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Question Number 194208    Answers: 0   Comments: 0

Question Number 194209    Answers: 0   Comments: 0

((1+kx))^(1/3) +x = 1 has two real roots . ⇒ k=? kx+1= 1−3x+3x^2 −x^( 3) x^( 3) −3x^( 2) + (k+3)x=0 x=0 x^( 2) −3x +k+3=0 1: Δ=0 9 − 4k −12=0 k= ((−3)/4) 2: k=−3 ⇒ x=0 is a root of x^( 2) −3x+k+3=0 ∴ k= ((−3)/4) or k=−3

$$ \\ $$$$\:\:\:\:\:\:\sqrt[{\mathrm{3}}]{\mathrm{1}+{kx}}\:+{x}\:=\:\mathrm{1}\:\:\:{has} \\ $$$$\:\:\:\:\:\:{two}\:{real}\:{roots}\:.\:\:\Rightarrow\:{k}=? \\ $$$$\:\:\:{kx}+\mathrm{1}=\:\mathrm{1}−\mathrm{3}{x}+\mathrm{3}{x}^{\mathrm{2}} −{x}^{\:\mathrm{3}} \\ $$$$\:\:\: \\ $$$$\:\:\:\:\:{x}^{\:\mathrm{3}} \:−\mathrm{3}{x}^{\:\mathrm{2}} +\:\left({k}+\mathrm{3}\right){x}=\mathrm{0} \\ $$$$\:\:\:\:\:{x}=\mathrm{0} \\ $$$$\:\:\:\:\:{x}^{\:\mathrm{2}} −\mathrm{3}{x}\:+{k}+\mathrm{3}=\mathrm{0} \\ $$$$\:\:\:\:\mathrm{1}:\:\:\Delta=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{9}\:−\:\mathrm{4}{k}\:−\mathrm{12}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:{k}=\:\frac{−\mathrm{3}}{\mathrm{4}} \\ $$$$\:\:\:\:\:\:\mathrm{2}:\:\:{k}=−\mathrm{3}\:\Rightarrow\:{x}=\mathrm{0}\:{is}\:{a}\:{root}\:{of} \\ $$$$\:\:\:\:\:\:{x}^{\:\mathrm{2}} −\mathrm{3}{x}+{k}+\mathrm{3}=\mathrm{0} \\ $$$$\:\:\:\:\therefore\:\:\:\:\:\:{k}=\:\frac{−\mathrm{3}}{\mathrm{4}}\:\:\:\:{or}\:\:\:{k}=−\mathrm{3} \\ $$

Question Number 194207    Answers: 0   Comments: 0

Solve for x x^(x−4) =(√(3 ))

$$\:\:\:\:\mathrm{Solve}\:\mathrm{for}\:\mathrm{x}\: \\ $$$$\:\:\:\:\mathrm{x}^{\mathrm{x}−\mathrm{4}} \:\:=\sqrt{\mathrm{3}\:} \\ $$

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(ax+b)+f(cx+d)=ex+g then f(x)=Ax^2 +Bx+C ¿

$${Explanation}\:{Why}: \\ $$$${While}\:{f}\left({ax}+{b}\right)+{f}\left({cx}+{d}\right)={ex}+{g} \\ $$$${then}\:{f}\left({x}\right)={Ax}^{\mathrm{2}} +{Bx}+{C}\:¿ \\ $$

Question Number 194185    Answers: 1   Comments: 0

lim_( x→ 0^( −) ) { (( x^( 2) +2cos(x) + ⌊−((tan(x))/x) ⌋)/(ax^( 4) )) } = 1 a = ? a: (1/(12)) b: −(1/2) c: 12 d: −12

$$ \\ $$$$\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\mathrm{lim}_{\:{x}\rightarrow\:\mathrm{0}^{\:−} } \:\left\{\:\frac{\:{x}^{\:\mathrm{2}} \:+\mathrm{2}{cos}\left({x}\right)\:+\:\lfloor−\frac{{tan}\left({x}\right)}{{x}}\:\rfloor}{{ax}^{\:\mathrm{4}} }\:\right\}\:=\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{a}\:=\:? \\ $$$$\:\:\:\:\:\:\:\:{a}:\:\:\:\frac{\mathrm{1}}{\mathrm{12}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{b}:\:\:−\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{c}:\:\:\:\mathrm{12}\:\:\:\:\:\:\:\:\:\:\:\:\:\:{d}:\:\:−\mathrm{12}\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\: \\ $$

Question Number 194183    Answers: 1   Comments: 0

Σ_(n=1) ^∞ (1/(n(n+15)(n+30)))

$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{{n}\left({n}+\mathrm{15}\right)\left({n}+\mathrm{30}\right)} \\ $$

Question Number 194176    Answers: 3   Comments: 0

Hello everyone I try to solve 4^(x+1) +2^(2−x) =65 Thx in advance

$${Hello}\:{everyone} \\ $$$${I}\:{try}\:{to}\:{solve}\:\mathrm{4}^{{x}+\mathrm{1}} +\mathrm{2}^{\mathrm{2}−{x}} =\mathrm{65} \\ $$$${Thx}\:{in}\:{advance} \\ $$

Question Number 194170    Answers: 0   Comments: 0

Question Number 194169    Answers: 0   Comments: 0

Question Number 194174    Answers: 1   Comments: 0

Know f(x^(−1) )=f^(−1) (x) (f^(−1) (x)=(1/(f(x)))) Find f(x)¿

$${Know}\:{f}\left({x}^{−\mathrm{1}} \right)={f}^{−\mathrm{1}} \left({x}\right)\:\left({f}^{−\mathrm{1}} \left({x}\right)=\frac{\mathrm{1}}{{f}\left({x}\right)}\right) \\ $$$${Find}\:{f}\left({x}\right)¿ \\ $$

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

Let a , b , c be positive real numbers prove that (a/b)+(b/c)+(c/a)+((3^3 (√(abc)))/(a+b+c))≥4

$$\boldsymbol{{Let}}\:\boldsymbol{{a}}\:,\:\boldsymbol{{b}}\:,\:\boldsymbol{{c}}\:\:\:\boldsymbol{{be}}\:\boldsymbol{{positive}}\:\boldsymbol{{real}}\:\boldsymbol{{numbers}} \\ $$$$\boldsymbol{{prove}}\:\boldsymbol{{that}} \\ $$$$\frac{\boldsymbol{{a}}}{\boldsymbol{{b}}}+\frac{\boldsymbol{{b}}}{\boldsymbol{{c}}}+\frac{\boldsymbol{{c}}}{\boldsymbol{{a}}}+\frac{\mathrm{3}^{\mathrm{3}} \sqrt{{abc}}}{\boldsymbol{{a}}+\boldsymbol{{b}}+\boldsymbol{{c}}}\geqslant\mathrm{4} \\ $$

Question Number 194159    Answers: 0   Comments: 0

...anybody tried question 193484? It′s not as hard as it may seem.

$$...\mathrm{anybody}\:\mathrm{tried}\:\mathrm{question}\:\mathrm{193484}? \\ $$$$\mathrm{It}'\mathrm{s}\:\mathrm{not}\:\mathrm{as}\:\mathrm{hard}\:\mathrm{as}\:\mathrm{it}\:\mathrm{may}\:\mathrm{seem}. \\ $$

Question Number 194158    Answers: 1   Comments: 0

Find all possible solutions: (1/s)+(1/t)+(1/u)+(1/v)=1 With s, t, u, v ∈N and s<t<u<v

$$\mathrm{Find}\:\mathrm{all}\:\mathrm{possible}\:\mathrm{solutions}: \\ $$$$\frac{\mathrm{1}}{{s}}+\frac{\mathrm{1}}{{t}}+\frac{\mathrm{1}}{{u}}+\frac{\mathrm{1}}{{v}}=\mathrm{1} \\ $$$$\mathrm{With}\:{s},\:{t},\:{u},\:{v}\:\in\mathbb{N}\:\mathrm{and}\:{s}<{t}<{u}<{v} \\ $$

Question Number 194147    Answers: 2   Comments: 0

lim_(n→+∞) ((1/1^2 )+(1/2^2 )+(1/3^2 )+...+(1/n^2 ))=?

$$\underset{\mathrm{n}\rightarrow+\infty} {\mathrm{lim}}\left(\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+...+\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} }\right)=? \\ $$

Question Number 194144    Answers: 5   Comments: 0

fill with different natural numbers: (1/(19))=(1/(( )))+(1/(( )))+(1/(( )))+(1/(( )))

$${fill}\:{with}\:{different}\:{natural}\:{numbers}: \\ $$$$\:\:\frac{\mathrm{1}}{\mathrm{19}}=\frac{\mathrm{1}}{\left(\:\:\right)}+\frac{\mathrm{1}}{\left(\:\:\right)}+\frac{\mathrm{1}}{\left(\:\:\right)}+\frac{\mathrm{1}}{\left(\:\:\right)} \\ $$

Question Number 194143    Answers: 0   Comments: 1

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