Question

Dados $x,\,y\in\mathbb{R},$ tais que

     $\left\{\begin{array}{l} x+xy+y=2+3\sqrt{2}\\\\ x^2+y^2=6\end{array}\right.$

encontre o valor de $|x+y+1|.$​

Answer 1

$\Large\boxed{\begin{array}{l}\begin{cases}\sf x+xy+y=2+3\sqrt{2}\\\sf x^2+y^2=6\end{cases}\\\sf x+xy+y=2+2\sqrt{2}+\sqrt{2}\\\sf x=2\\\sf y=\sqrt{2}\\\sf perceba\,que\\\sf x^2+y^2=2^2+(\sqrt{2})^2=4+2=6.\\\sf ou~seja\\\sf |x+y+1|=|2+\sqrt{2}+1|=|3+\sqrt{2}|\\\sf \end{array}}$

$\Large\boxed{\begin{array}{l}\sf Mas~|x|=\sqrt{x^2}\\\sf \sf |3+\sqrt{2}|=\sqrt{(3+\sqrt{2})^2}\\\sf |3+\sqrt{2}|=\sqrt{9+6\sqrt{2}+2}\\\sf |3+\sqrt{2}|=\sqrt{11+6\sqrt{2}}\\\sf |3+\sqrt{2}|=\sqrt{11+\sqrt{72}}\\\underline{\rm Radical\,duplo}\\\sf \sqrt{A\pm\sqrt{B}}=\sqrt{\dfrac{A+C}{2}}\pm\sqrt{\dfrac{A-C}{2}}\\\\\sf onde~C=\sqrt{A^2-B}\\\sf \sqrt{11+\sqrt{72}}=\sqrt{A+\sqrt{B}}\\\sf A=11~~B=72\\\sf C=\sqrt{A^2-B}\\\sf C=\sqrt{11^2-72}\\\sf C=\sqrt{121-72}\\\sf C=\sqrt{49}\\\sf C=7\end{array}}$

$\Large\boxed{\begin{array}{l}\sf\sqrt{11+\sqrt{72}}=\sqrt{\dfrac{11+7}{2}}+\sqrt{\dfrac{11-7}{2}}\\\\\sf\sqrt{11+\sqrt{72}}=\sqrt{\dfrac{18}{2}}+\sqrt{\dfrac{4}{2}}\\\\\sf\sqrt{11+\sqrt{72}}=\sqrt{9}+\sqrt{2}\\\sf \sqrt{11+\sqrt{72}}=3+\sqrt{2}\end{array}}$

$\Large\boxed{\begin{array}{l}\sf|3+\sqrt{2}|=\sqrt{(3+\sqrt{2})^2}=\sqrt{11+\sqrt{72}}\\\sf |3+\sqrt{2}|=3+\sqrt{2}\\\sf portanto\\\sf |x+y+1|=3+\sqrt{2}\checkmark\end{array}}$

Answer 2

$x + xy + y = 2 + 3 \sqrt{2} = > \\ \\ 2x + 2xy + 2y = 4 + 6 \sqrt{2} \\ x^{2} + y ^{2} \: \: \: \: \: ( +) \\ = = = = = = = = = = = = = = = = = = \\ x^{2} + y^{2} + 2x + 2xy + 2y = 10 + 6 \sqrt{2} \\ \\ x ^{2} + y^{2} + 1 + 2x + 2xy + 2y = 11 + 6 \sqrt{2} \\ \\ \\ (x + y + 1)^{2} = 11 + 6 \sqrt{2} \\ (x + y + 1) ^{2} = 2 + 9 + 2.3 \sqrt{2} \\ (x + y + 1)^{2} = (3 + \sqrt{2} ) ^{2} \\ \sqrt{(x + y + 1)^{2} } = \sqrt{(3 + \sqrt{2} } )^{2} \\ \\ |x + y + 1| = 3 + \sqrt{2} \\ \\ mario \: s.e$