Question

Cine mă ajuta as vrea rezolvarea cum e scris pe tabla nu același dar cum se face rezolvarea dar asemanator

Answer 1

     

$\displaystyle\bf\\f)\\\\\sqrt{x^2-3}=3-x~~\Big|~Ridicam~la~puterea~a~2-a.\\\\x^2-3=(3-x)^2\\\\x^2-3=9-2\cdot3\cdot x+x^2\\\\x^2-3=x^2+ 9-6x~~\Big|~-x^2\\\\-3=9-6x\\\\6x=9+3\\\\6x=12\\\\x=\frac{12}{6}\\\\\boxed{\bf x=2}$

.

$\displaystyle\bf\\g)\\\\\sqrt[\b3]{x^3-7}=x-1 ~~\Big|~Ridicam~la~puterea~a~3-a.\\\\x^3-7=(x-1)^3\\\\x^3-7=x^3-3x^2+3x-1~~\Big|~-x^3\\\\-7=-3x^2+3x-1\\\\3x^2-3x+1-7=0\\\\3x^2-3x-6=0~~\Big|~:3\\\\x^2-x-2=0\\\\x_{12}=\frac{1\pm\sqrt{1+4\cdot2}}{2}=\frac{1\pm\sqrt{9}}{2}=\frac{1\pm3}{2} x_1=\frac{1+3}{2}=\frac{4}{2} \\\\\boxed{\bf x_1=2}\\\\x_2=\frac{1-3}{2}=\frac{-2}{2}\\\\\boxed{\bf x_2=1}$