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AlgebraQuestion and Answers: Page 64
Question Number 194444 Answers: 1 Comments: 0
Question Number 194422 Answers: 1 Comments: 0
$${What}\:{books}\:{use}\:{for}\:{studying}\:{inequalities} \\ $$$${for}\:{beginners}\: \\ $$
Question Number 194389 Answers: 1 Comments: 0
$$\sqrt{\sqrt{\mathrm{49}+\mathrm{20}\sqrt{\mathrm{6}}}}=? \\ $$
Question Number 194383 Answers: 0 Comments: 0
Question Number 194373 Answers: 0 Comments: 0
$${Show}\:\:\: \\ $$$$\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{2}{i}−\mathrm{1}}−\:\frac{\mathrm{1}}{\mathrm{2}{i}}\right)=\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\:\frac{\mathrm{1}}{{n}+{i}}\: \\ $$$$ \\ $$
Question Number 194326 Answers: 1 Comments: 0
$$\:\:\sqrt{\frac{\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{66}^{\mathrm{2}} +\mathrm{x}}}{\mathrm{x}}\:}\:−\sqrt{\mathrm{x}\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{66}^{\mathrm{2}} }−\mathrm{x}^{\mathrm{2}} }\:=\:\mathrm{5}\: \\ $$
Question Number 194322 Answers: 0 Comments: 0
Question Number 194321 Answers: 1 Comments: 0
$$\mathrm{find}\:\mathrm{min}\:\:−\frac{\left({x}−{y}\right)\left(\frac{{xy}}{\mathrm{4}}−\mathrm{4}\right)^{\mathrm{2}} }{{xy}} \\ $$$$\mathrm{s}.\:\mathrm{t}.\:\:{x}>\mathrm{0}>{y} \\ $$
Question Number 194316 Answers: 0 Comments: 1
$${if}\:\:{x}\in{R}\:\:\&\:\:{x}^{{x}^{\mathrm{6}} } =\left(\sqrt{\mathrm{2}}\right)^{\sqrt{\mathrm{2}}} \:\Rightarrow\:\:{x}=? \\ $$
Question Number 194295 Answers: 1 Comments: 0
Question Number 194238 Answers: 1 Comments: 0
Question Number 194226 Answers: 3 Comments: 0
$$ \\ $$$$\mathrm{If}\:{x}^{\mathrm{2}} \:−\:\mathrm{65}{x}\:=\:\mathrm{64}\sqrt{{x}}\:\mathrm{then}\:\sqrt{{x}\:−\:\sqrt{{x}}\:}\:=\:? \\ $$
Question Number 194219 Answers: 1 Comments: 0
Question Number 194218 Answers: 1 Comments: 2
Question Number 194216 Answers: 1 Comments: 0
Question Number 194209 Answers: 0 Comments: 0
$$ \\ $$$$\:\:\:\:\:\:\sqrt[{\mathrm{3}}]{\mathrm{1}+{kx}}\:+{x}\:=\:\mathrm{1}\:\:\:{has} \\ $$$$\:\:\:\:\:\:{two}\:{real}\:{roots}\:.\:\:\Rightarrow\:{k}=? \\ $$$$\:\:\:{kx}+\mathrm{1}=\:\mathrm{1}−\mathrm{3}{x}+\mathrm{3}{x}^{\mathrm{2}} −{x}^{\:\mathrm{3}} \\ $$$$\:\:\: \\ $$$$\:\:\:\:\:{x}^{\:\mathrm{3}} \:−\mathrm{3}{x}^{\:\mathrm{2}} +\:\left({k}+\mathrm{3}\right){x}=\mathrm{0} \\ $$$$\:\:\:\:\:{x}=\mathrm{0} \\ $$$$\:\:\:\:\:{x}^{\:\mathrm{2}} −\mathrm{3}{x}\:+{k}+\mathrm{3}=\mathrm{0} \\ $$$$\:\:\:\:\mathrm{1}:\:\:\Delta=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{9}\:−\:\mathrm{4}{k}\:−\mathrm{12}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:{k}=\:\frac{−\mathrm{3}}{\mathrm{4}} \\ $$$$\:\:\:\:\:\:\mathrm{2}:\:\:{k}=−\mathrm{3}\:\Rightarrow\:{x}=\mathrm{0}\:{is}\:{a}\:{root}\:{of} \\ $$$$\:\:\:\:\:\:{x}^{\:\mathrm{2}} −\mathrm{3}{x}+{k}+\mathrm{3}=\mathrm{0} \\ $$$$\:\:\:\:\therefore\:\:\:\:\:\:{k}=\:\frac{−\mathrm{3}}{\mathrm{4}}\:\:\:\:{or}\:\:\:{k}=−\mathrm{3} \\ $$
Question Number 194207 Answers: 0 Comments: 0
$$\:\:\:\:\mathrm{Solve}\:\mathrm{for}\:\mathrm{x}\: \\ $$$$\:\:\:\:\mathrm{x}^{\mathrm{x}−\mathrm{4}} \:\:=\sqrt{\mathrm{3}\:} \\ $$
Question Number 194206 Answers: 1 Comments: 0
Question Number 194204 Answers: 1 Comments: 0
Question Number 194197 Answers: 1 Comments: 0
Question Number 194191 Answers: 1 Comments: 0
Question Number 194170 Answers: 0 Comments: 0
Question Number 194169 Answers: 0 Comments: 0
Question Number 194166 Answers: 1 Comments: 0
Question Number 194163 Answers: 0 Comments: 0
$$\boldsymbol{{Let}}\:\boldsymbol{{a}}\:,\:\boldsymbol{{b}}\:,\:\boldsymbol{{c}}\:\:\:\boldsymbol{{be}}\:\boldsymbol{{positive}}\:\boldsymbol{{real}}\:\boldsymbol{{numbers}} \\ $$$$\boldsymbol{{prove}}\:\boldsymbol{{that}} \\ $$$$\frac{\boldsymbol{{a}}}{\boldsymbol{{b}}}+\frac{\boldsymbol{{b}}}{\boldsymbol{{c}}}+\frac{\boldsymbol{{c}}}{\boldsymbol{{a}}}+\frac{\mathrm{3}^{\mathrm{3}} \sqrt{{abc}}}{\boldsymbol{{a}}+\boldsymbol{{b}}+\boldsymbol{{c}}}\geqslant\mathrm{4} \\ $$
Question Number 194159 Answers: 0 Comments: 0
$$...\mathrm{anybody}\:\mathrm{tried}\:\mathrm{question}\:\mathrm{193484}? \\ $$$$\mathrm{It}'\mathrm{s}\:\mathrm{not}\:\mathrm{as}\:\mathrm{hard}\:\mathrm{as}\:\mathrm{it}\:\mathrm{may}\:\mathrm{seem}. \\ $$
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