Question and Answers Forum

AllQuestion and Answers: Page 425

Question Number 176718 Answers: 1 Comments: 1

x^3 +y^3 =z^2 x^3 +z^3 =y^2 y^3 +z^3 =x^2 obviously x=y=z=0∨(1/2) trying to totally solve it let y=px∧z=qx (p^3 +1)x^3 =q^2 x^2 (q^3 +1)x^3 =p^2 x^2 (p^3 +q^3 )x^3 =x^2 ⇒ x=0 ⇒ y=0∧z=0 ★ x=(q^2 /(p^3 +1)) x=(p^2 /(q^3 +1)) x=(1/(p^3 +q^3 )) ⇒ (p^2 /(q^3 +1))=(q^2 /(p^3 +1)) (p^2 /(q^3 +1))=(1/(p^3 +q^3 )) ========== p^5 +p^2 −q^5 −q^2 =0 p^5 +p^2 q^3 −q^3 −1=0 subtracting both p^2 (q^3 −1)+q^5 −q^3 +q^2 −1=0 (q−1)(p^2 (q^2 +q+1)+q^4 +q^3 +q+1)=0 ⇒ q=1 ⇒ p^5 +p^2 −2=0 (p−1)(p^4 +p^3 +2p+2)=0 ⇒ p=1 ⇒ x=y=z=(1/2) ★ p^4 +p^3 +2p+2=0 no useable exact solutions p≈−.975564±.528237i ⇒ x≈.33635∓.515329i ∧ y≈−.0559113±.680406i ∧ z=x ★ p≈.475564±1.18273i ⇒ x≈−.586346±.562464i ∧ y≈−.944089∓.426001i ∧ z=x ★ p^2 (q^2 +q+1)+q^4 +q^3 +q+1=0 p^2 =−(((q+1)^2 (q^2 −q+1))/(q^2 +q+1)) we can be sure that (q≠−1⇒p=0) ⇒ p^2 <0 ⇒ p=±(√((q^2 −q+1)/(q^2 +q+1)))(q+1)i ...I′ll continue later

$${x}^{\mathrm{3}} +{y}^{\mathrm{3}} ={z}^{\mathrm{2}} \\ $$$${x}^{\mathrm{3}} +{z}^{\mathrm{3}} ={y}^{\mathrm{2}} \\ $$$${y}^{\mathrm{3}} +{z}^{\mathrm{3}} ={x}^{\mathrm{2}} \\ $$$$\mathrm{obviously}\:{x}={y}={z}=\mathrm{0}\vee\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{trying}\:\mathrm{to}\:\mathrm{totally}\:\mathrm{solve}\:\mathrm{it} \\ $$$$\mathrm{let}\:{y}={px}\wedge{z}={qx} \\ $$$$\left({p}^{\mathrm{3}} +\mathrm{1}\right){x}^{\mathrm{3}} ={q}^{\mathrm{2}} {x}^{\mathrm{2}} \\ $$$$\left({q}^{\mathrm{3}} +\mathrm{1}\right){x}^{\mathrm{3}} ={p}^{\mathrm{2}} {x}^{\mathrm{2}} \\ $$$$\left({p}^{\mathrm{3}} +{q}^{\mathrm{3}} \right){x}^{\mathrm{3}} ={x}^{\mathrm{2}} \\ $$$$\Rightarrow\:{x}=\mathrm{0}\:\Rightarrow\:{y}=\mathrm{0}\wedge{z}=\mathrm{0}\:\bigstar \\ $$$${x}=\frac{{q}^{\mathrm{2}} }{{p}^{\mathrm{3}} +\mathrm{1}} \\ $$$${x}=\frac{{p}^{\mathrm{2}} }{{q}^{\mathrm{3}} +\mathrm{1}} \\ $$$${x}=\frac{\mathrm{1}}{{p}^{\mathrm{3}} +{q}^{\mathrm{3}} } \\ $$$$\Rightarrow \\ $$$$\frac{{p}^{\mathrm{2}} }{{q}^{\mathrm{3}} +\mathrm{1}}=\frac{{q}^{\mathrm{2}} }{{p}^{\mathrm{3}} +\mathrm{1}} \\ $$$$\frac{{p}^{\mathrm{2}} }{{q}^{\mathrm{3}} +\mathrm{1}}=\frac{\mathrm{1}}{{p}^{\mathrm{3}} +{q}^{\mathrm{3}} } \\ $$$$========== \\ $$$${p}^{\mathrm{5}} +{p}^{\mathrm{2}} −{q}^{\mathrm{5}} −{q}^{\mathrm{2}} =\mathrm{0} \\ $$$${p}^{\mathrm{5}} +{p}^{\mathrm{2}} {q}^{\mathrm{3}} −{q}^{\mathrm{3}} −\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{subtracting}\:\mathrm{both} \\ $$$${p}^{\mathrm{2}} \left({q}^{\mathrm{3}} −\mathrm{1}\right)+{q}^{\mathrm{5}} −{q}^{\mathrm{3}} +{q}^{\mathrm{2}} −\mathrm{1}=\mathrm{0} \\ $$$$\left({q}−\mathrm{1}\right)\left({p}^{\mathrm{2}} \left({q}^{\mathrm{2}} +{q}+\mathrm{1}\right)+{q}^{\mathrm{4}} +{q}^{\mathrm{3}} +{q}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\:\:\:\:\:\Rightarrow\:{q}=\mathrm{1}\:\Rightarrow\:{p}^{\mathrm{5}} +{p}^{\mathrm{2}} −\mathrm{2}=\mathrm{0} \\ $$$$\:\:\:\:\:\left({p}−\mathrm{1}\right)\left({p}^{\mathrm{4}} +{p}^{\mathrm{3}} +\mathrm{2}{p}+\mathrm{2}\right)=\mathrm{0} \\ $$$$\:\:\:\:\:\Rightarrow\:{p}=\mathrm{1}\:\Rightarrow\:{x}={y}={z}=\frac{\mathrm{1}}{\mathrm{2}}\:\bigstar \\ $$$$\:\:\:\:\:{p}^{\mathrm{4}} +{p}^{\mathrm{3}} +\mathrm{2}{p}+\mathrm{2}=\mathrm{0} \\ $$$$\:\:\:\:\:\mathrm{no}\:\mathrm{useable}\:\mathrm{exact}\:\mathrm{solutions} \\ $$$$\:\:\:\:\:{p}\approx−.\mathrm{975564}\pm.\mathrm{528237i} \\ $$$$\:\:\:\:\:\Rightarrow\:{x}\approx.\mathrm{33635}\mp.\mathrm{515329i} \\ $$$$\:\:\:\:\:\wedge\:{y}\approx−.\mathrm{0559113}\pm.\mathrm{680406i} \\ $$$$\:\:\:\:\:\wedge\:{z}={x}\:\bigstar \\ $$$$\:\:\:\:\:{p}\approx.\mathrm{475564}\pm\mathrm{1}.\mathrm{18273i} \\ $$$$\:\:\:\:\:\Rightarrow\:{x}\approx−.\mathrm{586346}\pm.\mathrm{562464i} \\ $$$$\:\:\:\:\:\wedge\:{y}\approx−.\mathrm{944089}\mp.\mathrm{426001i} \\ $$$$\:\:\:\:\:\wedge\:{z}={x}\:\bigstar \\ $$$${p}^{\mathrm{2}} \left({q}^{\mathrm{2}} +{q}+\mathrm{1}\right)+{q}^{\mathrm{4}} +{q}^{\mathrm{3}} +{q}+\mathrm{1}=\mathrm{0} \\ $$$${p}^{\mathrm{2}} =−\frac{\left({q}+\mathrm{1}\right)^{\mathrm{2}} \left({q}^{\mathrm{2}} −{q}+\mathrm{1}\right)}{{q}^{\mathrm{2}} +{q}+\mathrm{1}} \\ $$$$\mathrm{we}\:\mathrm{can}\:\mathrm{be}\:\mathrm{sure}\:\mathrm{that}\:\left({q}\neq−\mathrm{1}\Rightarrow{p}=\mathrm{0}\right) \\ $$$$\Rightarrow\:{p}^{\mathrm{2}} <\mathrm{0}\:\Rightarrow\:{p}=\pm\sqrt{\frac{{q}^{\mathrm{2}} −{q}+\mathrm{1}}{{q}^{\mathrm{2}} +{q}+\mathrm{1}}}\left({q}+\mathrm{1}\right)\mathrm{i} \\ $$$$...\mathrm{I}'\mathrm{ll}\:\mathrm{continue}\:\mathrm{later} \\ $$

Question Number 176716 Answers: 1 Comments: 0

Question Number 176710 Answers: 3 Comments: 0

Question Number 176708 Answers: 1 Comments: 0

a_(n+2) −3a_(n+1) +2a_n =n+3^n please find a_n

$$\:{a}_{{n}+\mathrm{2}} −\mathrm{3}{a}_{{n}+\mathrm{1}} +\mathrm{2}{a}_{{n}} ={n}+\mathrm{3}^{{n}} \: \\ $$$${please}\:{find}\:{a}_{{n}} \\ $$$$ \\ $$

Question Number 176692 Answers: 1 Comments: 1

Question Number 176684 Answers: 1 Comments: 2

Question Number 176679 Answers: 2 Comments: 0

Eeasy integral.... 𝛀 = ∫_(−∫_0 ^( ∞) e^( −x^( 2) ) dx) ^( ∫_0 ^( ∞) e^( −x^( 2) ) dx) sin^( 2) (t).ln^( 3) ( t + (√(1+t^( 2) )))dt −−−m.n−−−

$$ \\ $$$$\:\:\:\:\:\:\:\:{Eeasy}\:\:{integral}.... \\ $$$$\:\:\:\:\:\:\boldsymbol{\Omega}\:=\:\int_{−\int_{\mathrm{0}} ^{\:\infty} {e}^{\:−{x}^{\:\mathrm{2}} } {dx}} ^{\:\int_{\mathrm{0}} ^{\:\infty} {e}^{\:−{x}^{\:\mathrm{2}} } {dx}} {sin}^{\:\mathrm{2}} \left({t}\right).{ln}^{\:\mathrm{3}} \left(\:{t}\:+\:\sqrt{\mathrm{1}+{t}^{\:\mathrm{2}} }\right){dt}\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:−−−{m}.{n}−−− \\ $$

Question Number 176676 Answers: 3 Comments: 0

If x^3 +(1/x^3 )=1, prove that x^5 +(1/x^5 )=−(x^4 +(1/x^4 ))

$$\mathrm{If}\:\:\mathrm{x}^{\mathrm{3}} +\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{3}} }=\mathrm{1}, \\ $$$$\mathrm{prove}\:\mathrm{that} \\ $$$$\mathrm{x}^{\mathrm{5}} +\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{5}} }=−\left(\mathrm{x}^{\mathrm{4}} +\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{4}} }\right) \\ $$

Question Number 176672 Answers: 1 Comments: 0

Question Number 176673 Answers: 1 Comments: 0

Question Number 176668 Answers: 3 Comments: 0

Question Number 176662 Answers: 3 Comments: 0

sequence V_(n+1) −V_n =n+3^n . Find V_n .

$$\:\:{sequence}\:{V}_{{n}+\mathrm{1}} −{V}_{{n}} ={n}+\mathrm{3}^{{n}} .\:{Find}\:{V}_{{n}} . \\ $$

Question Number 176660 Answers: 1 Comments: 1

Question Number 176652 Answers: 1 Comments: 0

f(x,y)= { ((e^(1/(r^2 −1)) if r<1, where r=∥(x,y)∥)),((0 if r≥1)) :} show that f(x,y) is continuous in R^2

$$\mathrm{f}\left(\mathrm{x},\mathrm{y}\right)=\begin{cases}{\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{r}^{\mathrm{2}} −\mathrm{1}}} \:\mathrm{if}\:\mathrm{r}<\mathrm{1},\:\mathrm{where}\:\mathrm{r}=\parallel\left(\mathrm{x},\mathrm{y}\right)\parallel}\\{\mathrm{0}\:\mathrm{if}\:\mathrm{r}\geqslant\mathrm{1}}\end{cases} \\ $$$$\mathrm{show}\:\mathrm{that}\:\mathrm{f}\left(\mathrm{x},\mathrm{y}\right)\:\mathrm{is}\:\mathrm{continuous}\:\mathrm{in}\:\mathbb{R}^{\mathrm{2}} \\ $$

Question Number 176648 Answers: 2 Comments: 1

∫ (dx/(a+bcosx)) ∫ (dx/(a−bsinx))

$$\int\:\frac{{dx}}{{a}+{bcosx}}\:\:\:\: \\ $$$$ \\ $$$$\int\:\frac{{dx}}{{a}−{bsinx}} \\ $$

Question Number 176643 Answers: 0 Comments: 2

Question Number 176638 Answers: 1 Comments: 0

lim_(x→0) ((sin^2 (x)−sin (x^2 ))/(x^2 (cos^2 (x)−cos (x^2 ))))=?

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:^{\mathrm{2}} \left(\mathrm{x}\right)−\mathrm{sin}\:\left(\mathrm{x}^{\mathrm{2}} \right)}{\mathrm{x}^{\mathrm{2}} \left(\mathrm{cos}\:^{\mathrm{2}} \left(\mathrm{x}\right)−\mathrm{cos}\:\left(\mathrm{x}^{\mathrm{2}} \right)\right)}=? \\ $$

Question Number 176637 Answers: 0 Comments: 0

Question Number 176636 Answers: 2 Comments: 2

{ ((p^3 +q^3 =r^2 )),((p^3 +r^3 =q^2 )),((q^3 +r^3 =p^2 )) :} ⇒20pqr =?

$$\:\begin{cases}{{p}^{\mathrm{3}} +{q}^{\mathrm{3}} ={r}^{\mathrm{2}} }\\{{p}^{\mathrm{3}} +{r}^{\mathrm{3}} ={q}^{\mathrm{2}} }\\{{q}^{\mathrm{3}} +{r}^{\mathrm{3}} ={p}^{\mathrm{2}} }\end{cases} \\ $$$$\:\Rightarrow\mathrm{20}{pqr}\:=? \\ $$

Question Number 176610 Answers: 1 Comments: 0

Question Number 176607 Answers: 1 Comments: 0

Question Number 176603 Answers: 2 Comments: 6

look the anser

$${look}\:{the}\:{anser} \\ $$

Question Number 176595 Answers: 2 Comments: 0

Question Number 176594 Answers: 0 Comments: 1

𝛗=∫_0 ^( 1) (( ( tanh^( −1) (x))^2 )/((1+x )^( 2) )) dx = ? ≺ solution ≻ note : tanh^( −1) (x)=− (1/2) ln(((1−x)/(1+x))) 𝛗= (1/4)∫_0 ^( 1) (( ln^( 2) (((1−x)/(1+x)) ))/((1+x )^( 2) )) dx =^(((1−x)/(1+x)) = t) (1/8)∫_0 ^( 1) ln^( 2) (t )dt =(1/8_ ) { [t.ln^( 2) (t)]_0 ^( 1) −2∫_0 ^( 1) ln(t)dt} =− (1/4) ∫_0 ^( 1) ln(t)dt= (1/4) ◂ m.n ▶

$$ \\ $$$$\:\:\:\:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:\left(\:{tanh}^{\:−\mathrm{1}} \left({x}\right)\right)^{\mathrm{2}} }{\left(\mathrm{1}+{x}\:\right)^{\:\mathrm{2}} }\:{dx}\:=\:?\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\prec\:\:\:{solution}\:\:\succ \\ $$$$\:\:\:\:\:{note}\::\:\:{tanh}^{\:−\mathrm{1}} \left({x}\right)=−\:\frac{\mathrm{1}}{\mathrm{2}}\:{ln}\left(\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\right) \\ $$$$\:\:\:\:\:\boldsymbol{\phi}=\:\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:{ln}^{\:\mathrm{2}} \left(\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\:\right)}{\left(\mathrm{1}+{x}\:\right)^{\:\mathrm{2}} }\:{dx} \\ $$$$\:\:\:\:\:\:\:\overset{\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\:=\:{t}} {=}\:\frac{\mathrm{1}}{\mathrm{8}}\int_{\mathrm{0}} ^{\:\mathrm{1}} {ln}^{\:\mathrm{2}} \left({t}\:\right){dt} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{8}_{\:} }\:\left\{\:\left[{t}.{ln}^{\:\mathrm{2}} \left({t}\right)\right]_{\mathrm{0}} ^{\:\mathrm{1}} −\mathrm{2}\int_{\mathrm{0}} ^{\:\mathrm{1}} {ln}\left({t}\right){dt}\right\} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=−\:\frac{\mathrm{1}}{\mathrm{4}}\:\int_{\mathrm{0}} ^{\:\mathrm{1}} {ln}\left({t}\right){dt}=\:\frac{\mathrm{1}}{\mathrm{4}}\:\:\:\:\:\:\:\blacktriangleleft\:{m}.{n}\:\blacktriangleright\: \\ $$

Question Number 176592 Answers: 2 Comments: 0

Solve the equation: 2^x + 3^x − 4^x + 6^x − 9^x = 1

$$\mathrm{Solve}\:\mathrm{the}\:\mathrm{equation}: \\ $$$$\mathrm{2}^{\boldsymbol{\mathrm{x}}} \:+\:\mathrm{3}^{\boldsymbol{\mathrm{x}}} \:−\:\mathrm{4}^{\boldsymbol{\mathrm{x}}} \:+\:\mathrm{6}^{\boldsymbol{\mathrm{x}}} \:−\:\mathrm{9}^{\boldsymbol{\mathrm{x}}} \:=\:\mathrm{1} \\ $$

Question Number 176589 Answers: 1 Comments: 0

Pg 420 Pg 421 Pg 422 Pg 423 Pg 424 Pg 425 Pg 426 Pg 427 Pg 428 Pg 429

Contact: info@tinkutara.com