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

`=> (2^x+1)^2 = (+-5)^2`

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

\(\Rightarrow\left[{}\begin{matrix}2^x=4\\2^x=-6\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=2\\x\in\varnothing\end{matrix}\right.\)

\(\left(x+6\right)\left(5^x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+6=0\\5^x-1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-6\\5^x=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-6\\x=0\end{matrix}\right.\)

\(\left(x-3\right)^{2023}=x-3\)

\(\Rightarrow\left(x-3\right)^{2023}-\left(x-3\right)=0\)

\(\Rightarrow\left(x-3\right)\left[\left(x-3\right)^{2022}-1\right]=0\)

\(\Rightarrow\left[{}\begin{matrix}x-3=0\\\left(x-3\right)^{2022}-1=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\\\left(x-3\right)^{2022}=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\\x-3=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)

`#3107.101107`

1.

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

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

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

`=>`\(\left[{}\begin{matrix}2^x=4\\2^x=-6\left(\text{vô lý}\right)\end{matrix}\right.\)

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

Vậy, `x =2.`

2.

`(x + 6)(5x - 1) = 0`

`=>`\(\left[{}\begin{matrix}x+6=0\\5x-1=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=-6\\5x=1\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=-6\\x=\dfrac{1}{5}\end{matrix}\right.\)

Vậy, `x \in {-6; 1/5}`

3.

`2*3^(x + 3) + 3^(2 + x) = 891`

`=> 2* 3^x * 3^3 + 3^2 * 3^x = 891`

`=> 54*3^x + 9*3^x = 891`

`=> 3^x * (54 + 9) = 891`

`=> 3^x * 63 = 891`

`=> 3^x = 891 \div 63`

`=> 3^x = 891/63`

Bạn xem lại đề.

4.

`(x - 3)^2023 = x - 3`

`=> (x - 3)^2023 - (x - 3) = 0`

`=> (x - 3) * [ (x - 3)^2022 - 1] = 0`

`=>`\(\left[{}\begin{matrix}x-3=0\\\left(x-3\right)^{2022}-1=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=3\\\left(x-3\right)^{2022}=1\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=3\\\left(x-3\right)^{2022}=\left(\pm1\right)^{2022}\end{matrix}\right.\)

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

`=>`\(\left[{}\begin{matrix}x=3\\x=4\\x=2\end{matrix}\right.\)

Vậy, `x \in {2; 3; 4}.`

a)

\(\text{( 25 – 2x )³ : 5 – 3^2 = 4^2}\)

\(\text{( 25 – 2x )³ : 5 – 9 = 16}\)

\(\text{( 25 – 2x )³ : 5 = 16 + 9}\)

\(\text{( 25 – 2x )³ : 5 = 25}\)

\(\text{( 25 – 2x )³ = 25 . 5}\)

\(\text{( 25 – 2x )³ = 125}\)

\(\text{( 25 – 2x )³ = 5³}\)

\(\text{25 – 2x = 5}\)

\(\text{2x = 25 – 5}\)

\(\text{2x = 20}\)

\(\text{x = 10}\)

\(\text{________________________________________}\)

b)

\(\text{2.3^x = 10.3^12 + 8.27^4}\)

\(\text{2.3^x = 10.3^12 + 8.(3^3)^4}\)

\(\text{2.3^x = 3^12 . (10+8)}\)

\(\text{2.3^x = 3^12 . 18}\)

\(\text{3^x = 3^12 . 18:2}\)

\(\text{3^x = 3^12 . 9}\)

\(\text{3^x = 3^12 . 3^2}\)

\(\text{3^x = 3^14}\)

\(\text{=> x=14}\)

a) a mũ 3 x a mũ 9 là : a³ x b⁹

b) ( a mũ 5 ) mũ 7 là : (a⁵)⁷

c) ( 2 mũ 3 ) mũ 5 x ( 2 mũ 3 ) mũ 3 là : (2³)⁵ x (2³)³

Hok tốt !

Đầu bài là thế ạ còn viết thì em ko biết anh giải giùm em ạ

Đề bài bn ghi thek thì ai làm nổi cho bn :V ?

mng giúp em với tối em nộp bài rồi a

cức + điên= lan ngọc cức điên

lớp 8 sao giống lớp 6 bọn em vậy

2.3^x= 10.3¹²+8.27⁴

2.3^x = 10.(3³)⁴+ (3³)⁴ . 8

2.3^x= (3³)⁴. 18

2.3^x= (3³)⁴.2.3²

2.3^x= 3¹²⁺².2

 2.3^x= 3¹⁴.2

=> x = 14