[tex]\displaystyle \\
\text{pentru puteri cu exponenti negativi aplicam formulele: } \\ \\
a^{-2} = \frac{1}{a^2} \\ \\
\left(\frac{ a}{b} \right)^{-2} = \left(\frac{b}{a} \right)^{2} \\ \\
\text{Rezolvare:} \\ \\
a) \\
(2-x^{-1})^{-2} =4 \\ \\
\Big(2- \frac{1}{x}\Big)^{-2} =4 \\ \\
\Big(\frac{2x-1}{x}\Big)^{-2} =4 \\ \\
\Big(\frac{x}{2x-1}\Big)^{2} =4 \\ \\
\frac{x^2}{(2x-1)^2} =4 \\ \\
x^2 = 4(2x-1)^2 \\
x^2 = 4(4x^2 - 4x +1) \\
x^2 = 16x^2-16x+4\\
15x^2-16x + 4 = 0[/tex]
[tex]\displaystyle
15x^2-16x + 4 = 0 \\ \\
x_{12} = \frac{16 \pm \sqrt{256 -240} }{30} = \frac{16 \pm \sqrt{16} }{30} = \frac{16 \pm 4}{30} = \frac{8 \pm 2}{15} \\ \\
x_1 = \frac{8 + 2}{15} =\frac{10}{15} = \boxed{\frac{2}{3} } \\ \\
x_2 = \frac{8 - 2}{15} =\frac{6}{15} = \boxed{\frac{2}{5} }[/tex]
[tex]\displaystyle
b)\\
\Big[ 17-(5x)^{-2}\Big]^{-1}=1\\\\
\Big[17-\Big(\frac{1}{5x}\Big)^{2}\Big]^{-1}=1\\\\
\Big[17-\frac{1}{25x^2}\Big]^{-1}=1\\\\
\Big[\frac{17\times 25x^2-1}{25x^2}\Big]^{-1}=1\\\\
\Big[\frac{425x^2 -1}{25x^2}\Big]^{-1}=1\\\\
\frac{25x^2}{425x^2-1}=1\\\\
25x^2=425x^2-1\\\\
425x^2-25x^2-1=0\\\\
400x^2-1=0\\\\
(20x-1)(20x+1)=0\\\\
20x-1=0~\Longrightarrow~x_1=\boxed{\frac{1}{20}}\\\\
20x+1=0~\Longrightarrow~x_2=\boxed{-\frac{1}{20}}[/tex]
