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
loga b -loga c=loga b/c
loga b+loga c=loga b*c
1/logb a=loga b
loga^n (b^n) =loga b
Решение :
log2 x-log2 3 +log2 3 +2=6
log2 ((x/3)*3)=6-2
log2 x =4
x=16