 
                                                ![\frac{10}{ \sqrt[3]{25} } = \frac{5*2}{25^{ \frac{1}{3} }} = \frac{5*2}{(5^2)^{ \frac{1}{3} }} = \frac{5*2}{5^{ \frac{2}{3} }}](/tpl/images/0402/9212/33344.png) 
                                                 
                                                ![\frac{10}{ \sqrt[3]{25}}= \frac{5*2}{ (5^{2})^{ \frac{1}{3} }}= \frac{5*2}{5^{ \frac{2}{3}}}=2*5^{ \frac{1}{3}}= 2 \sqrt[3]{5}](/tpl/images/0402/9212/7ad34.png) 
                                                 
                                                ответ:
d=b^2-4ac=(-1)^2-4*1*(-72)=1+288=\sqrt{289}
289
=17
х1=\frac{-b- \sqrt{d} }{2a} = \frac{1-17}{2} = \frac{-16}{2} =-8
2a
−b−
d
=
2
1−17
=
2
−16
=−8
х2=\frac{-b+ \sqrt{d} }{2a} = \frac{1+17}{2} = \frac{18}{2} = 9
2a
−b+
d
=
2
1+17
=
2
18
=9
ответ: -8 и 9
d=b^2-4ac=7^2-4*(-4)*(-3)=49-48=\sqrt{1} =1
1
=1
х1=\frac{-b- \sqrt{d} }{2a} = \frac{-7-1}{2*(-4)} = \frac{-8}{-8} =1
2a
−b−
d
=
2∗(−4)
−7−1
=
−8
−8
=1
х2=\frac{-b+ \sqrt{d} }{2a} = \frac{-7+1}{(-8)} = \frac{-6}{-8} =0,75
2a
−b+
d
=
(−8)
−7+1
=
−8
−6
=0,75
