Question

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é urgente

Answer 1

1) $\left(x-\,^{1}\!\!\!\diagup\!\!_{2} \right )^{2}=-\left(x-\,^{1}\!\!\!\diagup\!\!_{4} \right )$

$x^{2}-x+\,^{1}\!\!\!\diagup\!\!_{4}=-x+\,^{1}\!\!\!\diagup\!\!_{4}\\ \\ x^{2}-x+x-\,^{1}\!\!\!\diagup\!\!_{4}+\,^{1}\!\!\!\diagup\!\!_{4}=0\\ \\ x^{2}=0\\ \\ \boxed{x=0}$


2) $\left(2x+1 \right )^{2}=-\left(4x-1 \right )$

$4x^{2}+4x+1=-4x+1\\ \\ 4x^{2}+4x+4x+1-1=0\\ \\ 4x^{2}+8x=0\\ \\ 4x\cdot \left(x+2 \right )=0\\ \\ \begin{array}{rcl} 4x=0&\text{ ou }&x+2=0 \end{array}\\ \\ \boxed{ \begin{array}{rcl} x=0&\text{ ou }&x=-2 \end{array} }$


3) $\left(3-\,^{x}\!\!\!\diagup\!\!_{2} \right )\cdot\left(3+\,^{x}\!\!\!\diagup\!\!_{2} \right )=9-2x$

$9+\,^{3x}\!\!\!\diagup\!\!_{2}-\,^{3}\!\!\!\diagup\!\!_{2}-\,^{x^{2}}\!\!\!\diagup\!\!_{4}=9-2x\\ \\ 9-\,^{x^{2}}\!\!\!\diagup\!\!_{4}=9-2x\\ \\ \,^{x^{2}}\!\!\!\diagup\!\!_{4}-2x+9-9=0\\ \\ \,^{x^{2}}\!\!\!\diagup\!\!_{4}-2x=0\\ \\ x^{2}-8x=0\\ \\ x \cdot \left(x-8 \right )=0\\ \\ \begin{array}{rcl} x=0&\text{ ou }&x-8=0 \end{array}\\ \\ \boxed{\begin{array}{rcl} x=0&\text{ ou }&x=8 \end{array}}$