Răspuns: Ti-am dat cateva exemple ptr. a intelege metoda de rezolvare.
Mai departe va fi usor ptr. tine, bafta!
5. a)
[tex]\frac{1}{9} x^2= 16 => x^2 = \frac{16}{\frac{1}{9} } => x^2 = 16*9 = 144 => x1 = -12; x2= 12[/tex]
b) [tex]2x^2=3 => x^2 = \frac{3}{2} => x = \sqrt{\frac{3}{2}} (Rationalizam)\\[/tex] => [tex]\sqrt{\frac{3}{2}}*\sqrt{\frac{2}{2}} = \frac{\sqrt{6}}{2}[/tex]
[tex]x1 = +\frac{\sqrt{6} }{2} ; x2 = -\frac{\sqrt{6} }{2}[/tex]
c) [tex]9x^2-10 =0 => 9x^2=10 => x^2 = + sau -\frac{10}{9} => x1 = +\frac{\sqrt{10} }{\sqrt{9}}[/tex] = [tex]\frac{\sqrt{10} }{3}[/tex]
[tex]x2 = -\frac{\sqrt{10} }{3}[/tex]
d) [tex]4x^2=0.5 <=> 4x^2 = \frac{1}{2} => x^2 = + sau - \frac{1}{8} => x = \frac{1}{\sqrt{8} } = \frac{1}{2\sqrt{2} } = + sau - \frac{\sqrt{2} }{4}[/tex]
e) [tex]\frac{1}{100}*x^2 -\frac{1}{64}=0 => \frac{x^2}{100} = \frac{1}{64}=> x^2 = \frac{100}{64} = \frac{25}{16} => x = + sau - \frac{5}{4}[/tex]
f) [tex]3x^2=7 => x^2 = \frac{7}{3} => x = + sau - \frac{\sqrt{7} }{\sqrt{3} } = + sau -\frac{\sqrt{21} }{3}[/tex]
Explicație pas cu pas:
