Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 120110 by benjo_mathlover last updated on 29/Oct/20

{ ((((4x^2 )/(4x^2 +1)) = y)),((((4y^2 )/(4y^2 +1)) = z )),((((4z^2 )/(4z^2 +1)) = x)) :} where x,y,z ≠ 0

$$\begin{cases}{\frac{\mathrm{4}{x}^{\mathrm{2}} }{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{1}}\:=\:{y}}\\{\frac{\mathrm{4}{y}^{\mathrm{2}} }{\mathrm{4}{y}^{\mathrm{2}} +\mathrm{1}}\:=\:{z}\:}\\{\frac{\mathrm{4}{z}^{\mathrm{2}} }{\mathrm{4}{z}^{\mathrm{2}} +\mathrm{1}}\:=\:{x}}\end{cases} \\ $$$${where}\:{x},{y},{z}\:\neq\:\mathrm{0}\: \\ $$

Answered by bemath last updated on 29/Oct/20

⇒ { (((1/y) = 1+(1/(4x^2 )))),(((1/z) = 1+ (1/(4y^2 )))),(((1/x) = 1+(1/(4z^2 )))) :} (1+(1/(4x^2 ))−(1/x))+(1+(1/(4y^2 ))−(1/y))+(1+(1/(4z))−(1/z)) = 0 (1−(1/(2x)))^2 +(1−(1/(2y)))^2 + (1−(1/(2z)))^2 = 0 →(1/(2x)) = (1/(2y)) = (1/(2z)) = 1 , (x,y,z) = ((1/2),(1/2),(1/2))

$$\Rightarrow\:\begin{cases}{\frac{\mathrm{1}}{{y}}\:=\:\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}{x}^{\mathrm{2}} }}\\{\frac{\mathrm{1}}{{z}}\:=\:\mathrm{1}+\:\frac{\mathrm{1}}{\mathrm{4}{y}^{\mathrm{2}} }}\\{\frac{\mathrm{1}}{{x}}\:=\:\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}{z}^{\mathrm{2}} }}\end{cases} \\ $$$$\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}{x}^{\mathrm{2}} }−\frac{\mathrm{1}}{{x}}\right)+\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}{y}^{\mathrm{2}} }−\frac{\mathrm{1}}{{y}}\right)+\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}{z}}−\frac{\mathrm{1}}{{z}}\right)\:=\:\mathrm{0} \\ $$$$\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}{x}}\right)^{\mathrm{2}} +\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}{y}}\right)^{\mathrm{2}} +\:\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}{z}}\right)^{\mathrm{2}} \:=\:\mathrm{0} \\ $$$$\rightarrow\frac{\mathrm{1}}{\mathrm{2}{x}}\:=\:\frac{\mathrm{1}}{\mathrm{2}{y}}\:=\:\frac{\mathrm{1}}{\mathrm{2}{z}}\:=\:\mathrm{1}\:,\:\left({x},{y},{z}\right)\:=\:\left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$

Answered by mindispower last updated on 29/Oct/20

y=f(4x^2 ),f(t)=(t/(t+1)),f′(t)≥0,f increase x,y,z≥0 z=f0f(4x^2 ) x=fofof(4x^2 ), fof increase fonction 4x^2 >4y^2 ⇒f(4x^2 )>f(4y^2 )⇒y>z⇒4y^2 >4z^2 ⇒f(4y^2 )>f(4z^2 )⇔z>x z<y⇒x>y>z>x absurde x<y also⇒x>y so x=y⇒x=((4x^2 )/(4x^2 +1))⇔x(4x^2 −4x+1)=0 x(2x−1)^2 =0⇔x=0,x=(1/2) S=(x,y,z)∈{((1/2),(1/2),(1/2))}

$${y}={f}\left(\mathrm{4}{x}^{\mathrm{2}} \right),{f}\left({t}\right)=\frac{{t}}{{t}+\mathrm{1}},{f}'\left({t}\right)\geqslant\mathrm{0},{f}\:{increase} \\ $$$${x},{y},{z}\geqslant\mathrm{0} \\ $$$${z}={f}\mathrm{0}{f}\left(\mathrm{4}{x}^{\mathrm{2}} \right) \\ $$$${x}={fofof}\left(\mathrm{4}{x}^{\mathrm{2}} \right), \\ $$$${fof}\:{increase}\:{fonction} \\ $$$$\mathrm{4}{x}^{\mathrm{2}} >\mathrm{4}{y}^{\mathrm{2}} \Rightarrow{f}\left(\mathrm{4}{x}^{\mathrm{2}} \right)>{f}\left(\mathrm{4}{y}^{\mathrm{2}} \right)\Rightarrow{y}>{z}\Rightarrow\mathrm{4}{y}^{\mathrm{2}} >\mathrm{4}{z}^{\mathrm{2}} \\ $$$$\Rightarrow{f}\left(\mathrm{4}{y}^{\mathrm{2}} \right)>{f}\left(\mathrm{4}{z}^{\mathrm{2}} \right)\Leftrightarrow{z}>{x} \\ $$$${z}<{y}\Rightarrow{x}>{y}>{z}>{x}\:{absurde}\:{x}<{y}\:{also}\Rightarrow{x}>{y} \\ $$$${so}\:{x}={y}\Rightarrow{x}=\frac{\mathrm{4}{x}^{\mathrm{2}} }{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{1}}\Leftrightarrow{x}\left(\mathrm{4}{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{1}\right)=\mathrm{0} \\ $$$${x}\left(\mathrm{2}{x}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{0}\Leftrightarrow{x}=\mathrm{0},{x}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${S}=\left({x},{y},{z}\right)\in\left\{\left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{2}}\right)\right\} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com