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Question Number 164027    Answers: 1   Comments: 2

Question Number 164024    Answers: 1   Comments: 0

Question Number 164019    Answers: 0   Comments: 0

x^4 +ax^3 +bx^2 +cx+d=0 ⇒ x^2 +(d/x^2 )+ax+(c/x)+b=0 say (1/x)=t ⇒ x^2 +ax+b+dt^2 +ct=0 let x^2 +ax+p=0 & dt^2 +ct+q=0 p+q=b x=−(a/2)±(√((a^2 /4)−p)) t=−(c/(2d))±(√((c^2 /(4d^2 ))−(q/d))) tx=1 ⇒ ((ac)/(4d))∓(a/2)(√((c^2 /(4d^2 ))−(q/d)))∓(c/(2d))(√((a^2 /4)−p)) +(√((a^2 /4)−p))(√((c^2 /(4d^2 ))−(q/d))) =1 let p=(a^2 /4) ⇒ ((ac)/(4d))∓(a/2)(√((c^2 /(4d^2 ))−(b/d)+(a^2 /(4d))))=1 ⇒ (c^2 /(4d^2 ))−(b/d)+(a^2 /(4d))=(4/a^2 )+(c^2 /(4d^2 ))−((2c)/(ad)) ⇒ 4a^2 b−a^4 +16d−8ac=0 if x=s+h s^4 +4hs(s^2 +h^2 )+6h^2 s^2 +h^4 +a{s^3 +3hs(s+h)+h^2 } +b{s^2 +2hs+h^2 }+c(s+h)+d=0 A=4h+a B=6h^2 +3ah+b C=4h^3 +3ah^2 +2bh+c D=h^4 +ah^3 +bh^2 +ch+d ⇒ (4h+a)^2 (8h^2 +4ah+4b−a^2 ) +16(h^4 +ah^3 +bh^2 +ch+d) =8(4h+a)(4h^3 +3ah^2 +2bh+c) .....

$$\:\:{x}^{\mathrm{4}} +{ax}^{\mathrm{3}} +{bx}^{\mathrm{2}} +{cx}+{d}=\mathrm{0} \\ $$$$\:\Rightarrow\:\:{x}^{\mathrm{2}} +\frac{{d}}{{x}^{\mathrm{2}} }+{ax}+\frac{{c}}{{x}}+{b}=\mathrm{0} \\ $$$${say}\:\:\:\frac{\mathrm{1}}{{x}}={t}\:\:\Rightarrow \\ $$$$\:\:{x}^{\mathrm{2}} +{ax}+{b}+{dt}^{\mathrm{2}} +{ct}=\mathrm{0} \\ $$$${let}\:\:\:\:{x}^{\mathrm{2}} +{ax}+{p}=\mathrm{0} \\ $$$$\&\:\:\:\:\:\:{dt}^{\mathrm{2}} +{ct}+{q}=\mathrm{0} \\ $$$$\:\:\:\:{p}+{q}={b} \\ $$$${x}=−\frac{{a}}{\mathrm{2}}\pm\sqrt{\frac{{a}^{\mathrm{2}} }{\mathrm{4}}−{p}}\:\: \\ $$$${t}=−\frac{{c}}{\mathrm{2}{d}}\pm\sqrt{\frac{{c}^{\mathrm{2}} }{\mathrm{4}{d}^{\mathrm{2}} }−\frac{{q}}{{d}}} \\ $$$${tx}=\mathrm{1}\:\:\:\:\Rightarrow \\ $$$$\frac{{ac}}{\mathrm{4}{d}}\mp\frac{{a}}{\mathrm{2}}\sqrt{\frac{{c}^{\mathrm{2}} }{\mathrm{4}{d}^{\mathrm{2}} }−\frac{{q}}{{d}}}\mp\frac{{c}}{\mathrm{2}{d}}\sqrt{\frac{{a}^{\mathrm{2}} }{\mathrm{4}}−{p}} \\ $$$$\:\:\:\:\:\:\:\:+\sqrt{\frac{{a}^{\mathrm{2}} }{\mathrm{4}}−{p}}\sqrt{\frac{{c}^{\mathrm{2}} }{\mathrm{4}{d}^{\mathrm{2}} }−\frac{{q}}{{d}}}\:=\mathrm{1} \\ $$$${let}\:\:\:{p}=\frac{{a}^{\mathrm{2}} }{\mathrm{4}}\:\:\Rightarrow \\ $$$$\:\:\:\frac{{ac}}{\mathrm{4}{d}}\mp\frac{{a}}{\mathrm{2}}\sqrt{\frac{{c}^{\mathrm{2}} }{\mathrm{4}{d}^{\mathrm{2}} }−\frac{{b}}{{d}}+\frac{{a}^{\mathrm{2}} }{\mathrm{4}{d}}}=\mathrm{1} \\ $$$$\Rightarrow\:\:\cancel{\frac{{c}^{\mathrm{2}} }{\mathrm{4}{d}^{\mathrm{2}} }}−\frac{{b}}{{d}}+\frac{{a}^{\mathrm{2}} }{\mathrm{4}{d}}=\frac{\mathrm{4}}{{a}^{\mathrm{2}} }+\cancel{\frac{{c}^{\mathrm{2}} }{\mathrm{4}{d}^{\mathrm{2}} }}−\frac{\mathrm{2}{c}}{{ad}} \\ $$$$\Rightarrow\:\:\:\mathrm{4}{a}^{\mathrm{2}} {b}−{a}^{\mathrm{4}} +\mathrm{16}{d}−\mathrm{8}{ac}=\mathrm{0} \\ $$$${if}\:{x}={s}+{h} \\ $$$${s}^{\mathrm{4}} +\mathrm{4}{hs}\left({s}^{\mathrm{2}} +{h}^{\mathrm{2}} \right)+\mathrm{6}{h}^{\mathrm{2}} {s}^{\mathrm{2}} +{h}^{\mathrm{4}} \\ $$$$+{a}\left\{{s}^{\mathrm{3}} +\mathrm{3}{hs}\left({s}+{h}\right)+{h}^{\mathrm{2}} \right\} \\ $$$$\:\:+{b}\left\{{s}^{\mathrm{2}} +\mathrm{2}{hs}+{h}^{\mathrm{2}} \right\}+{c}\left({s}+{h}\right)+{d}=\mathrm{0} \\ $$$${A}=\mathrm{4}{h}+{a} \\ $$$${B}=\mathrm{6}{h}^{\mathrm{2}} +\mathrm{3}{ah}+{b} \\ $$$${C}=\mathrm{4}{h}^{\mathrm{3}} +\mathrm{3}{ah}^{\mathrm{2}} +\mathrm{2}{bh}+{c} \\ $$$${D}={h}^{\mathrm{4}} +{ah}^{\mathrm{3}} +{bh}^{\mathrm{2}} +{ch}+{d} \\ $$$$\Rightarrow \\ $$$$\left(\mathrm{4}{h}+{a}\right)^{\mathrm{2}} \left(\mathrm{8}{h}^{\mathrm{2}} +\mathrm{4}{ah}+\mathrm{4}{b}−{a}^{\mathrm{2}} \right) \\ $$$$+\mathrm{16}\left({h}^{\mathrm{4}} +{ah}^{\mathrm{3}} +{bh}^{\mathrm{2}} +{ch}+{d}\right) \\ $$$$\:\:\:\:\:\:=\mathrm{8}\left(\mathrm{4}{h}+{a}\right)\left(\mathrm{4}{h}^{\mathrm{3}} +\mathrm{3}{ah}^{\mathrm{2}} +\mathrm{2}{bh}+{c}\right) \\ $$$$..... \\ $$$$\:\:\:\:\:\: \\ $$

Question Number 164018    Answers: 1   Comments: 0

Question Number 164028    Answers: 1   Comments: 0

Question Number 164010    Answers: 0   Comments: 3

Question Number 164009    Answers: 2   Comments: 0

Question Number 164471    Answers: 1   Comments: 0

une primitive de ln(1−x^2 )dx puis la convergence de ∫_0 ^1 ln(1−x^2 )dx

$$\:{une}\:{primitive}\:{de}\:{ln}\left(\mathrm{1}−{x}^{\mathrm{2}} \right){dx} \\ $$$${puis}\:{la}\:{convergence}\:{de}\:\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\mathrm{1}−{x}^{\mathrm{2}} \right){dx} \\ $$

Question Number 164001    Answers: 1   Comments: 1

Question Number 163998    Answers: 0   Comments: 0

Hi... Calculate: ∫−(λ/(4πε_0 ))[((√(1+(πu/2)^2 ))/((1+u^2 )^(3/2) ))]du

$${Hi}... \\ $$$${Calculate}: \\ $$$$\int−\frac{\lambda}{\mathrm{4}\pi\varepsilon_{\mathrm{0}} }\left[\frac{\sqrt{\mathrm{1}+\left(\pi{u}/\mathrm{2}\right)^{\mathrm{2}} }}{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)^{\mathrm{3}/\mathrm{2}} }\right]{du} \\ $$

Question Number 163996    Answers: 1   Comments: 0

{ ((9^((a+1)/b) =125)),((5^(b/a) =3)) :} then faind the volve of ((25^b )/(2a^2 ))=?

$$\begin{cases}{\mathrm{9}^{\frac{{a}+\mathrm{1}}{{b}}} =\mathrm{125}}\\{\mathrm{5}^{\frac{{b}}{{a}}} =\mathrm{3}}\end{cases} \\ $$$${then}\:\:{faind}\:{the}\:{volve}\:{of}\:\:\frac{\mathrm{25}^{{b}} }{\mathrm{2}{a}^{\mathrm{2}} }=? \\ $$

Question Number 163993    Answers: 0   Comments: 5

Question Number 163988    Answers: 0   Comments: 1

if here are 20 apple in a box, how many are the significant figures in it?

$${if}\:{here}\:{are}\:\mathrm{20}\:{apple}\:{in}\:{a}\:{box},\:{how}\:{many}\: \\ $$$${are}\:{the}\:{significant}\:{figures}\:{in}\:{it}? \\ $$

Question Number 163985    Answers: 0   Comments: 0

Question Number 163986    Answers: 1   Comments: 0

((1000))^(1/3) ln^(80) (ln(8/5)−ln((8e)/5))^(80) is simplified form equl to=?

$$\sqrt[{\mathrm{3}}]{\mathrm{1000}}{ln}^{\mathrm{80}} \left({ln}\frac{\mathrm{8}}{\mathrm{5}}−{ln}\frac{\mathrm{8}{e}}{\mathrm{5}}\right)^{\mathrm{80}} \\ $$$${is}\:\:\:\:{simplified}\:\:{form}\:\:{equl}\:\:{to}=? \\ $$$$ \\ $$

Question Number 163978    Answers: 0   Comments: 5

Question Number 163976    Answers: 0   Comments: 0

plot the single-side and double-side line spectrum of g(t) g(t)=2sin^2 (10^3 πt)−10sin(10^3 πt−(π/6))

$${plot}\:{the}\:{single}-{side}\:{and}\:{double}-{side} \\ $$$${line}\:{spectrum}\:{of}\:{g}\left({t}\right) \\ $$$${g}\left({t}\right)=\mathrm{2}{sin}^{\mathrm{2}} \left(\mathrm{10}^{\mathrm{3}} \pi{t}\right)−\mathrm{10}{sin}\left(\mathrm{10}^{\mathrm{3}} \pi{t}−\frac{\pi}{\mathrm{6}}\right) \\ $$

Question Number 163974    Answers: 1   Comments: 1

is light a matter?

$${is}\:{light}\:{a}\:{matter}? \\ $$

Question Number 163972    Answers: 0   Comments: 1

Question Number 163968    Answers: 1   Comments: 0

(−2^2 )^3 =?

$$\left(−\mathrm{2}^{\mathrm{2}} \right)^{\mathrm{3}} =? \\ $$

Question Number 163966    Answers: 1   Comments: 0

lim_(x→(π/2)) (((√(2tan^2 x+3))−(√(2tan^2 x+10)))/(cot x)) =?

$$\:\:\:\:\:\:\underset{{x}\rightarrow\frac{\pi}{\mathrm{2}}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{2tan}\:^{\mathrm{2}} \mathrm{x}+\mathrm{3}}−\sqrt{\mathrm{2tan}\:^{\mathrm{2}} \mathrm{x}+\mathrm{10}}}{\mathrm{cot}\:\mathrm{x}}\:=? \\ $$

Question Number 163961    Answers: 1   Comments: 0

lim_(x→1^− ) (e^(1/(x^2 −1)) /(x−1))=?

$$\underset{\mathrm{x}\rightarrow\mathrm{1}^{−} } {\mathrm{lim}}\:\frac{\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} −\mathrm{1}}} }{\mathrm{x}−\mathrm{1}}=? \\ $$

Question Number 163954    Answers: 4   Comments: 0

Question Number 163960    Answers: 1   Comments: 0

lim_(x→0) (((√(2+7cot^2 x))−(√(7+7cot^2 x)))/(tan x)) =?

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{2}+\mathrm{7cot}\:^{\mathrm{2}} \mathrm{x}}−\sqrt{\mathrm{7}+\mathrm{7cot}\:^{\mathrm{2}} \mathrm{x}}}{\mathrm{tan}\:\mathrm{x}}\:=? \\ $$

Question Number 163959    Answers: 0   Comments: 0

Question Number 163958    Answers: 0   Comments: 0

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