534. Haãnure KOPHU ypaBH enua . 5x^2-11x+2=0 T) 35x^2+2x-1=0 2p^2+7p-30=0 11) 2y^2-y-5=0 B) 9y^2-30y+25=0 e) 16x^2-8x+1=0

Для решения квадратных уравнений можно использовать формулу дискриминанта:<br /><br />$D = b^2 - 4ac$<br /><br />где $a$, $b$ и $c$ - коэффициенты уравнения $ax^2 + bx + c = 0$.<br /><br />Давайте решим каждое уравнение:<br /><br />1) $5x^{2}-11x+2=0$<br /><br />$D = (-11)^2 - 4 \cdot 5 \cdot 2 = 121 - 40 = 81$<br /><br />$x = \frac{{-(-11) \pm \sqrt{81}}}{{2 \cdot 5}} = \frac{{11 \pm 9}}{{10}}$<br /><br />$x_1 = \frac{{11 + 9}}{{10}} = 2$, $x_2 = \frac{{11 - 9}}{{10}} = \frac{1}{5}$<br /><br />2) $35x^{2}+2x-1=0$<br /><br />$D = 2^2 - 4 \cdot 35 \cdot (-1) = 4 + 140 = 144$<br /><br />$x = \frac{{-2 \pm \sqrt{144}}}{{2 \cdot 35}} = \frac{{-2 \pm 12}}{{70}}$<br /><br />$x_1 = \frac{{-2 + 12}}{{70}} = \frac{1}{7}$, $x_2 = \frac{{-2 - 12}}{{70}} = -\frac{1}{5}$<br /><br />3) $2p^{2}+7p-30=0$<br /><br />$D = 7^2 - 4 \cdot 2 \cdot (-30) = 49 + 240 = 289$<br /><br />$p = \frac{{-7 \pm \sqrt{289}}}{{2 \cdot 2}} = \frac{{-7 \pm 17}}{{4}}$<br /><br />$p_1 = \frac{{-7 + 17}}{{4}} = \frac{5}{2}$, $p_2 = \frac{{-7 - 17}}{{4}} = -6$<br /><br />4) $2y^{2}-y-5=0$<br /><br />$D = (-1)^2 - 4 \cdot 2 \cdot (-5) = 1 + 40 = 41$<br /><br />$y = \frac{{-(-1) \pm \sqrt{41}}}{{2 \cdot 2}} = \frac{{1 \pm \sqrt{41}}}{{4}}$<br /><br />$y_1 = \frac{{1 + \sqrt{41}}}{{4}}$, $y_2 = \frac{{1 - \sqrt{41}}}{{4}}$<br /><br />5) $9y^{2}-30y+25=0$<br /><br />$D = (-30)^2 - 4 \cdot 9 \cdot 25 = 900 - 900 = 0$<br /><br />$y = \frac{{-(-30) \pm \sqrt{0}}}{{2 \cdot 9}} = \frac{{30}}{{18}} = \frac{5}{3}$<br /><br />$y_1 = y_2 = \frac{5}{3}$<br /><br />6) $16x^{2}-8x+1=0$<br /><br />$D = (-8)^2 - 4 \cdot 16 \cdot 1 = 64 - 64 = 0$<br /><br />$x = \frac{{-(-8) \pm \sqrt{0}}}{{2 \cdot 16}} = \frac{{8}}{{32}} = \frac{1}{4}$<br /><br />$x_1 = x_2 = \frac{1}{4}$<br /><br />Таким образом, решения уравнений:<br /><br />1) $x_1 = 2$, $x_2 = \frac{1}{5}$<br />2) $x_1 = \frac{1}{7}$, $x_2 = -\frac{1}{5}$<br />3) $p_1 = \frac{5}{2}$, $p_2 = -6$<br />4) $y_1 = \frac{{1 + \sqrt{41}}}{{4}}$, $y_2 = \frac{{1 - \sqrt{41}}}{{4}}$<br />5) $y_1 = y_2 = \frac{5}{3}$<br />6) $x_1 = x_2 = \frac{1}{4}$