a) \(10^n+1-6\cdot10^n=\left(1-6\right)10^n+1=-5\cdot10^n+1\)

b)  \(90\cdot10^n-10^2-2+10^n+1=\left(90-1+1\right)\cdot10^n-2+1=90\cdot10^n-1\)

c)  \(2,5\cdot56^n-3=\frac{5}{2}\cdot56^n-3\)

kkkkkk

\(2,5.5^{n-3}.10+5^n-6.5^{n-1}\)

=25.\(5^n\):3+\(5^n\)\(-\)6.\(5^n\):5

=\(\dfrac{25}{3}\).\(5^n\)+\(5^n\)\(-\)\(\dfrac{6}{5}\).\(5^n\)

=\(5^n\).\(\left(\dfrac{25}{3}+1-\dfrac{6}{5}\right)\)

=\(5^n\).\(\dfrac{158}{15}\)

1/ 

= -10 - ( -10) - 75 + 4

= 0 - 75 + 4

= -71

2/ (-5)^2 : (-5) = -5

3/ \(\Leftrightarrow\orbr{\begin{cases}n+1< 0\\n+3< 0\end{cases}}\Leftrightarrow\orbr{\begin{cases}n>-1\\n>-3\end{cases}}\)

a) -10 - (-10) + 75 : (-1)³ + (-2)³ : (-2)

= -10 + 10 + 75 : (-1) + (-8) : (-2)

= 0 + (-75) + 4

= 0 - 75 + 4

= -71

b) E = (-5²) : (-5)

E = (-25) : (-5)

E = 5

c) (n + 1)(n + 3) < 0

=> \(\hept{\begin{cases}n+1< 0\\n+3>0\end{cases}}\Rightarrow\hept{\begin{cases}n< -1\\n>-3\end{cases}}\Rightarrow-3< n< -1\)

Hoặc \(\hept{\begin{cases}n+1>0\\n+3< 0\end{cases}}\Rightarrow\hept{\begin{cases}n>-1\\n< -3\end{cases}}\)(Loại)

Vậy -3 < n < -1