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AlgebraQuestion and Answers: Page 5
Question Number 222141 Answers: 2 Comments: 0
$${x}^{\mathrm{2}} +\left(\frac{{x}}{\mathrm{2}{x}−\mathrm{1}}\right)^{\mathrm{2}} =\mathrm{12} \\ $$
Question Number 222125 Answers: 4 Comments: 2
$$\mathrm{If}:\:\:\:\mathrm{log}_{\mathrm{3}} \left(\mathrm{5}^{\boldsymbol{\mathrm{x}}} \:+\:\frac{\mathrm{1}}{\mathrm{5}^{\boldsymbol{\mathrm{x}}} }\:+\:\mathrm{7}\right)\:\:\:\Rightarrow\:\:\:\mathrm{min}\:=\:? \\ $$
Question Number 222076 Answers: 2 Comments: 0
Question Number 222066 Answers: 0 Comments: 4
Question Number 222031 Answers: 3 Comments: 0
$${x}^{{x}} =−\mathrm{1} \\ $$$${Number}\:{of}\:{solutions}?? \\ $$
Question Number 222026 Answers: 2 Comments: 0
$${If}\:\left(\mathrm{1}.\mathrm{234}\right)^{{a}} =\left(\mathrm{0}.\mathrm{1234}\right)^{{b}} =\mathrm{10}^{{c}} \\ $$$${prove}\:{that}\:\frac{\mathrm{1}}{{a}}−\frac{\mathrm{1}}{{c}}=\frac{\mathrm{1}}{{b}} \\ $$
Question Number 222003 Answers: 2 Comments: 0
$${If}\:{a}+{b}+{c}=\mathrm{0}\:{then}\:{prove}\:{that} \\ $$$$\frac{\mathrm{1}}{{x}^{{b}} +{x}^{−{c}} +\mathrm{1}}+\frac{\mathrm{1}}{{x}^{{c}} +{x}^{−{a}} +\mathrm{1}}+\frac{\mathrm{1}}{{x}^{{a}} +{x}^{−{b}} +\mathrm{1}}=\mathrm{1} \\ $$
Question Number 222001 Answers: 1 Comments: 6
$$\left(\frac{\mathrm{4}^{{m}+\frac{\mathrm{1}}{\mathrm{4}}} ×\sqrt{\mathrm{2}.\mathrm{2}^{{m}} }}{\mathrm{2}.\sqrt{\mathrm{2}^{−{m}} }}\right)^{\frac{\mathrm{1}}{{m}}} =?? \\ $$
Question Number 221991 Answers: 1 Comments: 4
$$\mathrm{Simplify}:\:\:\:\mathrm{2}^{\mathrm{2}} \:\centerdot\:\mathrm{2}^{\mathrm{2}^{\frac{\mathrm{70}\:−\:\boldsymbol{\mathrm{t}}_{\mathrm{1}} }{\mathrm{10}}} \:\:\:=\:\:\:?} \\ $$
Question Number 221968 Answers: 3 Comments: 0
$$\left({a}+{b}+{c}\right)^{\mathrm{3}} \\ $$
Question Number 221944 Answers: 1 Comments: 1
Question Number 221943 Answers: 1 Comments: 0
Question Number 221863 Answers: 1 Comments: 1
$${If}\:{a}\:{and}\:{b}\:{are}\:{whole}\:{numbers}\:{such}\:{a}^{{b}} =\mathrm{121} \\ $$$${then}\:{find}\:{the}\:{value}\:{of}\:\left({a}−\mathrm{1}\right)^{{b}+\mathrm{1}} \\ $$
Question Number 221856 Answers: 0 Comments: 3
Question Number 221848 Answers: 0 Comments: 2
Question Number 221829 Answers: 4 Comments: 0
$$\sqrt{\mathrm{70}.\mathrm{71}.\mathrm{72}.\mathrm{73}+\mathrm{1}} \\ $$
Question Number 221782 Answers: 0 Comments: 0
$$\left(\sqrt[{\sqrt[{\mathrm{3}^{{a}} }]{\mathrm{27}}}]{\mathrm{81}}\right)^{\sqrt{\mathrm{4}}} \\ $$$${where}\:{a}=\mathrm{4}^{\mathrm{0}^{\mathrm{4}^{\mathrm{3}} } } {and}\:{b}=\underset{{n}=\mathrm{1}} {\overset{\mathrm{6}} {\sum}}{n} \\ $$
Question Number 221778 Answers: 0 Comments: 0
$$\mathrm{For}\:\forall{n}\in\boldsymbol{{N}}^{\ast} ,{n}\geq\mathrm{3}\:\mathrm{Prove}: \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \left(\underset{\mathrm{2}<{p}\leq\mathrm{2}{n}} {\sum}{e}^{\mathrm{2}\pi{ip}+\alpha} \right)^{\mathrm{2}} {e}^{−\mathrm{4}\pi{in}\alpha} {d}\alpha>\mathrm{0},{p}\:\mathrm{is}\:\mathrm{a}\:\mathrm{prime}\:\mathrm{number} \\ $$
Question Number 221754 Answers: 1 Comments: 0
$$\left.\sqrt[{\sqrt[{\sqrt[{\:\mathrm{3}^{\mathrm{4}^{\mathrm{0}^{\mathrm{4}^{\mathrm{3}} } } } }]{\mathrm{27}}}]{\mathrm{64}}}]{\mathrm{81}}\right)^{\sqrt{\mathrm{4}}} \\ $$
Question Number 221647 Answers: 0 Comments: 1
$${solve}\:{for}\:{x}. \\ $$$${x}^{\mathrm{1}} \:+\:{x}^{\mathrm{2}} \:+\:{x}^{\mathrm{3}} \:\:=\:\:\mathrm{4096} \\ $$
Question Number 221618 Answers: 3 Comments: 0
$${solve}\:{for}\:{x} \\ $$$$\mathrm{2}^{{x}} +\mathrm{4}^{{x}} =\mathrm{8}^{{x}} \\ $$
Question Number 221669 Answers: 2 Comments: 3
Question Number 221585 Answers: 6 Comments: 1
$${solve}\:{for}\:{x}\:\in{R} \\ $$$$\left({x}^{\mathrm{3}} −\mathrm{6}\right)^{\mathrm{3}} ={x}+\mathrm{6} \\ $$
Question Number 221544 Answers: 1 Comments: 0
$$\:\:\sqrt[{{x}+\mathrm{1}}]{{x}−\mathrm{1}}\:=\:\sqrt[{{x}−\mathrm{1}}]{{x}+\mathrm{1}}\:,\:{x}\:{real}\: \\ $$
Question Number 221504 Answers: 1 Comments: 0
$${Find}\:{x}\:{if}\:\:\:\:\frac{{x}^{\mathrm{4}} }{\mathrm{4}^{{x}} }=\frac{\mathrm{4}^{{x}} }{{x}^{\mathrm{4}} }\:\:.\:\:\:{x}\in\mathbb{R} \\ $$
Question Number 221501 Answers: 3 Comments: 0
$${solve}\:{for}\:\mathrm{x}: \\ $$$$\sqrt{\mathrm{a}−\sqrt{\mathrm{a}+\mathrm{x}}}+\sqrt{\mathrm{a}+\sqrt{\mathrm{a}−\mathrm{x}}}=\mathrm{2x} \\ $$$${it}'{s}\:{possible}\:{to}\:{solve}\:{for}\:\mathrm{a}\:{but}\:\mathrm{x}\:{seems} \\ $$$${impossible}\:{to}\:{me} \\ $$
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