Question

Resolva a equação
$\frac{x}{x\²-x+2} +\frac{2x}{x\²+2x+2}=-\frac{9}{4}$

Answer 1

$\boxed{\begin{array}{l}\rm\dfrac{x}{x^2-x+2}+\dfrac{2x}{x^2+2x+2}=-\dfrac{9}{4}\\\\\rm\dfrac{4x(x^2+2x+2)+8x(x^2-x+2)}{4(x^2-x+2)(x^2+2x+2)}=-\dfrac{9(x^2-x+2)(x^2+2x+2)}{4(x^2-x+2)(x^2+2x+2)}\\\\\rm 4x^3+8x^2+8x+8x^3-8x^2+16x\\\rm =-9(x^4+2x^3+2x^2-x^3-2x^2-2x+2x^2+4x+4)\\\rm 12x^3+24x=-9(x^4+x^3+2x^2+2x+4)\\\rm 12x^3+24x=-9x^4-9x^3-18x^2-18x-36\\\rm9x^4+12x^3+9x^3+18x^2+24x+18x+36=0\\\rm9x^4+21x^3+18x^2+42x+36=0\div3\\\rm 3x^4+7x^3+6x^2+14x+12=0 \end{array}}$

$\large\boxed{\begin{array}{l}\underline{\sf Teorema\,das\,ra\acute izes\,racionais}\\\rm Seja\\\rm a_nx^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\dotsc+a_1x+a_0=0\\\rm uma\,equac_{\!\!,}\tilde ao\,com\,coeficientes\,inteiros.\\\rm Se\,\dfrac{p}{q}~(frac_{\!\!,}\tilde ao\,irredut\acute ivel)\,\acute e\,uma\,ra\acute iz,ent\tilde ao\\\\\rm p\,\acute e\,divisor\,de\,a_0\,e\,q\,\acute e\,divisor\,de\,a_n.\end{array}}$

$\boxed{\begin{array}{l}\rm valores\,de\,p:\pm1,\pm2,\pm3,\pm4,\pm6,\pm12\\\rm valores\,de\,q:\pm1,\pm3\\\rm \dfrac{1}{1}=1~~\dfrac{1}{-1}=-1~~\dfrac{1}{3}~~-\dfrac{1}{3}\\\rm\dfrac{2}{1}=2~~\dfrac{2}{-1}=-2~~\dfrac{2}{3}~~-\dfrac{2}{3}\\\\\rm\dfrac{3}{1}=3~~\dfrac{3}{-1}=-3~~\dfrac{3}{3}=1~~\dfrac{3}{-3}=-1\\\\\rm\dfrac{4}{1}=4~~\dfrac{4}{-1}=-4~~\dfrac{4}{3}~-\dfrac{4}{3}\\\\\rm\dfrac{6}{1}=6~~~\dfrac{6}{-1}=-6~~\dfrac{6}{3}=2~~\dfrac{6}{-3}=-2\end{array}}$

$\large\boxed{\begin{array}{l}\rm\dfrac{12}{1}=12~~\dfrac{12}{-1}=-12~~\dfrac{12}{3}=4~~\dfrac{12}{-3}=-4\\\rm candidatos\,a\,ra\acute iz:+1,-1,+2,-2,+3,-3,+4,-4,+6,-6\\\rm +12,-12,+\dfrac{1}{3},-\dfrac{1}{3},+\dfrac{2}{3},-\dfrac{2}{3},+\dfrac{4}{3},-\dfrac{4}{3}\end{array}}$

$\large\boxed{\begin{array}{l}\rm testando\,alguns\,valores:\\\rm3x^4+7x^3+6x^2+14x+12=0,para\,x=-1\\\rm3(-1)^4+7\cdot(-1)^3+6\cdot(-1)^2+14\cdot(-1)+12\\\rm3-7+6-14+12=0\implies -1\,\acute e\,ra\acute iz\\\rm 3x^4+7x^3+6x^2+!4x+12=0,para\,x=-2\\\rm 3\cdot(-2)^4+7\cdot(-2)^3+6\cdot(-2)^2+14\cdot(-2)+12\\\rm48-56+24-28+12=0\implies -2\,\acute e\,ra\acute iz\end{array}}$

$\large\boxed{\begin{array}{l}\rm3x^2+7x^3+6x^2+14x+12=(x+1)(x+2)(3x^2-2x+6)\\\rm (x+1)(x+2)(3x^2-2x+6)=0\\\rm 3x^2-2x+6=0\\\rm\Delta=b^2-4ac\\\rm\Delta=(-2)^2-4\cdot3\cdot6\\\rm\Delta=4-72\\\rm\Delta=-68\\\rm x=\dfrac{-b\pm\sqrt{\Delta}}{2a}\\\\\rm x=\dfrac{2\pm\sqrt{-68}}{2\cdot3}\\\\\rm x=\dfrac{2\pm2\sqrt{17}i}{2\cdot3}\\\\\rm x=\dfrac{1\pm\sqrt{17}i}{3}\begin{cases} \rm x_1=\dfrac{1+\sqrt{17}i}{3}\\\\\rm x_2=\dfrac{1-\sqrt{17}i}{3}\end{cases}\end{array}}$

$\large\boxed{\begin{array}{l}\rm A\,soluc_{\!\!,}\tilde ao\,da\,equac_{\!\!,}\tilde ao\,\acute e\\\rm S=\bigg\{-2,-1,\dfrac{1-\sqrt{17}i}{3},\dfrac{1+\sqrt{17}i}{3}\bigg\}\end{array}}$