2>\(\sqrt{\pi}\)>\(\sqrt{3}\)

Đặt \(A=2^2+2^3+...+2^{2021}\)

=>\(2A=2^3+2^4+...+2^{2022}\)

=>\(2A-A=2^3+2^4+...+2^{2022}-2^2-2^3-...-2^{2021}\)

=>\(A=2^{2022}-4\)

\(\left(x-2\right)^6=4+2^2+2^3+...+2^{2021}\)

=>\(\left(x-2\right)^6=4+2^{2022}-4=2^{2022}\)

=>\(\left(x-2\right)^6=\left(2^{337}\right)^6\)

=>\(\left[{}\begin{matrix}x-2=2^{337}\\x-2=-2^{337}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2+2^{337}\\x=-2^{337}+2\end{matrix}\right.\)

a+1 là bội của a-1

=>\(a+1⋮a-1\)

=>\(a-1+2⋮a-1\)

=>\(2⋮a-1\)

=>\(a-1\in\left\{1;-1;2;-2\right\}\)

=>\(a\in\left\{2;0;3;-1\right\}\)

mà a là số tự nhiên

nên \(a\in\left\{2;0;3\right\}\)

Ta có: \(28-\left(2x-1\right)^3=2^2\cdot5\)

=>\(28-\left(2x-1\right)^3=20\)

=>\(\left(2x-1\right)^3=8=2^3\)

=>2x-1=2

=>2x=3

=>\(x=\dfrac{3}{2}\)

`28 - (2x-1)^3 = 2^2 xx 5`

`=> 28 - (2x -1)^3 = 20`

`=>        (2x -1)^3 = 28 - 20`

`=> (2x-1)^3 = 8`

`=> (2x - 1)^3 = 2^3`

`=> 2x - 1= 2`

`=> 2x =2+1`

`=> 2x = 3`

`=> x = 3/2`

Vậy `x =3/2`

\(\dfrac{x+1}{x-2}\) = \(\dfrac{3}{5}\)

(\(x+1\)).5= (\(x-2\)).3

5\(x+5\) = 3\(x\) - 6

5\(x-3x\) = - 6 - 5

    2\(x\) = -11

       \(x=-\dfrac{11}{2}\)

Vậy \(x=-\dfrac{11}{2}\) 

Ta có: \(\dfrac{x+1}{x-2}=\dfrac{3}{5}\)

=>5(x+1)=3(x-2)

=>5x+5=3x-6

=>5x-3x=-6-5

=>2x=-11

=>\(x=-\dfrac{11}{2}\)

    -|2,68 - 2\(x\)| - 5,9

Vì |2,68 - 2\(x\)| ≥ 0 ⇒ -|2,68 - 2\(x\)| ≤ 0 ⇒ - |2,68 - 2\(x\)| - 5,9 ≤ -5,9

Dấu bằng xảy ra khi:

2,68 - 2\(x\)  = 0 ⇒ 2\(x\) = 2,68 ⇒ \(x\) = 2,68 : 2 ⇒ \(x=1,34\)

Vậy giá trị lớn nhất của biểu thức:

- |2,68 - 2\(x\)| - 5,9 là -5,9 xảy ra khi \(x=1,34\)