Question

Помогите,всё на фото

Answer 1

1) $\frac{5a^{2}+19a-4}{1-25a^{2}} =\frac{5*(a-4)(a+0,2)}{(1-5a)(1+5a)}=\frac{(a-4)(5a+1)}{(1-5a)(1+5a)} =\frac{a-4}{1-5a}$

5a²+19a-4=0

D = 19²-4*5*(-4) = 361+80 = 441 = 21²

$a_{12}=\frac{-19б21}{2*5}$

$a_{1}=\frac{19+21}{2*5}=\frac{40}{10}=4$

$a_{1}=\frac{19-21}{2*5}=-\frac{2}{10}=-0,2$

2) x²-8x+7=0

D = (-8)²-4*1*7=64-28=36=6²

$x_{12}=\frac{8б6}{2*1}$

$x_{1} =\frac{8+6}{2*1}=\frac{14}{2}=7$

$x_{1} =\frac{8-6}{2*1}=\frac{2}{2}=1$

Ответ: 1