5⁸.5² = 5⁸⁺²

= 5¹⁰

= 9765625

-----------

4⁹ : 64²

= 4⁹ : (4³)²

= 4⁹ : 4⁶

= 4⁹⁻⁶

= 4³

= 64

------------

2²⁵ : 32⁴

= 2²⁵ : (2⁵)4

= 2²⁵ : 2²⁰

= 2²⁵⁻²⁰

= 2⁵

= 32

------------

125³ : 25⁴

= (5³)³ : (5²)⁴

= 5⁹ : 5⁸

= 5⁹⁻⁸

= 5

\(\left(x+1\right)^2=16\\ \Rightarrow\left[{}\begin{matrix}x+1=4\\x+1=-4\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)

\(\left(x+1\right)^2=16\\ \Rightarrow\left[{}\begin{matrix}x+1=4\\x+1=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)

4 số đó lần lượt là 16;17;18;19

4 SỐ ĐÓ LÀ 16,17,18,19

\(Q=2+2^2+2^3+...+2^{99}\)

\(=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{98}+2^{99}\right)\)

\(=6+2^2.\left(2+2^2\right)+...+2^{97}.\left(2+2^2\right)\)

\(=6+2^2.6+...+2^{97}.6\)

\(=6.\left(1+2^2+...+2^{97}\right)\)

Vì \(6⋮3\) nên \(6.\left(1+2^2+...+2^{97}\right)⋮3\)

Vậy \(Q⋮3\)

\(#WendyDang\)

Ta có:

\(6=2.3\)

\(18=2.3^2\)

\(1=1\)

\(BCNN\left(6;18;1\right)=1.2.3^2=18\)

`6 = 2.3`

`18 = 2.3^2`

`1 = 1`

`BCNN (6; 18; 1) = 1.2.3^2 = 18`

\(2^x=2^5\)

\(\Rightarrow x=5\)

\(2^x=2^5\)

\(\Rightarrow2^x=32\)

\(\Rightarrow2^5=32\)

\(\Rightarrow x=5\)