Question

va rog sa ma ajute cineva dar sa fie rezolvat clar si explicativ la nivel de clasa a 10

Answer 1

   
[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]