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Question Number 194226 Answers: 3 Comments: 0

If x^2 − 65x = 64(√x) then (√(x − (√x) )) = ?

$$ \\ $$$$\mathrm{If}\:{x}^{\mathrm{2}} \:−\:\mathrm{65}{x}\:=\:\mathrm{64}\sqrt{{x}}\:\mathrm{then}\:\sqrt{{x}\:−\:\sqrt{{x}}\:}\:=\:? \\ $$

Question Number 194219 Answers: 1 Comments: 0

Question Number 194218 Answers: 1 Comments: 2

Question Number 194216 Answers: 1 Comments: 0

Question Number 194211 Answers: 1 Comments: 0

(1/x^2 ) +(1/y^2 ) = (1/3) (d^2 y/dx^2 ) =?

$$\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\:+\frac{\mathrm{1}}{\mathrm{y}^{\mathrm{2}} }\:=\:\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\:\:\:\:\:\frac{\mathrm{d}^{\mathrm{2}} \mathrm{y}}{\mathrm{dx}^{\mathrm{2}} }\:=? \\ $$

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

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

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