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Question Number 194208 Answers: 0 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 194201 Answers: 1 Comments: 3
Question Number 194197 Answers: 1 Comments: 0
Question Number 194194 Answers: 0 Comments: 0
Question Number 194193 Answers: 0 Comments: 0
Question Number 194191 Answers: 1 Comments: 0
Question Number 194190 Answers: 0 Comments: 1
$${Explanation}\:{Why}: \\ $$$${While}\:{f}\left({ax}+{b}\right)+{f}\left({cx}+{d}\right)={ex}+{g} \\ $$$${then}\:{f}\left({x}\right)={Ax}^{\mathrm{2}} +{Bx}+{C}\:¿ \\ $$
Question Number 194185 Answers: 1 Comments: 0
$$ \\ $$$$\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\mathrm{lim}_{\:{x}\rightarrow\:\mathrm{0}^{\:−} } \:\left\{\:\frac{\:{x}^{\:\mathrm{2}} \:+\mathrm{2}{cos}\left({x}\right)\:+\:\lfloor−\frac{{tan}\left({x}\right)}{{x}}\:\rfloor}{{ax}^{\:\mathrm{4}} }\:\right\}\:=\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{a}\:=\:? \\ $$$$\:\:\:\:\:\:\:\:{a}:\:\:\:\frac{\mathrm{1}}{\mathrm{12}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{b}:\:\:−\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{c}:\:\:\:\mathrm{12}\:\:\:\:\:\:\:\:\:\:\:\:\:\:{d}:\:\:−\mathrm{12}\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\: \\ $$
Question Number 194183 Answers: 1 Comments: 0
$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{{n}\left({n}+\mathrm{15}\right)\left({n}+\mathrm{30}\right)} \\ $$
Question Number 194176 Answers: 3 Comments: 0
$${Hello}\:{everyone} \\ $$$${I}\:{try}\:{to}\:{solve}\:\mathrm{4}^{{x}+\mathrm{1}} +\mathrm{2}^{\mathrm{2}−{x}} =\mathrm{65} \\ $$$${Thx}\:{in}\:{advance} \\ $$
Question Number 194170 Answers: 0 Comments: 0
Question Number 194169 Answers: 0 Comments: 0
Question Number 194174 Answers: 1 Comments: 0
$${Know}\:{f}\left({x}^{−\mathrm{1}} \right)={f}^{−\mathrm{1}} \left({x}\right)\:\left({f}^{−\mathrm{1}} \left({x}\right)=\frac{\mathrm{1}}{{f}\left({x}\right)}\right) \\ $$$${Find}\:{f}\left({x}\right)¿ \\ $$
Question Number 194166 Answers: 1 Comments: 0
Question Number 194172 Answers: 2 Comments: 1
Question Number 194165 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
$$...\mathrm{anybody}\:\mathrm{tried}\:\mathrm{question}\:\mathrm{193484}? \\ $$$$\mathrm{It}'\mathrm{s}\:\mathrm{not}\:\mathrm{as}\:\mathrm{hard}\:\mathrm{as}\:\mathrm{it}\:\mathrm{may}\:\mathrm{seem}. \\ $$
Question Number 194158 Answers: 1 Comments: 0
$$\mathrm{Find}\:\mathrm{all}\:\mathrm{possible}\:\mathrm{solutions}: \\ $$$$\frac{\mathrm{1}}{{s}}+\frac{\mathrm{1}}{{t}}+\frac{\mathrm{1}}{{u}}+\frac{\mathrm{1}}{{v}}=\mathrm{1} \\ $$$$\mathrm{With}\:{s},\:{t},\:{u},\:{v}\:\in\mathbb{N}\:\mathrm{and}\:{s}<{t}<{u}<{v} \\ $$
Question Number 194147 Answers: 2 Comments: 0
$$\underset{\mathrm{n}\rightarrow+\infty} {\mathrm{lim}}\left(\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+...+\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} }\right)=? \\ $$
Question Number 194144 Answers: 5 Comments: 0
$${fill}\:{with}\:{different}\:{natural}\:{numbers}: \\ $$$$\:\:\frac{\mathrm{1}}{\mathrm{19}}=\frac{\mathrm{1}}{\left(\:\:\right)}+\frac{\mathrm{1}}{\left(\:\:\right)}+\frac{\mathrm{1}}{\left(\:\:\right)}+\frac{\mathrm{1}}{\left(\:\:\right)} \\ $$
Question Number 194143 Answers: 0 Comments: 1
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