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Question Number 143822 Answers: 0 Comments: 0

x^4 +ax^2 +bx+c=0 let x=(t/s) ⇒ t^4 +as^2 t^2 +bs^3 t+cs^4 =0 let t=p+h ⇒ p^4 +4hp^3 +6h^2 p^2 +4h^3 p+h^4 +as^2 (p^2 +2hp+h^2 ) +bs^3 (p+h)+cs^4 =0 ⇒ p^4 +4hp^3 +(6h^2 +as^2 )p^2 +(4h^3 +2ahs^2 +bs^3 )p +(h^4 +as^2 h^2 +bs^3 h+cs^4 )=0 let ((√2)p^2 +Ap+B)^2 =(p^2 +mp+k)^2 ⇒ 2(√2)A−2m=4h 2(√2)B+A^2 −m^2 −2k = 6h^2 +as^2 2AB−2mk=4h^3 +2ah+bs^3 B^2 −k^2 =h^4 +as^2 h^2 +bs^3 h+cs^4 let m=0 ⇒ A=(√2)h 2(√2)B=4h^2 +as^2 +2k 4h^3 +ahs^2 +2hk=4h^3 +2ah+bs^3 ⇒ (as^2 +2k−2a)h=bs^3 (4h^3 +as^2 +2k)^2 −8k^2 =8(h^4 +as^2 h^2 +bs^3 h+cs^4 ) let k=a ⇒ ah=bs ⇒ (((4b^2 h)/a^2 )+a+((2a^3 h^2 )/b^2 ))^2 −((8b^4 h^4 )/a^2 ) =8((b^4 /a^4 )+((2b^2 )/a)+c) ⇒ ((16b^4 h^2 )/a^4 )+((4a^6 h^4 )/b^4 ) +((8b^2 h)/a)+32ah^3 +((4a^4 h^2 )/b^2 ) −((8b^4 h^4 )/a^2 )=λ ⇒ (((4a^6 )/b^4 )−((8b^4 )/a^2 ))h^4 +32ah^3 +(((16b^4 )/a^4 )+((4a^4 )/b^2 ))h^2 +((8b^2 h)/a)−λ=0 ...

$$\:{x}^{\mathrm{4}} +{ax}^{\mathrm{2}} +{bx}+{c}=\mathrm{0} \\ $$$${let}\:\:{x}=\frac{{t}}{{s}} \\ $$$$\Rightarrow\:{t}^{\mathrm{4}} +{as}^{\mathrm{2}} {t}^{\mathrm{2}} +{bs}^{\mathrm{3}} {t}+{cs}^{\mathrm{4}} =\mathrm{0} \\ $$$${let}\:{t}={p}+{h}\:\:\Rightarrow \\ $$$${p}^{\mathrm{4}} +\mathrm{4}{hp}^{\mathrm{3}} +\mathrm{6}{h}^{\mathrm{2}} {p}^{\mathrm{2}} +\mathrm{4}{h}^{\mathrm{3}} {p}+{h}^{\mathrm{4}} \\ $$$$+{as}^{\mathrm{2}} \left({p}^{\mathrm{2}} +\mathrm{2}{hp}+{h}^{\mathrm{2}} \right) \\ $$$$+{bs}^{\mathrm{3}} \left({p}+{h}\right)+{cs}^{\mathrm{4}} =\mathrm{0} \\ $$$$\Rightarrow \\ $$$${p}^{\mathrm{4}} +\mathrm{4}{hp}^{\mathrm{3}} +\left(\mathrm{6}{h}^{\mathrm{2}} +{as}^{\mathrm{2}} \right){p}^{\mathrm{2}} \\ $$$$+\left(\mathrm{4}{h}^{\mathrm{3}} +\mathrm{2}{ahs}^{\mathrm{2}} +{bs}^{\mathrm{3}} \right){p} \\ $$$$+\left({h}^{\mathrm{4}} +{as}^{\mathrm{2}} {h}^{\mathrm{2}} +{bs}^{\mathrm{3}} {h}+{cs}^{\mathrm{4}} \right)=\mathrm{0} \\ $$$${let}\:\:\left(\sqrt{\mathrm{2}}{p}^{\mathrm{2}} +{Ap}+{B}\right)^{\mathrm{2}} =\left({p}^{\mathrm{2}} +{mp}+{k}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\:\mathrm{2}\sqrt{\mathrm{2}}{A}−\mathrm{2}{m}=\mathrm{4}{h} \\ $$$$\mathrm{2}\sqrt{\mathrm{2}}{B}+{A}^{\mathrm{2}} −{m}^{\mathrm{2}} −\mathrm{2}{k} \\ $$$$\:\:\:\:\:\:\:=\:\mathrm{6}{h}^{\mathrm{2}} +{as}^{\mathrm{2}} \\ $$$$\mathrm{2}{AB}−\mathrm{2}{mk}=\mathrm{4}{h}^{\mathrm{3}} +\mathrm{2}{ah}+{bs}^{\mathrm{3}} \\ $$$${B}^{\mathrm{2}} −{k}^{\mathrm{2}} ={h}^{\mathrm{4}} +{as}^{\mathrm{2}} {h}^{\mathrm{2}} +{bs}^{\mathrm{3}} {h}+{cs}^{\mathrm{4}} \\ $$$${let}\:\:{m}=\mathrm{0}\:\:\Rightarrow \\ $$$${A}=\sqrt{\mathrm{2}}{h} \\ $$$$\mathrm{2}\sqrt{\mathrm{2}}{B}=\mathrm{4}{h}^{\mathrm{2}} +{as}^{\mathrm{2}} +\mathrm{2}{k} \\ $$$$\mathrm{4}{h}^{\mathrm{3}} +{ahs}^{\mathrm{2}} +\mathrm{2}{hk}=\mathrm{4}{h}^{\mathrm{3}} +\mathrm{2}{ah}+{bs}^{\mathrm{3}} \\ $$$$\Rightarrow\:\left({as}^{\mathrm{2}} +\mathrm{2}{k}−\mathrm{2}{a}\right){h}={bs}^{\mathrm{3}} \\ $$$$\left(\mathrm{4}{h}^{\mathrm{3}} +{as}^{\mathrm{2}} +\mathrm{2}{k}\right)^{\mathrm{2}} \\ $$$$\:\:−\mathrm{8}{k}^{\mathrm{2}} =\mathrm{8}\left({h}^{\mathrm{4}} +{as}^{\mathrm{2}} {h}^{\mathrm{2}} +{bs}^{\mathrm{3}} {h}+{cs}^{\mathrm{4}} \right) \\ $$$${let}\:\:{k}={a} \\ $$$$\Rightarrow\:\:{ah}={bs} \\ $$$$\Rightarrow\:\left(\frac{\mathrm{4}{b}^{\mathrm{2}} {h}}{{a}^{\mathrm{2}} }+{a}+\frac{\mathrm{2}{a}^{\mathrm{3}} {h}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\right)^{\mathrm{2}} −\frac{\mathrm{8}{b}^{\mathrm{4}} {h}^{\mathrm{4}} }{{a}^{\mathrm{2}} } \\ $$$$\:\:=\mathrm{8}\left(\frac{{b}^{\mathrm{4}} }{{a}^{\mathrm{4}} }+\frac{\mathrm{2}{b}^{\mathrm{2}} }{{a}}+{c}\right) \\ $$$$\Rightarrow\:\frac{\mathrm{16}{b}^{\mathrm{4}} {h}^{\mathrm{2}} }{{a}^{\mathrm{4}} }+\frac{\mathrm{4}{a}^{\mathrm{6}} {h}^{\mathrm{4}} }{{b}^{\mathrm{4}} } \\ $$$$+\frac{\mathrm{8}{b}^{\mathrm{2}} {h}}{{a}}+\mathrm{32}{ah}^{\mathrm{3}} +\frac{\mathrm{4}{a}^{\mathrm{4}} {h}^{\mathrm{2}} }{{b}^{\mathrm{2}} } \\ $$$$\:\:−\frac{\mathrm{8}{b}^{\mathrm{4}} {h}^{\mathrm{4}} }{{a}^{\mathrm{2}} }=\lambda \\ $$$$\Rightarrow\:\left(\frac{\mathrm{4}{a}^{\mathrm{6}} }{{b}^{\mathrm{4}} }−\frac{\mathrm{8}{b}^{\mathrm{4}} }{{a}^{\mathrm{2}} }\right){h}^{\mathrm{4}} +\mathrm{32}{ah}^{\mathrm{3}} \\ $$$$\:\:\:+\left(\frac{\mathrm{16}{b}^{\mathrm{4}} }{{a}^{\mathrm{4}} }+\frac{\mathrm{4}{a}^{\mathrm{4}} }{{b}^{\mathrm{2}} }\right){h}^{\mathrm{2}} +\frac{\mathrm{8}{b}^{\mathrm{2}} {h}}{{a}}−\lambda=\mathrm{0} \\ $$$$... \\ $$

Question Number 144084 Answers: 1 Comments: 0

if a+(1/a)=−1 a^(1234567891011) +(1/a^(1110987654321) )=??

$${if}\:\:{a}+\frac{\mathrm{1}}{{a}}=−\mathrm{1} \\ $$$$\:{a}^{\mathrm{1234567891011}} +\frac{\mathrm{1}}{{a}^{\mathrm{1110987654321}} }=?? \\ $$$$ \\ $$

Question Number 144086 Answers: 1 Comments: 0

If sin t + cos t = (2/3) then cosec t+sec t =?

$$\mathrm{If}\:\mathrm{sin}\:\mathrm{t}\:+\:\mathrm{cos}\:\:\mathrm{t}\:=\:\frac{\mathrm{2}}{\mathrm{3}}\:\mathrm{then}\:\mathrm{cosec}\:\mathrm{t}+\mathrm{sec}\:\mathrm{t}\:=? \\ $$

Question Number 144085 Answers: 1 Comments: 1

Question Number 143814 Answers: 1 Comments: 0

∀a;b;c∈R , find all f:R→R , such that f(a)f(bc)+9≤f(ab)+5f(ac)

$$\forall{a};{b};{c}\in\mathbb{R}\:,\:{find}\:{all}\:{f}:\mathbb{R}\rightarrow\mathbb{R}\:,\:{such}\:{that} \\ $$$${f}\left({a}\right){f}\left({bc}\right)+\mathrm{9}\leqslant{f}\left({ab}\right)+\mathrm{5}{f}\left({ac}\right) \\ $$

Question Number 143812 Answers: 1 Comments: 0

log _a (ax).log _x (ax)=log _a^2 ((1/a)) a>0 , a≠1 . So x = ?

$$\:\mathrm{log}\:_{\mathrm{a}} \left(\mathrm{ax}\right).\mathrm{log}\:_{\mathrm{x}} \left(\mathrm{ax}\right)=\mathrm{log}\:_{\mathrm{a}^{\mathrm{2}} } \left(\frac{\mathrm{1}}{\mathrm{a}}\right) \\ $$$$\:\mathrm{a}>\mathrm{0}\:,\:\mathrm{a}\neq\mathrm{1}\:.\:\mathrm{So}\:\mathrm{x}\:=\:? \\ $$

Question Number 143811 Answers: 1 Comments: 0

{ ((5(log _y (x)+log _x (y))=26)),(( xy = 64)) :}then x^2 +y^2 +xy =?

$$\:\begin{cases}{\mathrm{5}\left(\mathrm{log}\:_{\mathrm{y}} \left(\mathrm{x}\right)+\mathrm{log}\:_{\mathrm{x}} \left(\mathrm{y}\right)\right)=\mathrm{26}}\\{\:\mathrm{xy}\:=\:\mathrm{64}}\end{cases}\mathrm{then} \\ $$$$\:\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} +\mathrm{xy}\:=? \\ $$

Question Number 143810 Answers: 2 Comments: 0

∫cos(cosx)dx=?

$$\:\:\:\:\int\mathrm{cos}\left(\mathrm{cosx}\right)\mathrm{dx}=? \\ $$

Question Number 144132 Answers: 2 Comments: 0

Question Number 143827 Answers: 1 Comments: 0

Question Number 143790 Answers: 0 Comments: 2

Π_(n=1) ^∞ (1+(x^3 /n^3 ))

$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\left(\mathrm{1}+\frac{{x}^{\mathrm{3}} }{{n}^{\mathrm{3}} }\right) \\ $$

Question Number 143786 Answers: 1 Comments: 1

∫_(−∞) ^( ∞) (e^(iax) /(1+x^2 ))dx how can it solve this

$$\int_{−\infty} ^{\:\infty} \frac{{e}^{{iax}} }{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:\:\:\:\:\:{how}\:{can}\:{it}\:{solve}\:{this} \\ $$

Question Number 143776 Answers: 1 Comments: 0

x×y′′−y=x^3

$$\mathrm{x}×\mathrm{y}''−\mathrm{y}=\mathrm{x}^{\mathrm{3}} \\ $$

Question Number 143775 Answers: 1 Comments: 0

x×y′′−y=x^ 3

$$\mathrm{x}×\mathrm{y}''−\mathrm{y}=\hat {\mathrm{x}3} \\ $$

Question Number 143774 Answers: 1 Comments: 2

Is this statement true or not? ∃ A∈M_3 (R) ∣ tr(A)=0 and A^2 +^t A=I_3

$$\mathrm{Is}\:\mathrm{this}\:\mathrm{statement}\:{true}\:\mathrm{or}\:\mathrm{not}? \\ $$$$\exists\:\mathrm{A}\in\mathscr{M}_{\mathrm{3}} \left(\mathbb{R}\right)\:\mid\:\mathrm{tr}\left(\mathrm{A}\right)=\mathrm{0}\:\mathrm{and}\:\mathrm{A}^{\mathrm{2}} +^{{t}} \mathrm{A}=\mathrm{I}_{\mathrm{3}} \\ $$

Question Number 143783 Answers: 4 Comments: 0

Ω :=∫_(−∞) ^( ∞) ((log(2+x^( 2) ))/(4+x^( 2) ))dx=?

$$ \\ $$$$\:\:\:\:\:\:\:\:\Omega\::=\int_{−\infty} ^{\:\infty} \frac{{log}\left(\mathrm{2}+{x}^{\:\mathrm{2}} \right)}{\mathrm{4}+{x}^{\:\mathrm{2}} }{dx}=? \\ $$$$ \\ $$

Question Number 143781 Answers: 3 Comments: 0

Question Number 143808 Answers: 2 Comments: 0

Question Number 143769 Answers: 3 Comments: 0

Question Number 143766 Answers: 1 Comments: 1

x×y′′−y=x^ 3

$${x}×{y}''−{y}=\hat {{x}}\mathrm{3} \\ $$

Question Number 143765 Answers: 1 Comments: 0

∫∫x+2dx

$$\int\int{x}+\mathrm{2}{dx} \\ $$

Question Number 143764 Answers: 1 Comments: 0

can anyone tell me,how can I bring everything in this app to the new phone. And after bringing it to the new phone, I will be able to edit everything again.

$${can}\:{anyone}\:{tell}\:{me},{how}\:{can}\:{I} \\ $$$${bring}\:{everything}\:{in}\:{this}\:{app}\:{to} \\ $$$${the}\:{new}\:{phone}. \\ $$$${And}\:{after}\:{bringing}\:{it}\:{to}\:{the}\:{new} \\ $$$${phone},\:{I}\:{will}\:{be}\:{able}\:{to}\:{edit}\:{everything} \\ $$$${again}. \\ $$

Question Number 143763 Answers: 0 Comments: 0

$$ \: \: \: \: \: \: \: \\ $$$$ \: \: \: \: \: \: \: \: \\ $$$$ \: \: \: \: \: \: \: \: \: \\ $$$$ \: \: \\ $$

Question Number 143758 Answers: 0 Comments: 1

Question Number 143755 Answers: 1 Comments: 0

Study the convergence with respect to α and β the improper integral below; ∫_0 ^∞ (dx/(x^α (lnx)^β ))

$$\mathrm{Study}\:\mathrm{the}\:\mathrm{convergence}\:\mathrm{with}\:\mathrm{respect}\:\mathrm{to} \\ $$$$\alpha\:\mathrm{and}\:\beta\:\mathrm{the}\:\mathrm{improper}\:\mathrm{integral}\:\mathrm{below}; \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{dx}}{\mathrm{x}^{\alpha} \left(\mathrm{lnx}\right)^{\beta} } \\ $$

Question Number 143751 Answers: 1 Comments: 0

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