`=40`

\(=\dfrac{3\cdot2^{x+1}}{2^{x+3}}=\dfrac{3}{2^2}=\dfrac{3}{4}\)

\(\Leftrightarrow\dfrac{3\cdot2^{x+1}}{2^{x+3}}=40\\ \Leftrightarrow\dfrac{3}{2^2}=40\left(vô.lí\right)\Leftrightarrow x\in\varnothing\)

6².2².x-1=288

(2.3)².2².x-1=288

2².3².2².x-1=288

(2².2²).3².x-1=288

(2⁴.3²).x-1=288

144.x-1=288

144.x=288+1

144.x=289

x=289:144

x=289/144

1. 

$(3^2-2^3)x+3^2.2^2=4^2.3$

$\Leftrightarrow x+36=48$

$\Leftrightarrow x=48-36=12$

2.

$x^5-x^3=0$

$\Leftrightarrow x^3(x^2-1)=0$

$\Leftrightarrow x^3(x-1)(x+1)=0$

$\Leftrightarrow x^3=0$ hoặc $x-1=0$ hoặc $x+1=0$

$\Leftrightarrow x=0$ hoặc $x=\pm 1$
3.

$(x-1)^2+(-3)^2=5^2(-1)^{100}$

$\Leftrightarrow (x-1)^2+9=25$

$\Leftrightarrow (x-1)^2=25-9=16=4^2=(-4)^2$

$\Rightarrow x-1=4$ hoặc $x-1=-4$

$\Leftrightarrow x=5$ hoặc $x=-3$

4.

$(2x-1)^2-(2x-1)=0$

$\Leftrightarrow (2x-1)(2x-1-1)=0$

$\Leftrightarrow (2x-1)(2x-2)=0$

$\Leftrightarrow 2x-1=0$ hoặc $2x-2=0$

$\Leftrightarrow x=\frac{1}{2}$ hoặc $x=1$

$\Lef

`@` `\text {Ans}`

`\downarrow`

\((3^2-2^3)x+3^2.2^2=4^2.3\)

`=> x + (3*2)^2 = 48`

`=> x+6^2 = 48`

`=> x + 36 = 48`

`=> x = 48 - 36`

`=> x=12`

Vậy, `x=12`

\(x^5-x^3=0\)

`=> x^3(x^2 - 1)=0`

`=>`\(\left[{}\begin{matrix}x^3=0\\x^2-1=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=0\\x^2=1\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=0\\x=\pm1\end{matrix}\right.\)

Vậy, `x \in {0; +- 1 }`

\(\left(x-1\right)^2+\left(-3\right)^2=5^2\cdot\left(-1\right)^{100}\)

`=> (x-1)^2 + 9 = 25*1`

`=> (x-1)^2 + 9 = 25`

`=> (x-1)^2 = 25 - 9`

`=> (x-1)^2 = 16`

`=> (x-1)^2 = (+-4)^2`

`=>`\(\left[{}\begin{matrix}x-1=4\\x-1=-4\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=4+1\\x=-4+1\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\)

Vậy, `x \in {5; -3}`

\((2x-1)^2-(2x-1)=0\)

`=> (2x-1)(2x-1) - (2x-1)=0`

`=> (2x-1)(2x-1-1)=0`

`=>`\(\left[{}\begin{matrix}2x-1=0\\2x-2=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}2x=1\\2x=2\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=1\end{matrix}\right.\)

Vậy, `x \in {1; 1/2}`

Đặt A = \(\frac{1}{2}+\frac{1}{2.2}+\frac{1}{2.2.2}+...+\frac{1}{2.2.2.....2}\)

          = \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{50}}\)

=> 2A = \(2.\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{50}}\right)\)

           = \(2\times\frac{1}{2}+2\times\frac{1}{2^2}+2\times\frac{1}{2^3}+...+2\times\frac{1}{2^{50}}\)

           = \(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{49}}\)

 Lấy 2A - A = \(\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{49}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{50}}\right)\)

               A  = \(1+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-...-\frac{1}{2^{50}}\)

                   = \(1-\frac{1}{2^{50}}\)

Vậy  \(\frac{1}{2}+\frac{1}{2.2}+\frac{1}{2.2.2}+...+\frac{1}{2.2.2.....2}\)=  \(1-\frac{1}{2^{50}}\)

a, A = 2 + 2.2 + 2.2.2 + 2.2.2.2 + ... + 2.2...2 ( 22...2 có 16 số 2)

A = 2 + 2² + 2³ + 2⁴ + ... + 2¹⁶

2A = 2² + 2³ + 2⁴ + 2⁵ + ... + 2¹⁷

2A - A = ( 2² + 2³ + 2⁴ + 2⁵ + ... + 2¹⁷) - ( 2 + 2² + 2³ + 2⁴ + ... + 2¹⁶)

A = 2¹⁷ - 2

b, B = 1 + 1/3 + 1/6 + 1/10 + 1/15 + 1/21 + 1/28

1/2 x B = 1/2 + 1/6 + 1/12 + ... + 1/56

1/2 x B = 1/1x2 + 1/2x3 + 1/3x4 + ... + 1/7x8

1/2 x B = 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/7 - 1/8

1/2 x B = 1 - 1/8 = 7/8

B = 7/8 : 1/2

B = 7/8 x 2 = 7/4

a, A = 2 + 2.2 + 2.2.2 + 2.2.2.2 + ... + 2.2...2 ( 22...2 có 16 số 2)

A = 2 + 2² + 2³ + 2⁴ + ... + 2¹⁶

2A = 2² + 2³ + 2⁴ + 2⁵ + ... + 2¹⁷

2A - A = ( 2² + 2³ + 2⁴ + 2⁵ + ... + 2¹⁷) - ( 2 + 2² + 2³ + 2⁴ + ... + 2¹⁶)

A = 2¹⁷ - 2

b, B = 1 + 1/3 + 1/6 + 1/10 + 1/15 + 1/21 + 1/28

1/2 x B = 1/2 + 1/6 + 1/12 + ... + 1/56

1/2 x B = 1/1x2 + 1/2x3 + 1/3x4 + ... + 1/7x8

1/2 x B = 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/7 - 1/8

1/2 x B = 1 - 1/8 = 7/8

B = 7/8 : 1/2

B = 7/8 x 2 = 7/4

\(a;\left(x+1\right)\times\left(x+3\right)-x\times\left(x+2\right)=7\)

\(\Leftrightarrow x^2+4x+3-x^2-2x=7\)

\(\Leftrightarrow2x+3=7\Rightarrow x=\frac{7-3}{2}=2\)

            Vậy x=2

\(b;2x\left(3x+5\right)-x\left(6x-1\right)=33\)

\(\Leftrightarrow6x^2+10x-6x^2+x=33\)

\(\Leftrightarrow11x=33\Leftrightarrow x=3\)

        Vậy x=3