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Question Number 64061    Answers: 1   Comments: 0

(2x+3)^2 +25/(x+3)^2 =(√2)

$$\left(\mathrm{2}{x}+\mathrm{3}\right)^{\mathrm{2}} +\mathrm{25}/\left({x}+\mathrm{3}\right)^{\mathrm{2}} =\sqrt{\mathrm{2}} \\ $$

Question Number 63958    Answers: 0   Comments: 1

x^6 −3x^5 +4x^4 −6x^3 +5x^2 −3x+2=0

$${x}^{\mathrm{6}} −\mathrm{3}{x}^{\mathrm{5}} +\mathrm{4}{x}^{\mathrm{4}} −\mathrm{6}{x}^{\mathrm{3}} +\mathrm{5}{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{2}=\mathrm{0} \\ $$

Question Number 63945    Answers: 0   Comments: 1

solve at Z^2 x^2 −2y^2 +xy +2 =0

$${solve}\:{at}\:{Z}^{\mathrm{2}} \:\:{x}^{\mathrm{2}} −\mathrm{2}{y}^{\mathrm{2}} \:+{xy}\:+\mathrm{2}\:=\mathrm{0} \\ $$

Question Number 63930    Answers: 0   Comments: 0

Question Number 63893    Answers: 0   Comments: 1

1) simplify W_n (z)=(1+z)(1+z^2 )....(1+z^2^n ) (z from C) 2) simplify P_n (θ) =(1+e^(iθ) )(1+e^(2iθ) ).....(1+e^(i2^n θ) ) and sove P_n (θ)=0

$$\left.\mathrm{1}\right)\:{simplify}\:{W}_{{n}} \left({z}\right)=\left(\mathrm{1}+{z}\right)\left(\mathrm{1}+{z}^{\mathrm{2}} \right)....\left(\mathrm{1}+{z}^{\mathrm{2}^{{n}} } \right)\:\left({z}\:{from}\:{C}\right) \\ $$$$\left.\mathrm{2}\right)\:{simplify}\:{P}_{{n}} \left(\theta\right)\:=\left(\mathrm{1}+{e}^{{i}\theta} \right)\left(\mathrm{1}+{e}^{\mathrm{2}{i}\theta} \right).....\left(\mathrm{1}+{e}^{{i}\mathrm{2}^{{n}} \theta} \right)\:{and}\:{sove} \\ $$$${P}_{{n}} \left(\theta\right)=\mathrm{0} \\ $$

Question Number 63858    Answers: 1   Comments: 0

if a_1 , a_2 , a_3 , a_4 are the coefficient of any four four consecutive terms in the expansion of (1+x)^n then (a_1 /(a_2 +a_1 ))+(a_3 /(a_3 +a_4 )) is equal to...

$$\mathrm{if}\:\mathrm{a}_{\mathrm{1}} ,\:\mathrm{a}_{\mathrm{2}} ,\:\mathrm{a}_{\mathrm{3}} ,\:\mathrm{a}_{\mathrm{4}} \:\mathrm{are}\:\mathrm{the}\:\mathrm{coefficient} \\ $$$$\mathrm{of}\:\mathrm{any}\:\mathrm{four}\:\mathrm{four}\:\mathrm{consecutive} \\ $$$$\mathrm{terms}\:\mathrm{in}\:\mathrm{the}\:\mathrm{expansion}\:\mathrm{of}\:\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{n}} \\ $$$$\mathrm{then}\:\frac{\mathrm{a}_{\mathrm{1}} }{\mathrm{a}_{\mathrm{2}} +\mathrm{a}_{\mathrm{1}} }+\frac{\mathrm{a}_{\mathrm{3}} }{\mathrm{a}_{\mathrm{3}} +\mathrm{a}_{\mathrm{4}} }\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to}... \\ $$

Question Number 63803    Answers: 1   Comments: 4

Question Number 63784    Answers: 0   Comments: 6

question 63639 again prove: ∀z∈C: ∣z+1∣+∣z^2 +z+1∣+∣z^3 +1∣≥1

$$\mathrm{question}\:\mathrm{63639}\:\mathrm{again} \\ $$$$\mathrm{prove}: \\ $$$$\forall{z}\in\mathbb{C}:\:\mid{z}+\mathrm{1}\mid+\mid{z}^{\mathrm{2}} +{z}+\mathrm{1}\mid+\mid{z}^{\mathrm{3}} +\mathrm{1}\mid\geqslant\mathrm{1} \\ $$

Question Number 63678    Answers: 0   Comments: 1

Question Number 63642    Answers: 2   Comments: 1

Question Number 63639    Answers: 0   Comments: 0

Question Number 63602    Answers: 2   Comments: 0

32x^3 −48x^2 −22x−3=0

$$\mathrm{32}{x}^{\mathrm{3}} −\mathrm{48}{x}^{\mathrm{2}} −\mathrm{22}{x}−\mathrm{3}=\mathrm{0} \\ $$

Question Number 63574    Answers: 0   Comments: 12

prove that Σ_(k = 1) ^∞ (1/(k(2k + 1))) = 2 − 2ln(2)

$$\mathrm{prove}\:\mathrm{that}\:\:\:\underset{\mathrm{k}\:=\:\mathrm{1}} {\overset{\infty} {\sum}}\:\:\frac{\mathrm{1}}{\mathrm{k}\left(\mathrm{2k}\:+\:\mathrm{1}\right)}\:\:=\:\:\mathrm{2}\:−\:\mathrm{2ln}\left(\mathrm{2}\right) \\ $$

Question Number 63573    Answers: 0   Comments: 0

Question Number 63565    Answers: 0   Comments: 0

Question Number 63522    Answers: 0   Comments: 2

Question Number 63485    Answers: 1   Comments: 0

f(x−3)+f(x)=2x−3 F(2)=0. F(−2)=?

$$\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}−\mathrm{3}\right)+\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}\right)=\mathrm{2}\boldsymbol{\mathrm{x}}−\mathrm{3} \\ $$$$\boldsymbol{\mathrm{F}}\left(\mathrm{2}\right)=\mathrm{0}. \\ $$$$\boldsymbol{\mathrm{F}}\left(−\mathrm{2}\right)=? \\ $$

Question Number 63474    Answers: 1   Comments: 0

let P(x)=x^2 +(1/2)x+b and Q(x)=x^2 +cx+d be to polynomials with real coefficient such that P(x) Q(x)=Q(P(x)) find all the real roots of P(Q(x))=0

$${let}\:{P}\left({x}\right)={x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}{x}+{b} \\ $$$$ \\ $$$${and}\:{Q}\left({x}\right)={x}^{\mathrm{2}} +{cx}+{d} \\ $$$$ \\ $$$${be}\:{to}\:{polynomials}\:{with}\:{real}\:{coefficient}\:{such}\:{that} \\ $$$$ \\ $$$${P}\left({x}\right)\:{Q}\left({x}\right)={Q}\left({P}\left({x}\right)\right) \\ $$$$ \\ $$$${find}\:{all}\:{the}\:{real}\:{roots}\:{of}\:{P}\left({Q}\left({x}\right)\right)=\mathrm{0} \\ $$

Question Number 63449    Answers: 0   Comments: 2

Do you remember that π≈3.1415926535897932⋱ 38462643383279502884⋱ 197...

$${Do}\:{you}\:{remember}\:{that} \\ $$$$\pi\approx\mathrm{3}.\mathrm{1415926535897932}\ddots \\ $$$$\mathrm{38462643383279502884}\ddots \\ $$$$\mathrm{197}... \\ $$

Question Number 63614    Answers: 0   Comments: 0

solve the equation with using (FPI_Fixed Point Iteration) x−tan(x)=0 , in[4,5]

$${solve}\:{the}\:{equation}\:{with}\:{using}\:\left({FPI\_Fixed}\:{Point}\:{Iteration}\right) \\ $$$$ \\ $$$${x}−{tan}\left({x}\right)=\mathrm{0}\:,\:{in}\left[\mathrm{4},\mathrm{5}\right] \\ $$

Question Number 63383    Answers: 1   Comments: 1

solve this equation in all part of complex number: (√((x^9 −3x^2 +1)(x−6)+4))=(x^9 −3x^2 +1)(x−6)−16

$${solve}\:{this}\:{equation}\:{in}\:{all}\: \\ $$$${part}\:{of}\:{complex}\:{number}: \\ $$$$\sqrt{\left({x}^{\mathrm{9}} −\mathrm{3}{x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}−\mathrm{6}\right)+\mathrm{4}}=\left({x}^{\mathrm{9}} −\mathrm{3}{x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}−\mathrm{6}\right)−\mathrm{16} \\ $$

Question Number 63381    Answers: 0   Comments: 2

solve for both x and n in equation: x^n =216 in all part of integer A {_(n=3) ^(x=6) B{_(n=4) ^(x=5) C{_(n=5) ^(x=4) D {_(n=6) ^(x=3)

$${solve}\:{for}\:{both}\:{x}\:{and}\:{n} \\ $$$${in}\:{equation}:\:{x}^{{n}} =\mathrm{216}\:{in}\:{all} \\ $$$${part}\:{of}\:{integer} \\ $$$$\mathscr{A}\:\underset{{n}=\mathrm{3}} {\overset{{x}=\mathrm{6}} {\left\{}}\right. \\ $$$$\mathscr{B}\underset{{n}=\mathrm{4}} {\overset{{x}=\mathrm{5}} {\left\{}}\right. \\ $$$$\mathscr{C}\underset{{n}=\mathrm{5}} {\overset{{x}=\mathrm{4}} {\left\{}}\right. \\ $$$$\mathscr{D}\:\underset{{n}=\mathrm{6}} {\overset{{x}=\mathrm{3}} {\left\{}}\right. \\ $$

Question Number 63373    Answers: 0   Comments: 2

just found this on the web I thought it might help in some cases where quartics appear i.e. Sir Aifour′s geometric questions. sometimes we know the nature of the roots, but how to use this information? ax^4 +bx^3 +cx^2 +dx+e=0 1. divide by a 2. x=z−(b/(4a)) this leads to the reduced z^4 +pz^2 +qz+r=0 now we find the nature of the roots: T_1 =16p^4 r−4p^3 q^2 −128p^2 r^2 +144pq^2 r−27q^4 +256r^3 T_2 =p^2 +12r T_3 =−p^2 +4r T_1 <0 ⇒ 2 distinct real and 2 conjugated complex roots T_1 >0∧(p<0∧T_3 <0) ⇒ 4 distinct real roots T_1 >0∧(p>0∨T_3 >0) ⇒ 2 pairs of conjugated complex roots T_1 =0∧(p<0∧T_3 <0∧T_2 ≠0) ⇒ 1 real double and 2 real simple roots T_1 =0∧(T_3 >0∨(p>0∧(T_3 ≠0∨q≠0))) ⇒ 1 real double and 2 conjugated complex roots T_1 =0∧(T_2 =0∧T_3 ≠0) ⇒ 1 real triple and 1 real simple roots T_1 =0∧(T_3 =0∧p<0) ⇒ 2 real double roots T_1 =0∧(T_3 =0∧p>0∧q=0) ⇒ 2 conjugated complex double roots T_1 =0∧T_2 =0 ⇒ all roots are equal

$$\mathrm{just}\:\mathrm{found}\:\mathrm{this}\:\mathrm{on}\:\mathrm{the}\:\mathrm{web} \\ $$$$\mathrm{I}\:\mathrm{thought}\:\mathrm{it}\:\mathrm{might}\:\mathrm{help}\:\mathrm{in}\:\mathrm{some}\:\mathrm{cases}\:\mathrm{where} \\ $$$$\mathrm{quartics}\:\mathrm{appear}\:\mathrm{i}.\mathrm{e}.\:\mathrm{Sir}\:\mathrm{Aifour}'\mathrm{s}\:\mathrm{geometric} \\ $$$$\mathrm{questions}.\:\mathrm{sometimes}\:\mathrm{we}\:\mathrm{know}\:\mathrm{the}\:\mathrm{nature}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{roots},\:\mathrm{but}\:\mathrm{how}\:\mathrm{to}\:\mathrm{use}\:\mathrm{this}\:\mathrm{information}? \\ $$$$ \\ $$$${ax}^{\mathrm{4}} +{bx}^{\mathrm{3}} +{cx}^{\mathrm{2}} +{dx}+{e}=\mathrm{0} \\ $$$$\mathrm{1}.\:\mathrm{divide}\:\mathrm{by}\:{a} \\ $$$$\mathrm{2}.\:{x}={z}−\frac{{b}}{\mathrm{4}{a}} \\ $$$$\mathrm{this}\:\mathrm{leads}\:\mathrm{to}\:\mathrm{the}\:\mathrm{reduced} \\ $$$$ \\ $$$${z}^{\mathrm{4}} +{pz}^{\mathrm{2}} +{qz}+{r}=\mathrm{0} \\ $$$$ \\ $$$$\mathrm{now}\:\mathrm{we}\:\mathrm{find}\:\mathrm{the}\:\mathrm{nature}\:\mathrm{of}\:\mathrm{the}\:\mathrm{roots}: \\ $$$${T}_{\mathrm{1}} =\mathrm{16}{p}^{\mathrm{4}} {r}−\mathrm{4}{p}^{\mathrm{3}} {q}^{\mathrm{2}} −\mathrm{128}{p}^{\mathrm{2}} {r}^{\mathrm{2}} +\mathrm{144}{pq}^{\mathrm{2}} {r}−\mathrm{27}{q}^{\mathrm{4}} +\mathrm{256}{r}^{\mathrm{3}} \\ $$$${T}_{\mathrm{2}} ={p}^{\mathrm{2}} +\mathrm{12}{r} \\ $$$${T}_{\mathrm{3}} =−{p}^{\mathrm{2}} +\mathrm{4}{r} \\ $$$${T}_{\mathrm{1}} <\mathrm{0}\:\Rightarrow\:\mathrm{2}\:\mathrm{distinct}\:\mathrm{real}\:\mathrm{and}\:\mathrm{2}\:\mathrm{conjugated}\:\mathrm{complex}\:\mathrm{roots} \\ $$$${T}_{\mathrm{1}} >\mathrm{0}\wedge\left({p}<\mathrm{0}\wedge{T}_{\mathrm{3}} <\mathrm{0}\right)\:\Rightarrow\:\mathrm{4}\:\mathrm{distinct}\:\mathrm{real}\:\mathrm{roots} \\ $$$${T}_{\mathrm{1}} >\mathrm{0}\wedge\left({p}>\mathrm{0}\vee{T}_{\mathrm{3}} >\mathrm{0}\right)\:\Rightarrow\:\mathrm{2}\:\mathrm{pairs}\:\mathrm{of}\:\mathrm{conjugated}\:\mathrm{complex}\:\mathrm{roots} \\ $$$${T}_{\mathrm{1}} =\mathrm{0}\wedge\left({p}<\mathrm{0}\wedge{T}_{\mathrm{3}} <\mathrm{0}\wedge{T}_{\mathrm{2}} \neq\mathrm{0}\right)\:\Rightarrow\:\mathrm{1}\:\mathrm{real}\:\mathrm{double}\:\mathrm{and}\:\mathrm{2}\:\mathrm{real}\:\mathrm{simple}\:\mathrm{roots} \\ $$$${T}_{\mathrm{1}} =\mathrm{0}\wedge\left({T}_{\mathrm{3}} >\mathrm{0}\vee\left({p}>\mathrm{0}\wedge\left({T}_{\mathrm{3}} \neq\mathrm{0}\vee{q}\neq\mathrm{0}\right)\right)\right)\:\Rightarrow\:\mathrm{1}\:\mathrm{real}\:\mathrm{double}\:\mathrm{and}\:\mathrm{2}\:\mathrm{conjugated}\:\mathrm{complex}\:\mathrm{roots} \\ $$$${T}_{\mathrm{1}} =\mathrm{0}\wedge\left({T}_{\mathrm{2}} =\mathrm{0}\wedge{T}_{\mathrm{3}} \neq\mathrm{0}\right)\:\Rightarrow\:\mathrm{1}\:\mathrm{real}\:\mathrm{triple}\:\mathrm{and}\:\mathrm{1}\:\mathrm{real}\:\mathrm{simple}\:\mathrm{roots} \\ $$$${T}_{\mathrm{1}} =\mathrm{0}\wedge\left({T}_{\mathrm{3}} =\mathrm{0}\wedge{p}<\mathrm{0}\right)\:\Rightarrow\:\mathrm{2}\:\mathrm{real}\:\mathrm{double}\:\mathrm{roots} \\ $$$${T}_{\mathrm{1}} =\mathrm{0}\wedge\left({T}_{\mathrm{3}} =\mathrm{0}\wedge{p}>\mathrm{0}\wedge{q}=\mathrm{0}\right)\:\Rightarrow\:\mathrm{2}\:\mathrm{conjugated}\:\mathrm{complex}\:\mathrm{double}\:\mathrm{roots} \\ $$$${T}_{\mathrm{1}} =\mathrm{0}\wedge{T}_{\mathrm{2}} =\mathrm{0}\:\Rightarrow\:\mathrm{all}\:\mathrm{roots}\:\mathrm{are}\:\mathrm{equal} \\ $$

Question Number 63247    Answers: 0   Comments: 1

Prove that (√(abc)) + (√((1−a)(1−b)(1−c))) ≤ 1 for 0 ≤ a,b,c ≤ 1

$$\mathrm{Prove}\:\mathrm{that} \\ $$$$\sqrt{{abc}}\:+\:\sqrt{\left(\mathrm{1}−{a}\right)\left(\mathrm{1}−{b}\right)\left(\mathrm{1}−{c}\right)}\:\leqslant\:\mathrm{1} \\ $$$$\mathrm{for}\:\mathrm{0}\:\leqslant\:{a},{b},{c}\:\leqslant\:\mathrm{1} \\ $$

Question Number 63645    Answers: 0   Comments: 4

n integr natural prove that 5 divide n^5 −n

$${n}\:{integr}\:{natural}\:{prove}\:{that}\:\mathrm{5}\:{divide}\:{n}^{\mathrm{5}} −{n} \\ $$

Question Number 63291    Answers: 0   Comments: 3

find some of all real x such that ((4x^2 +15x+17)/(x^2 +4x+12)) = ((5x^2 +16x+18)/(2x^2 +5x+13))

$${find}\:{some}\:{of}\:{all}\:{real}\:{x}\:{such}\:{that} \\ $$$$ \\ $$$$\frac{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{15}{x}+\mathrm{17}}{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{12}}\:=\:\frac{\mathrm{5}{x}^{\mathrm{2}} +\mathrm{16}{x}+\mathrm{18}}{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{5}{x}+\mathrm{13}} \\ $$

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