2( \(\dfrac{1}{2}\) x - \(\dfrac{1}{3}\)) - \(\dfrac{3}{2}\) = \(\dfrac{1}{4}\)

2(\(\dfrac{1}{2}\)x - \(\dfrac{1}{3}\))  = \(\dfrac{1}{4}\) + \(\dfrac{3}{2}\)

2(\(\dfrac{1}{2}\)x - \(\dfrac{1}{3}\)) = \(\dfrac{7}{4}\)

\(\dfrac{1}{2}\)x = \(\dfrac{7}{4}\) : 2 + \(\dfrac{1}{3}\)

\(\dfrac{1}{2}\)x = \(\dfrac{29}{24}\)

x = \(\dfrac{29}{24}\) x2

x = \(\dfrac{29}{12}\)

`2.(1/2 x - 1/3) - 3/2 = 1/4`

`2.(1/2 x - 1/3) = 1/4 + 3/2`

`2.(1/2 x - 1/3) = 1/4 + 6/4`

`2.(1/2 x - 1/3) = 7/4`

`1/2x - 1/3 = 7/4 : 2`

`1/2 x - 1/3 = 7/4 xx 1/2`

`1/2 x - 1/3 = 7/8`

`1/2 x  = 7/8 + 1/3`

`1/2 x = 21/28 + 7/28`

`1/2 x = 29/24`

`x=29/24 : 1/2`

`x=29/24 xx 2`

`x=29/12`

Vậy `x = 29/12`

          A   = 1 +  2 + 2² + 2³+......+2⁹

         2A  =       2+ 2² + 2³ +.......+2⁹ + 2¹⁰

    2A - A  =      2¹⁰ -1

          A  =     2¹⁰ - 1

11+12+123=123553444

Ta có:

218 = 23.215 = 8.215 > 7.215

=> 218 > 7.215

67+80=169789765

x = 6

2^2x+1 + 2^2x+2 = 3.2¹¹

2^2x+1 (1 +2) = 3.2¹¹

2^2x+1..3 = 3,2¹¹

2^2x+1 = 2¹¹

2x + `1 = 11

2x = 10

x = 10: 2

x = 5

đặt tổng trên là A

\(2A=2+2^2+2^3+...+2^{10}\)

\(A=2A-A=2^{10}-1\)

(351 - 875) - (125 - 149)

= 351 - 875 - 125 + 149

= (351 + 149) - (875 + 125)

= 500 - 1000

= -500

`(351 - 875) - (125 - 149)`

`= 351 - 875 - 125 + 149`

`= (351 + 149) - (875 + 125)`

`=500 -  (875 + 125)`

`=500 - 1000`

`=-500`

\(3A=1.2.3+2.3.3+3.4.3+...+99.100.3=\)

\(=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+99.100.\left(101-98\right)=\)

\(=1.2.3-1.2.3+2.3.4-2.3.4+3.4.5-...-98.99.100+99.100.101=\)

\(=99.100.101\Rightarrow A=33.100.101\)

          A = 1.2 + 2.3 + 3.4 +........+99.100

3 x    A = 1.2.3+2.3.3 +3.4.3+....99.100.3

3xA = 1.2.3+2.3.(4-1) + 3.4.(5-2) +4.5.(6-3)+....+99.100(101-98)

3xA = 1.2.3 -1.2.3 +2.3.4 - 2.3.4 +3.4.5 - 3.4.5+....+99.100.101

3xA = 99.100.101

A = 99.100.101 : 3

A = 33.100.101

\(3^4\cdot3^2=3^6=729\)

729

X = 0 và X = 1 

đúng 100% :3

\(x^3-x=0\)

\(\Leftrightarrow x\left(x^2-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\pm1\end{matrix}\right.\)

Vậy \(S=\left\{0;\sqrt{1}\right\}\)