Barbeau - polynomials (PBM)

Tema 4.7 p. 150

5. For which values of \(a\), \(b\), \(c\) does the equation \( \displaystyle\sqrt{x+a\sqrt{x}+b } + \sqrt{x} =c \) have infinitely many real zeros.
\[ \sqrt{x+a\sqrt{x}+b } + \sqrt{x} =c \] \[ \sqrt{x+a\sqrt{x}+b } = c - \sqrt{x} \] \[ \left(\sqrt{x+a\sqrt{x}+b }\right)^2 = \left(c - \sqrt{x}\right)^2 \] \[ x+a\sqrt{x}+b = c^2 - 2c\sqrt{x} + \left(\sqrt{x}\right)^2 \] \[ x+a\sqrt{x}+b = c^2 - 2c\sqrt{x} + x \] \[ a\sqrt{x}+b = c^2 - 2c\sqrt{x} \] \[ \left(a+2c\right)\sqrt{x} = c^2 - b \] \[ \sqrt{x} = \frac{c^2 - b}{a+2c} \] \[ x = \left(\frac{c^2 - b}{a+2c}\right)^2 \]

6. Solve for \(k\) real \( \displaystyle\sqrt{x+\sqrt{2x-1}} + \sqrt{x-\sqrt{2x-1}} = k \)
\[ \left(\sqrt{x+\sqrt{2x-1}} + \sqrt{x-\sqrt{2x-1}}\right)^2 = k^2 \] \[ \left(\sqrt{x+\sqrt{2x-1}}\right)^2 + 2\sqrt{x+\sqrt{2x-1}}\sqrt{x-\sqrt{2x-1}} + \left(\sqrt{x-\sqrt{2x-1}}\right)^2 = k^2 \] \[ x+\sqrt{2x-1} + 2\sqrt{\left(x+\sqrt{2x-1}\right)\left(x-\sqrt{2x-1}\right)} + x-\sqrt{2x-1} = k^2 \] \[ 2x + 2\sqrt{x^2-\left(2x-1\right)} = k^2 \] \[ 2x + 2\sqrt{x^2-2x+1} = k^2 \] \[ 2x + 2\sqrt{\left(x-1\right)^2} = k^2 \] \[ 2x + 2\left(x-1\right) = k^2 \] \[ 2x + 2x - 2 = k^2 \] \[ 4x = k^2 + 2\] \[ x = \frac{k^2 + 2}{4}\]

Tema 4.7 p. 151

7. Solve    \( \left\{ \eqalign{x^2 - (y-z)^2 &= a^2 \\ y^2 - (z-x)^2 &= b^2 \\ z^2 - (x-y)^2 &= c^2 } \right. \)    in which \(a\), \(b\), \(c\) area constants different from zero.


Si    \( \left\{ \eqalign{x^2 - (y-z)^2 &= a^2 \\ y^2 - (z-x)^2 &= b^2 \\ z^2 - (x-y)^2 &= c^2 } \right. \)    podem fer    \( \left\{ \eqalign{(x+y-z)\cdot(x-y+z) &= a^2 \\ (y+z-x)\cdot(y-z+x) &= b^2 \\ (z+x-y)\cdot(z-x+y) &= c^2 } \right. \)    llavors    \( \left\{ \eqalign{(x+y-z)\cdot(x-y+z) &= a^2 \\ (-x+y+z)\cdot(x+y-z) &= b^2 \\ (x-y+z)\cdot(-x+y+z) &= c^2 } \right. \)
Podem traduir el problema en    \( \left\{ \eqalign{x+y-z &= u \\ x-y+z &= v \\ -x+y+z &= w } \right. \) amb les solucions de    \( \left\{ \eqalign{u\cdot v &= a^2 \\ w\cdot u &= b^2 \\ v\cdot w &= c^2 } \right. \)
Primer \( \left\{ \eqalign{u\cdot v &= a^2 \\ w\cdot u &= b^2 \\ v\cdot w &= c^2 } \right. \)       \( \left\{ \eqalign{ u\cdot v &= a^2 \\ w\cdot u &= b^2 \\ w &= \frac{c^2}{v} } \right.\)    \( \left\{ \eqalign{ u\cdot v &= a^2 \\ v &= \frac{c^2}{b^2}\cdot u \\ w &= \frac{c^2}{v} } \right.\)    \( \left\{ \eqalign{ u^2 &= \frac{a^2b^2}{c^2} \\ v &= \frac{c^2}{b^2}\cdot u \\ w &= \frac{c^2}{v} } \right.\)    \( \left\{ \eqalign{ u^2 &= \frac{a^2b^2}{c^2} \\ v &= \frac{c^2}{b^2}\cdot u \\ w &= \frac{b^2}{u} } \right.\)    llavors       \( \left\{ \eqalign{ u_1 &= +\frac{a\cdot b}{c} \\ v_1 &= +\frac{a\cdot c}{b} \\ w_1 &= +\frac{b\cdot c}{a} } \right.\)    \( \left\{ \eqalign{ u_2 &= -\frac{a\cdot b}{c} \\ v_2 &= -\frac{a\cdot c}{b} \\ w_2 &= -\frac{b\cdot c}{a} } \right.\)
Segon \( \left\{ \eqalign{x+y-z &= u \\ x-y+z &= v \\ -x+y+z &= w } \right. \)       \( \eqalign{\mbox{1a + 2a eq.}\\ \mbox{1a + 3a eq.}\\ \mbox{2a + 3a eq.}} \)    \( \left\{ \eqalign{ 2x&=u+v \\ 2y&=u+w \\ 2z &= u+w } \right. \)    \( \left\{ \eqalign{ x&=\frac{u+v}{2} \\ y&=\frac{u+w}{2} \\ z&=\frac{u+w}{2} } \right. \)    \( \left\{ \eqalign{ x&=\pm\frac{a}{2}\left(\frac{b}{c}+\frac{c}{b}\right) \\ y&=\pm\frac{b}{2}\left(\frac{c}{d}+\frac{d}{c}\right) \\ z&=\pm\frac{c}{2}\left(\frac{a}{b}+\frac{b}{a}\right) } \right. \)    \( \left\{ \eqalign{ x&=\pm \frac{ a\left(b^2+c^2\right) }{2bc} \\ y&=\pm \frac{ b\left(c^2+d^2\right) }{2cd} \\ z&=\pm \frac{ c\left(a^2+b^2\right) }{2ab} } \right. \)
\( \left\{ \eqalign{ x&=\pm\frac{a}{2}\left(\frac{b}{c}+\frac{c}{b}\right) \\ y&=\pm\frac{b}{2}\left(\frac{c}{d}+\frac{d}{c}\right) \\ z&=\pm\frac{c}{2}\left(\frac{a}{b}+\frac{b}{a}\right) } \right. \) \( \left\{ \eqalign{ x&=\pm \frac{ a\left(b^2+c^2\right) }{2bc} \\ y&=\pm \frac{ b\left(c^2+d^2\right) }{2cd} \\ z&=\pm \frac{ c\left(a^2+b^2\right) }{2ab} } \right. \)
10. Solve for real \(x\):    \( \displaystyle\frac{x-\sqrt{x^2-1}}{\sqrt{ x+\sqrt{x^2-1} } } = \sqrt[4]{x^2-1}\left[\sqrt{x^2+x} - \sqrt{x^2-x}\right] \)
Malgrat l′aparatositat de l′equació es resol aplicant els metodes usuals. Primer simplifiquem el membre esquerra racionalitzant:
\[ \eqalign{ \frac{x-\sqrt{x^2-1}}{\sqrt{ x+\sqrt{x^2-1} } } &= \frac{x-\sqrt{x^2-1}}{\sqrt{ x+\sqrt{x^2-1} } }\cdot\displaystyle\frac{\sqrt{ x-\sqrt{x^2-1} }}{\sqrt{ x-\sqrt{x^2-1} } } \\ &= \frac{\left(x-\sqrt{x^2-1}\right)^{\frac{3}{2}}}{\sqrt{ x^2-\left(\sqrt{x^2-1}\right)^2 }} \\ &= \frac{\left(x-\sqrt{x^2-1}\right)^{\frac{3}{2}}}{\sqrt{ x^2-x^2+1}} \\ &= \frac{\left(x-\sqrt{x^2-1}\right)^{\frac{3}{2}}}{\sqrt{+1}} \\ &= \left(x-\sqrt{x^2-1}\right)^{\frac{3}{2}} } \] La nostra equació s′ha reduit:
\[ \left(x-\sqrt{x^2-1}\right)^{\frac{3}{2}} = \sqrt[4]{x^2-1}\left[\sqrt{x^2+x} - \sqrt{x^2-x}\right] \] Si, seguint el metode usual l′elevem al quadrat quedarà com:
\[ \eqalign{ \left(x-\sqrt{x^2-1}\right)^3 &= \sqrt{x^2-1}\left[\sqrt{x^2+x} - \sqrt{x^2-x}\right]^2 \\ &= \sqrt{x^2-1}\left[x^2+x + x^2-x - 2\sqrt{x^2+x}\sqrt{x^2-x}\right] \\ &= \sqrt{x^2-1}\left[ 2x^2 - 2\sqrt{(x^2+x)(x^2-x)}\right] \\ &= \sqrt{x^2-1}\left[ 2x^2 - 2\sqrt{x^2(x+1)(x-1)}\right] \\ &= \sqrt{x^2-1}\left[ 2x^2 - 2x\sqrt{x^2-1}\right] \\ &= 2x\sqrt{x^2-1}\left[ x -\sqrt{x^2-1}\right] \\ }\] L′equació s′ha convertit
\[ \left(x-\sqrt{x^2-1}\right)^3 = 2x\sqrt{x^2-1}\left( x - \sqrt{x^2-1} \right) \] simplificant per \( \left(x-\sqrt{x^2-1}\right) \) ens queda
\[ \left(x-\sqrt{x^2-1}\right)^2 = 2x\sqrt{x^2-1} \] així \[ x^2+x^2-1-2x\sqrt{x^2-1} = 2x\sqrt{x^2-1} \] \[ 2x^2-1 = 4x\sqrt{x^2-1} \] tornem a elevar al quadrat \[ 4x^4-4x^2+1 = 16x^2(x^2-1) \] \[ 4x^4-4x^2+1 = 16x^4-16x^2 \] \[ 12x^4-12x^2-1 = 0 \] \( \Delta = (-12)^2-4\cdot12\cdot(-1)=144+48=192 \) \[x^2 = \frac{-(-12)\pm\sqrt{192}}{2\cdot12} = \frac{12\pm\sqrt{8^2\cdot3}}{24} = \frac{12\pm8\sqrt{3}}{24} = \frac{3\pm2\sqrt{3}}{6} = \frac{1}{2}\pm\frac{\sqrt{3}}{3} = \frac{1}{2}\pm\frac{1}{\sqrt{3}} \] Les solucions      \(x = \pm\sqrt{\displaystyle\frac{1}{2}-\frac{1}{\sqrt{3}} }\)      son imaginaries pures.
Les solucions reals son:      \(x = \pm\sqrt{\displaystyle\frac{1}{2}+\frac{1}{\sqrt{3}} }\) L′unica que fa positiva l′expressió \( \left(x-\sqrt{x^2-1}\right) \) és \[x = \sqrt{\frac{1}{2}+\frac{1}{\sqrt{3}} }\]


\( 11. \left\{ \eqalign{ x^2+y^2&= 13 \\ x^3+y^3&=35 } \right. \)


\[ \left\{ \eqalign{ (x+y)^2-2xy&= 13 \\ (x+y)^3-3xy(x+y)&=35 } \right. \] \( \left\{ \eqalign{ x+y&= u \\ xy &= v } \right. \)    \(y=u-x\)    \(x(u-x)\)    \(x^2-ux+v=0\)    \( \eqalign{ x=\frac{u\pm\sqrt{u^2-4v}}{2}\\ y=\frac{u\mp\sqrt{u^2-4v}}{2} } \)    \[ \left\{ \eqalign{ u^2-2v&= 13 \\ u^3-3uv&=35 } \right. \]    \(v=\displaystyle\frac{u^2-13}{2}\)    \(u^3-\displaystyle\frac{3}{2}u(u^2-13)-35=0\)    \(2u^3-3u(u^2-13)-70=0\)    \(2u^3 - 3u^3 + 39u - 70=0\)    \(- u^3 + 39u - 70=0\)    \(u^3 - 39u + 70=0\)
5 1 0 -39 70
5 25 -70
2 1 5 -14 0
2 +14
-7 1 7 0
-7
1 0
\(u^3 - 39u + 70=(u-2)\cdot(u-5)\cdot(u+7)=0\)

\( \left\{ \eqalign{ \displaystyle\frac{x}{a} + \frac{b}{y} + \frac{c}{z} &= 1\\ \displaystyle\frac{a}{x} + \frac{y}{b} + \frac{c}{z} &= 1\\ \displaystyle\frac{a}{x} + \frac{b}{y} + \frac{z}{c} &= 1 } \right. \)


(page 153) 34. \( \left\{ \eqalign{ \displaystyle y\cdot z &= a\cdot\left(y+z\right) \\ \displaystyle x\cdot z &= b\cdot\left(x+z\right) \\ \displaystyle x\cdot y &= c\cdot\left(x+y\right) } \right. \)


\( \left\{ \eqalign{ \displaystyle y\cdot z &= a\cdot\left(y+z\right) \\ \displaystyle x\cdot z &= b\cdot\left(x+z\right) \\ \displaystyle x\cdot y &= c\cdot\left(x+y\right) } \right. \)     \( \left\{ \eqalign{ \displaystyle y\cdot z &= a\cdot y + a\cdot z \\ \displaystyle x\cdot z &= b\cdot x + b\cdot z \\ \displaystyle x\cdot y &= c\cdot x + c\cdot y } \right. \)     \( \left\{ \eqalign{ \displaystyle y\cdot\left( z-a \right)&= a\cdot z \\ \displaystyle x\cdot\left( z-b \right)&= b\cdot z \\ \displaystyle y\cdot\left( x-c \right)&= c\cdot x } \right. \)     \( \left\{ \eqalign{ y&= \displaystyle\frac{a\cdot z}{z-a} \\ x&= \displaystyle\frac{b\cdot z}{z-b} \\ y&= \displaystyle\frac{c\cdot x}{x-c} } \right. \)
Igualant la 1a i la 3a     \( \displaystyle\frac{a\cdot z}{z-a} = \displaystyle\frac{c\cdot x}{x-c} \)     i substituint la segona     \( \displaystyle\frac{a\cdot z}{z-a} = \displaystyle\frac{c\cdot \displaystyle\frac{b\cdot z}{z-b}}{\displaystyle\frac{b\cdot z}{z-b}-c} \)
\( \displaystyle\frac{a\cdot z}{z-a} = \displaystyle\frac{c\cdot \displaystyle\frac{b\cdot z}{z-b}}{\displaystyle\frac{b\cdot z}{z-b}-c} = \displaystyle\frac{\displaystyle\frac{b\cdot c\cdot z}{z-b}}{\displaystyle\frac{b\cdot z - c\cdot z +b\cdot c}{z-b}} \)         \( \displaystyle\frac{a\cdot z}{z-a} = \displaystyle\frac{b\cdot c\cdot z}{( b - c)\cdot z +b\cdot c} \) \[ a\cdot z(( b - c)\cdot z +b\cdot c) = b\cdot c\cdot z\cdot (z-a) \] \[ a\cdot ( b - c)\cdot z^2 + a\cdot b\cdot c\cdot z = b\cdot c\cdot z^2 - a\cdot b\cdot c\cdot z \] \[ (a\cdot b - a\cdot c - b\cdot c)\cdot z^2 + 2\cdot a\cdot b\cdot c\cdot z = 0 \] \[ z\cdot \left(\left(a\cdot b - a\cdot c - b\cdot c\right)\cdot z + 2\cdot a\cdot b\cdot c \right) = 0 \] Hi ha una solució trivial que es    \(z=0\)    amb la que    \(y=0\)    i    \(x=0\)    i l˜altre \[z = -\frac{2\cdot a\cdot b\cdot c}{a\cdot b - a\cdot c - b\cdot c}\] \[ y = \displaystyle\frac{a\cdot z}{z-a} = -\displaystyle\frac{a\cdot\left(-\displaystyle\frac{2\cdot a\cdot b\cdot c}{a\cdot b - a\cdot c - b\cdot c}\right)}{-\displaystyle\frac{2\cdot a\cdot b\cdot c}{a\cdot b - a\cdot c - b\cdot c}-a} = \displaystyle\frac{\displaystyle\frac{2\cdot a\cdot b\cdot c}{a\cdot b - a\cdot c - b\cdot c}}{\displaystyle\frac{2\cdot b\cdot c}{a\cdot b - a\cdot c - b\cdot c}+1} = \displaystyle\frac{\displaystyle\frac{2\cdot a\cdot b\cdot c}{a\cdot b - a\cdot c - b\cdot c}}{\displaystyle\frac{2\cdot b\cdot c+a\cdot b - a\cdot c - b\cdot c}{a\cdot b - a\cdot c - b\cdot c}} \] llavors \[y = \displaystyle\frac{2\cdot a\cdot b\cdot c}{a\cdot b - a\cdot c + b\cdot c} \] \[ x= \displaystyle\frac{b\cdot z}{z-b} = -\displaystyle\frac{b\cdot\left(-\displaystyle\frac{2\cdot a\cdot b\cdot c}{a\cdot b - a\cdot c - b\cdot c}\right)}{-\displaystyle\frac{2\cdot a\cdot b\cdot c}{a\cdot b - a\cdot c - b\cdot c}-b} = \displaystyle\frac{\displaystyle\frac{2\cdot a\cdot b\cdot c}{a\cdot b - a\cdot c - b\cdot c}}{\displaystyle\frac{2\cdot a\cdot c}{a\cdot b - a\cdot c - b\cdot c}+1} = \displaystyle\frac{\displaystyle\frac{2\cdot a\cdot b\cdot c}{a\cdot b - a\cdot c - b\cdot c}}{\displaystyle\frac{2\cdot a\cdot c+a\cdot b - a\cdot c - b\cdot c}{a\cdot b - a\cdot c - b\cdot c}} \] llavors \[x = \displaystyle\frac{2\cdot a\cdot b\cdot c}{a\cdot b + a\cdot c - b\cdot c} \]
\(x = 0\)      \(y = 0\)      \(z = 0\)
\(x = \displaystyle\frac{2\cdot a\cdot b\cdot c}{ a\cdot b + a\cdot c - b\cdot c}\)      \(y = \displaystyle\frac{2\cdot a\cdot b\cdot c}{ a\cdot b - a\cdot c + b\cdot c}\)      \(z = \displaystyle\frac{2\cdot a\cdot b\cdot c}{-a\cdot b + a\cdot c + b\cdot c}\)


(page 236) 47. Suppose that \(x+y+z=x^{-1}+y^{-1}+z^{-1}=0\). Show that \[\frac{x^{6}+y^{6}+z^{6}}{x^{3}+y^{3}+z^{3}}=xyz.\]
Tenim que \(x+y+z=0\) per tant podem intentar \[ \eqalign{ x^{3}+y^{3}+z^{3}= x^{3}+y^{3}+(-(x+y))^{3}&=x^{3}+y^{3}-(x^{3}+3x^{2}y+3xy^{2}+y^{3}) \\ &=-3x^{2}y-3xy^{2} \\ &=-3xy(x+y) \\ &=-3xy(-z) \\ &=3xyz } \]
Per altra part \(x^{-1}+y^{-1}+z^{-1}=0\) que és equivalent a \(y\cdot z + x\cdot z + x\cdot y =0\)
\[ xyz\cdot \left(x^{-1}+y^{-1}+z^{-1}\right)= xyz\cdot \left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)= \left(\frac{xyz}{x}+\frac{xyz}{y}+\frac{xyz}{z}\right)= y\cdot z + x\cdot z + x\cdot y =0 \]
Aquesta agrupació fa recordar el quadrat d'un trinomi: \((x+y+z)^2=x^2+y^2+z^2+2xy + 2xz + 2yz\)
Per tant \[ \eqalign{ (x+y+z)^2&=x^2+y^2+z^2+2\cdot(xy + xz + yz) \\ (0)^2&=x^2+y^2+z^2+2\cdot(0) } \]
Resumint, de moment, hem trobat \[ \eqalign{ x+y+z&=0 \\ x^2+y^2+z^2&=0\\ x^{3}+y^{3}+z^{3}&=3xyz } \]
Repetint l'esquema
\[ \eqalign{ x^{6}+y^{6}+z^{6}= x^{6}+y^{6}+(z^2)^{3}&=x^{6}+y^{6}+(-(x^{2}+y^{2}))^{3} \\ &=x^{6}+y^{6}-(x^{6}+3x^{4}y^{2}+3x^{2}y^{4}+y^{6}) \\ &=-3x^{4}y^{2}-3x^{2}y^{4} \\ &=-3x^{2}y^{2}(x^{2}+y^{2}) \\ &=-3x^{2}y^{2}(-z^{2}) \\ &=3x^{2}y^{2}z^{2} } \]
Finalment es fácil trobar
\( \eqalign{ x^{6}+y^{6}+z^{6} & =3x^{2}y^{2}z^{2}\\ x^{3}+y^{3}+z^{3}&=3xyz } \)          \( \frac{\displaystyle x^{6}+y^{6}+z^{6}}{\displaystyle x^{3}+y^{3}+z^{3}}=xyz \)




(page 237) 54. Show that \[ 2\left[ (x-y)(x-z) + (y-z)(y-x) + (z-x)(z-y) \right] \] Can be expressed as the sum of three squares.
Si fem els canvis \[ \eqalign{ x-y=&u \\ x-z=&v \\ y-z=&w } \]
l'expressió pren una forma més reconeixible: \[ \eqalign{ (x-y)(x-z) + (y-z)(y-x) + (z-x)(z-y) =& u\cdot v+ w\cdot (-u) + (-v)\cdot (-w)\\ =& u\cdot v- w\cdot u + v\cdot w } \]
Aquesta agrupació fa recordar el quadrat d'un cert trinomi: \[ \eqalign{ (u+v+w)^2&=u^2+v^2+w^2+2uv + 2uw + 2vw\\ (u-v+w)^2&=u^2+v^2+w^2-2uv + 2uw - 2vw } \]
és important veure que \[ u-v+w = (x-y) - (x-z) + (y-z) = x-y-x+z+y-z=0 \]
Llavors \[ \eqalign{ (0)^2&=u^2+v^2+w^2-2uv + 2uw - 2vw\\ 2uv - 2uw + 2vw &=u^2+v^2+w^2 } \] Finalment
\[ 2\left[ (x-y)(x-z) + (y-z)(y-x) + (z-x)(z-y) \right]= (x-y)^2 + (x-z)^2 + (y-z)^2 \]




(page 235) 41. Suppose that \[ \frac{a^2+b^2-c^2-d^2}{a-b+c-d} = \frac{a^2-b^2-c^2+d^2}{a+b+c+d} \] Show that \[ \frac{ab-cd}{a-b+c-d} = \frac{bc-ad}{a+b+c+d} \] Find a set of integers a, b, c, d which satisfy the previous equations.
\[ \eqalign{u^2+v^2 &= (u-v)^2 +2uv\\ s^2-t^2 &= (s+t)(s-t)} \]
\[ \eqalign{ a^2+b^2-c^2-d^2 &= (a-b)^2 +2ab - \left[(c-d)^2+2cd\right] \\ &= (a-b)^2 - (c-d)^2 +2ab-2cd \\ &= (a-b+c-d)(a-b - c+d) +2(ab-cd) } \] Pertant \[ \frac{a^2+b^2-c^2-d^2}{a-b+c-d} = a-b - c+d + 2\frac{ab-cd}{a-b+c-d} \] També
\[ \eqalign{ u^2+v^2 &= (u+v)^2 -2uv\\ s^2-t^2 &= (s+t)(s-t) } \]
\[ \eqalign{ a^2-b^2-c^2+d^2 &= a^2+d^2-b^2-c^2 \\ &= (a+d)^2 -2ad - \left[(b+c)^2-2bc\right] \\ &= (a+d)^2 - (b+c)^2 +2bc-2ad \\ &= (a+b + c+d)(a+d-b-c) +2(bc-ad) \\ &= (a+b + c+d)(a-b-c+d) +2(bc-ad) } \] Pertant \[ \frac{a^2-b^2-c^2+d^2}{a+b+c+d} = a-b-c+d + 2\frac{bc-ad}{a+b+c+d} \] Finalment
\[ \frac{a^2+b^2-c^2-d^2}{a-b+c-d} = \frac{a^2-b^2-c^2+d^2}{a+b+c+d} \Leftrightarrow \frac{ab-cd}{a-b+c-d} = \frac{bc-ad}{a+b+c+d} \]










(page 237) 56. Reduce to lowest terms \[ \frac{(ab-x^2)^2+(ax+bx-2x^2)(ax+abx-2ab)}{(ab+x^2)^2-x^2(a+b)^2} \]
\[ \eqalign{ (ab-x^2)^2+(ax+bx-2x^2)(ax+bx-2ab) &= (x^2-ab)^2 + (ax+bx)^2 + (ax+bx)(-2ab-2x^2) + 4abx^2 \\ &= (x^2-ab)^2 + (ax+bx)^2 - 2(ax+bx)(x^2+ab) + 4abx^2 \\ &= x^4 -2abx^2 + (ab)^2 + (ax+bx)^2 - 2(ax+bx)(x^2+ab) + 4abx^2 \\ &= x^4 +2abx^2 + (ab)^2 + (ax+bx)^2 - 2(ax+bx)(x^2+ab) \\ &= (x^2+ab)^2 + (ax+bx)^2 - 2(ax+bx)(x^2+ab) \\ &= (x^2+ab -[ax+bx])^2 \\ &= ((x-a)(x-b))^2 } \] Resumint \[ (ab-x^2)^2+(ax+bx-2x^2)(ax+bx-2ab) = (x-a)^2(x-b)^2 \] Al denominador \[ \eqalign{ (ab+x^2)^2-x^2(a+b)^2 &= [ab+x^2]^2-[x(a+b)]^2 \\ &= [ab+x^2+x(a+b)][ab+x^2-x(a+b)] \\ &= [x^2+x(a+b)+ab][x^2-x(a+b)+ab] \\ &= (x+a)(x+b)(x-a)(x-b) } \] Llavors el denominador \[ (ab+x^2)^2-x^2(a+b)^2 = (x+a)(x+b)(x-a)(x-b) \] Finalment
\[ \frac{(ab-x^2)^2+(ax+bx-2x^2)(ax+abx-2ab)}{(ab+x^2)^2-x^2(a+b)^2} = \frac{(x-a)(x-b)}{(x+a)(x+b)}\]







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