{cases} \\ $$
Question Number 104782 Answers: 1 Comments: 0
Question Number 104758 Answers: 0 Comments: 0
Question Number 104811 Answers: 2 Comments: 0
$${Given}\:\begin{cases}{{a}+{b}\sqrt{\mathrm{3}}−\mathrm{2}{c}\:=\:\mathrm{1}}\\{\mathrm{3}{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \:=\:\mathrm{2}{a}^{\mathrm{2}} }\\{{a}^{\mathrm{2}} +\mathrm{4}{ac}\:=\:\mathrm{5}{c}^{\mathrm{2}} }\end{cases} \\ $$$${find}\:{b} \\ $$
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