Решить биквадратное уравнение: б) а^4 -9а^2 =0; г) с^4 + 1/4с^2 =0; е) 9r^4 +8r^2-1=0; з)...

$x ^{4} - 9 {x}^{2} = 0$
возьмём за y x^2. получим

x = \sqrt{y} \\ x = 0 \\ x = \pm 3" alt=" {y}^{2} - 9y = 0 \\ y(y - 9) = 0 \\ y = 0 \\ y = 9 \\ \\ y = {x}^{2} = > x = \sqrt{y} \\ x = 0 \\ x = \pm 3" align="absmiddle" class="latex-formula">

${x}^{4} + \frac{1}{4} {x}^{2} = 0$
возьмём за y x^2. Получим:

x = \sqrt{y} \\ x = 0" alt=" {y}^{2} + 0.25y = 0 \\ y(y + \frac{1}{4} ) = 0 \\ y = 0 \\ y = - \frac{1}{4} \\ y = {x}^{2} = > x = \sqrt{y} \\ x = 0" align="absmiddle" class="latex-formula">

$9 {x}^{4} + 8 {x}^{2} - 1 = 0$
возьмём за y x^2

x = \sqrt{y} \\ x = \pm \frac{1}{3} " alt="9 {y}^{2} + 8y - 1 = 0 \\ y = \frac{ - 8 \pm \sqrt{64 + 36} }{2 \times 9} \\ y = \frac{ - 8 \pm \sqrt{100} }{18} \\ y = - 1 \\ y = \frac{1}{9} \\ y = {x}^{2} = > x = \sqrt{y} \\ x = \pm \frac{1}{3} " align="absmiddle" class="latex-formula">