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AlgebraQuestion and Answers: Page 252
Question Number 107313 Answers: 2 Comments: 0
$$\begin{cases}{{x}+{y}\sqrt{{x}}\:=\:\frac{\mathrm{95}}{\mathrm{8}}}\\{{y}+{x}\sqrt{{y}}\:=\:\frac{\mathrm{93}}{\mathrm{8}}}\end{cases}\:.\:\mathcal{F}{ind}\:\sqrt{{xy}} \\ $$
Question Number 107262 Answers: 0 Comments: 0
$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{requirement}\:\mathrm{of}\:\mathrm{last} \\ $$$$\mathrm{axioms}\:\mathrm{i}.\mathrm{e}.\:\mathrm{1}{v}={v}\:\forall\:{v}\in{V}\:\mathrm{in}\:\mathrm{the} \\ $$$$\mathrm{definition}\:\mathrm{of}\:\mathrm{vector}\:\mathrm{space}? \\ $$
Question Number 107242 Answers: 4 Comments: 0
$$\:\:\:\:\:...{question}... \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{prove}\:{that}: \\ $$$$\:{if}\:\:{a},{b},{c}\in\mathbb{R}^{+} \:{then}: \\ $$$$\:\:\:\:\:\clubsuit\:\:\:\sqrt{{a}}\:+\sqrt{{b}}+\sqrt{{c}}>\:\sqrt{{a}+{b}+{c}}\:\clubsuit\: \\ $$$$\:\:\:\:\:\:\:....{sincerly}\:{yours}... \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:...\:\mathscr{M}.\mathscr{N}... \\ $$$$\:\: \\ $$
Question Number 107234 Answers: 1 Comments: 0
Question Number 107117 Answers: 2 Comments: 0
Question Number 107073 Answers: 0 Comments: 0
Question Number 107033 Answers: 0 Comments: 0
Question Number 106941 Answers: 1 Comments: 0
$${Find}\:{the}\:{maximum}\:{value}\:{of}\:\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{sin}^{\mathrm{5}} \:\theta_{{i}} \\ $$$${with}\:\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{sin}\:\theta_{{i}} =\mathrm{0}. \\ $$
Question Number 106943 Answers: 3 Comments: 1
Question Number 106907 Answers: 1 Comments: 0
Question Number 106906 Answers: 2 Comments: 0
Question Number 106816 Answers: 2 Comments: 1
$$\mathrm{prove}\:\mathrm{by}\:\mathrm{mathematical}\:\mathrm{induction} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{n}^{\mathrm{2}} \:\leqslant\:\mathrm{2}^{\mathrm{n}} \:;\:\mathrm{n}\:\geqslant\:\mathrm{4} \\ $$$$\left(\mathrm{2}\right)\:\left(\mathrm{n}+\mathrm{1}\right)^{\mathrm{2}} \:<\:\mathrm{2n}^{\mathrm{2}} \:;\:\mathrm{n}\:\geqslant\:\mathrm{3} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{2}^{\mathrm{n}} −\mathrm{3}\:\geqslant\:\mathrm{2}^{\mathrm{n}−\mathrm{2}} \:;\:\mathrm{n}\:\geqslant\:\mathrm{5} \\ $$$$\:\:\:\:\:\:@\mathrm{bemath}@ \\ $$
Question Number 106794 Answers: 3 Comments: 0
$$\mathrm{repost}\:\mathrm{old}\:\mathrm{question}\:\mathrm{unanswer} \\ $$$$\mathcal{G}\mathrm{iven}\:\rightarrow\begin{cases}{\frac{\mathrm{4x}^{\mathrm{2}} }{\mathrm{1}+\mathrm{4x}^{\mathrm{2}} }\:=\:\mathrm{y}}\\{\frac{\mathrm{4y}^{\mathrm{2}} }{\mathrm{1}+\mathrm{4y}^{\mathrm{2}} }\:=\:\mathrm{z}}\\{\frac{\mathrm{4z}^{\mathrm{2}} }{\mathrm{1}+\mathrm{4z}^{\mathrm{2}} }\:=\:\mathrm{x}}\end{cases} \\ $$
Question Number 106745 Answers: 2 Comments: 8
Question Number 106730 Answers: 1 Comments: 0
Question Number 106727 Answers: 2 Comments: 0
Question Number 106774 Answers: 4 Comments: 0
$$\:\:\:\:\overset{@\mathrm{bemath}@} {\:} \\ $$$$\:\left(\mathrm{1}\right)\:\:\:\:\mathrm{3}^{{x}} \:+\:\mathrm{3}^{\sqrt{{x}}\:} =\:\mathrm{90}.\:\mathrm{find}\:{x}\:?\: \\ $$$$\:\:\left(\mathrm{2}\right)\:\mathrm{x}\:\frac{\mathrm{dy}}{\mathrm{dx}}−\left(\mathrm{1}+\mathrm{x}\right)\mathrm{y}\:=\:\mathrm{xy}^{\mathrm{2}} \\ $$
Question Number 106659 Answers: 1 Comments: 0
$$\mathrm{How}\:\mathrm{many}\:\mathrm{terms}\:\mathrm{has}\:\mathrm{the}\:\mathrm{polynomial}; \\ $$$$\left(\mathrm{X}_{\mathrm{1}} +\mathrm{2X}_{\mathrm{2}} −\mathrm{X}_{\mathrm{3}} +\mathrm{34}\right)^{\mathrm{10}} \:? \\ $$
Question Number 106655 Answers: 0 Comments: 1
Question Number 106378 Answers: 2 Comments: 0
Question Number 106364 Answers: 1 Comments: 2
$$\mathrm{0}^{{i}} =????? \\ $$
Question Number 106314 Answers: 3 Comments: 0
Question Number 106281 Answers: 2 Comments: 0
$$\sqrt{\mathrm{x}+\mathrm{50}}+\sqrt{\mathrm{y}+\mathrm{100}}+\sqrt{\mathrm{z}+\mathrm{150}}=\frac{\mathrm{x}+\mathrm{y}+\mathrm{z}}{\mathrm{4}}+\mathrm{78} \\ $$$$\mathrm{find}\:\mathrm{x}+\mathrm{y}−\mathrm{z}\: \\ $$
Question Number 106272 Answers: 1 Comments: 1
$$\sqrt{\mathrm{1}\:+\:\sqrt{\mathrm{5}\:+\sqrt{\mathrm{11}\:+\:\sqrt{\mathrm{19}\:+\:\sqrt{\mathrm{29}\:+\:\sqrt{\ldots}}}}}}\:=\:? \\ $$
Question Number 106259 Answers: 2 Comments: 1
Question Number 106233 Answers: 2 Comments: 0
$$\left({x}^{\mathrm{2}} \:+\:\mathrm{1}\right)\left({x}\:−\:\mathrm{1}\right)^{\mathrm{2}} \:=\:\mathrm{2017}{yz} \\ $$$$\left({y}^{\mathrm{2}} \:+\:\mathrm{1}\right)\left({y}\:−\:\mathrm{1}\right)^{\mathrm{2}} \:=\:\mathrm{2017}{xz} \\ $$$$\left({z}^{\mathrm{2}} \:+\:\mathrm{1}\right)\left({z}\:−\:\mathrm{1}\right)^{\mathrm{2}} \:=\:\mathrm{2017}{xy} \\ $$$${x}\:\geqslant\:\mathrm{1}\:;\:{y}\:\geqslant\:\mathrm{1}\:;\:{z}\:\geqslant\:\mathrm{1} \\ $$$${prove}\:{that}\:\:\:{x}\:=\:{y}\:=\:{z}\:= \\ $$$$\frac{\mathrm{1}+\:\sqrt{\mathrm{2018}}\:+\:\sqrt{\mathrm{2015}+\:\mathrm{2}\sqrt{\mathrm{2018}}}}{\mathrm{2}} \\ $$
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