\(\left(\dfrac{1}{5}\right)^{25}\cdot\left(\dfrac{1}{5}\right)^{30}\)

\(=\left(\dfrac{1}{5}\right)^{25+30}\)

\(=\left(\dfrac{1}{5}\right)^{55}\)

\(\left(\dfrac{1}{16}\right)^3:\left(\dfrac{1}{8}\right)^2\)

\(=\left[\left(\dfrac{1}{2}\right)^4\right]^3:\left[\left(\dfrac{1}{2}\right)^3\right]^2\)

\(=\left(\dfrac{1}{2}\right)^{12}:\left(\dfrac{1}{2}\right)^6\)

\(=\left(\dfrac{1}{2}\right)^{12-6}=\left(\dfrac{1}{2}\right)^6\)

\(\left(x^3\right)^2:\left(x^2\right)^3\)

\(=x^{3\cdot2-2\cdot3}\)

\(=x^0=1\)

\(\left(\dfrac{1}{3}+\dfrac{1}{6}\right)\cdot2^{x+4}-2^x=2^{13}-2^{10}\)

\(\Rightarrow\dfrac{1}{2}\cdot2^{x+4}-2^x=2^{13}-2^{10}\)

\(\Rightarrow2^{x+3}-2^x=2^{13}-2^{10}\)

\(\Rightarrow x+3=13;x+0=10\)

\(\Rightarrow x=10\)

(\(\dfrac{1}{3}\) +\(\dfrac{1}{6}\) ) . 2x+4 - 2x = 213 - 210

(\(\dfrac{2}{6}\)  + \(\dfrac{1}{6}\)) .   \(2^{x+4}\) -   \(2^x\) = 8192 - 1024

\(\dfrac{3}{6}\) . 2x . \(2^4\) -\(2^x\) = 7168

8 . 2x  - 2x . 1   = 7168

 2x . ( 8 - 1 ) = 7168

 2x . 7 = 7168

 2x = 7168 : 7

 2x = 1024

 2x = \(2^{10}\)

⇒ x = 10

`@` `\text {Ans}`

`\downarrow`

\((6x-5)(x+8)-(3x-1)(2x+3)-9(4x-3)\)

`= 6x(x+8) - 5(x+8) - [ 3x(2x+3) - 2x - 3] - 36x + 27`

`= 6x^2 + 48x - 5x - 40 - (6x^2 + 9x - 2x - 3) - 36x + 27`

`= 6x^2 + 48x - 5x - 40 - (6x^2 + 7x - 3) - 36x + 27`

`= 6x^2 + 48x - 5x - 40 - 6x^2 - 7x + 3 - 36x + 27`

`= (6x^2 - 6x^2) + (48x - 5x - 7x - 36x) + (-40 + 3 + 27)`

`= 0 + 0 - 10`

`= - 10`

Vậy, giá trị của biểu thức không phụ thuộc vào giá trị của biến

32 < 211 < 128 ( xem lại đề bài em nhé)

4< 2N < 216

\(\dfrac{4}{2}< \) N < \(\dfrac{216}{2}\)

2 < N < 108

vì N \(\in\) Z nên N \(\in\) { 3; 4; 5; 6; 7;...; 107}

`@` `\text {Ans}`

`\downarrow`

`-1/27 = ( (-1)^3/(-3)^3) =(-1/3)^3`

`8/729 = ( (2^3)/(9^3) ) = (2/9)^3`

`16/685 = ( (4^2)/( \sqrt {685}) ) = (4/(\sqrt {685}) )^2`

(\(x\) + 1)² = \(\dfrac{4}{25}\)

(\(x+1\))² = (\(\dfrac{2}{5}\))²

\(\left[{}\begin{matrix}x+1=-\dfrac{2}{5}\\x+1=\dfrac{2}{5}\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=\dfrac{2}{5}-1\\x=-\dfrac{2}{5}-1\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=-\dfrac{3}{5}\\x=-\dfrac{7}{5}\end{matrix}\right.\)

Vậy \(x\in\){  \(-\dfrac{7}{5}\) ; - \(\dfrac{3}{5}\)} 

`@` `\text {Ans}`

`\downarrow`

`(x+1)^2 = 4/25`

`=> (x+1)^2 = (+-2/5)^2`

`=>`\(\left[{}\begin{matrix}x+1=\dfrac{2}{5}\\x+1=-\dfrac{2}{5}\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=\dfrac{2}{5}-1\\x=-\dfrac{2}{5}-1\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=-\dfrac{3}{5}\\x=-\dfrac{7}{5}\end{matrix}\right.\)

Vậy, `x \in {-3/5; -7/5}.`

(2 x X - 1)mũ 3 = (2 x X - 2)mũ 2

\(\left(x^2-1\right)\left(x^2-5\right)< 0\)

\(\Rightarrow\left\{{}\begin{matrix}x^2-1< 0\\x^2-5>0\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}x^2-1>0\\x^2-5< 0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x^2< 1\\x^2>5\end{matrix}\right.\)(vô lí) hoặc \(\left\{{}\begin{matrix}x^2>1\\x^2< 5\end{matrix}\right.\)

\(\Rightarrow1< x< \sqrt{5}\) hoặc \(-\sqrt{5}< x< -1\)

Vậy \(-\sqrt{5}< x< -1\) hoặc \(1< x< \sqrt{5}\)