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


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
${ ((((4x^2 )/(4x^2 +1)) = y)),((((4y^2 )/(4y^2 +1)) = z )),((((4z^2 )/(4z^2 +1)) = x)) :} where x,y,z ≠ 0$
${\begin{cases} \frac{4 x^{2}}{4 x^{2} + 1} = y \\ \frac{4 y^{2}}{4 y^{2} + 1} = z \\ \frac{4 z^{2}}{4 z^{2} + 1} = x \end{cases}$ $w h e r e x, y, z \neq 0$
	Answered by bemath last updated on 29/Oct/20
	$⇒ { (((1/y) = 1+(1/(4x^2 )))),(((1/z) = 1+ (1/(4y^2 )))),(((1/x) = 1+(1/(4z^2 )))) :} (1+(1/(4x^2 ))−(1/x))+(1+(1/(4y^2 ))−(1/y))+(1+(1/(4z))−(1/z)) = 0 (1−(1/(2x)))^2 +(1−(1/(2y)))^2 + (1−(1/(2z)))^2 = 0 →(1/(2x)) = (1/(2y)) = (1/(2z)) = 1 , (x,y,z) = ((1/2),(1/2),(1/2))$
	$\Rightarrow {\begin{cases} \frac{1}{y} = 1 + \frac{1}{4 x^{2}} \\ \frac{1}{z} = 1 + \frac{1}{4 y^{2}} \\ \frac{1}{x} = 1 + \frac{1}{4 z^{2}} \end{cases}$ $(1 + \frac{1}{4 x^{2}} - \frac{1}{x}) + (1 + \frac{1}{4 y^{2}} - \frac{1}{y}) + (1 + \frac{1}{4 z} - \frac{1}{z}) = 0$ ${(1 - \frac{1}{2 x})}^{2} + {(1 - \frac{1}{2 y})}^{2} + {(1 - \frac{1}{2 z})}^{2} = 0$ $\to \frac{1}{2 x} = \frac{1}{2 y} = \frac{1}{2 z} = 1, (x, y, z) = (\frac{1}{2}, \frac{1}{2}, \frac{1}{2})$
	Answered by mindispower last updated on 29/Oct/20
	$y=f(4x^2 ),f(t)=(t/(t+1)),f′(t)≥0,f increase x,y,z≥0 z=f0f(4x^2 ) x=fofof(4x^2 ), fof increase fonction 4x^2 >4y^2 ⇒f(4x^2 )>f(4y^2 )⇒y>z⇒4y^2 >4z^2 ⇒f(4y^2 )>f(4z^2 )⇔z>x z<y⇒x>y>z>x absurde x<y also⇒x>y so x=y⇒x=((4x^2 )/(4x^2 +1))⇔x(4x^2 −4x+1)=0 x(2x−1)^2 =0⇔x=0,x=(1/2) S=(x,y,z)∈{((1/2),(1/2),(1/2))}$
	$y = f (4 x^{2}), f (t) = \frac{t}{t + 1}, f^{'} (t) ⩾ 0, f i n c r e a s e$ $x, y, z ⩾ 0$ $z = f 0 f (4 x^{2})$ $x = f o f o f (4 x^{2}),$ $f o f i n c r e a s e f o n c t i o n$ $4 x^{2} > 4 y^{2} \Rightarrow f (4 x^{2}) > f (4 y^{2}) \Rightarrow y > z \Rightarrow 4 y^{2} > 4 z^{2}$ $\Rightarrow f (4 y^{2}) > f (4 z^{2}) \Leftrightarrow z > x$ $z < y \Rightarrow x > y > z > x a b s u r d e x < y a l s o \Rightarrow x > y$ $s o x = y \Rightarrow x = \frac{4 x^{2}}{4 x^{2} + 1} \Leftrightarrow x (4 x^{2} - 4 x + 1) = 0$ $x {(2 x - 1)}^{2} = 0 \Leftrightarrow x = 0, x = \frac{1}{2}$ $S = (x, y, z) \in {(\frac{1}{2}, \frac{1}{2}, \frac{1}{2})}$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum