Question

As raízes de x2/3-3-x2/6-1/2 são:​

Answer 1

Resposta:

1/3- ( 3-) . 1/6 = 1/2  

2x² -3 + x²= 3  

3x²=6

x²=2  

X = +√2 X= - √2

Raiz de 2

Answer 2

Temos uma equação do 2º grau com frações. Para a resolução, a idéia é igualar a equação à zero, simplificar as frações, isolar a incógnita e extrair a raiz quadrada dos dois membros, assim vamos obter as raízes (valores para x):

$\begin{array}{l}\\\sf \dfrac{x^2}{3}-\dfrac{3-x^2}{6}-\dfrac{1}{2}\\\\\sf\dfrac{x^2}{3}-\dfrac{3-x^2}{6}-\dfrac{1}{2}=0\\\\\sf \!\!\bigg(\dfrac{x^2}{3}-\dfrac{3-x^2}{6}-\dfrac{1}{2}\bigg)\cdot6=(0)\cdot6\\\\\sf \dfrac{\diagdown\!\!\!\!6\cdot x^2}{\diagdown\!\!\!\!3}-\dfrac{\diagdown\!\!\!\!6\cdot(3-x^2)}{\diagdown\!\!\!\!6}-\dfrac{\diagdown\!\!\!\!6\cdot1}{\diagdown\!\!\!\!2}=0\\\\\sf 2x^2-(3-x^2)-3=0\\\\\sf 2x^2-3+x^2-3=0\\\\\sf 3x^2-6=0\\\\\sf 3x^2=6\\\\\sf x^2=\dfrac{6}{3}\\\\\sf x^2=2\\\\\sf \sqrt{x^2}=\sqrt{2}\\\\ \begin{vmatrix}\sf\, x\,\end{vmatrix}\sf =\sqrt{2}\\\\\sf x=\pm~\sqrt{2}~,~~\therefore~x'=-\sqrt{2}\quad e\quad x''=\sqrt{2}\end{array}$

Assim o conjunto solução é:

$\large\begin{array}{l}\sf S=\left\{\, - \: \sqrt{2}~ \: ~;~~\sqrt{2}\,\right\}\end{array}$

ㅤ

Att. Nasgovaskov

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