Номер 278,...

Решите задачу:

$y^4-6y^2+8=0 \\\ y^2=a \geq 0 \\\ a^2-6a+8=0 \\\ D_1=(-3)^2-1\cdot8=1 \\\ a=3+1=4\Rightarrow y^2=4; \ y=\pm2 \\\ a=3-1=2\Rightarrow y^2=2; \ y=\pm \sqrt{2}$

$t^4-10t^2+25=0 \\\ (t^2)^2-2\cdot t^2\cdot5+5^2=0 \\\ (t^2-5)^2=0 \\\ t^2-5=0 \\\ t^2=5 \\\ t=\pm \sqrt{5}$

$9x^4-9x^2+2=0 \\\ x^2=a \geq 0 \\\ 9a^2-9a+2=0 \\\ D=(-9)^2-4\cdot9\cdot2=81-72=9 \\\ a= \frac{9+3}{18} =\frac{2}{3} \Rightarrow x^2=\frac{2}{3} ; \ x=\pm \sqrt{\frac{2}{3}} =\pm \frac{ \sqrt{6} }{3} \\\ a= \frac{9-3}{18} =\frac{1}{3} \Rightarrow x^2=\frac{1}{3} ; \ x=\pm \sqrt{\frac{1}{3}} =\pm \frac{ \sqrt{3} }{3}$