\(x^3\) + 125 + (\(x\) + 5)(\(x\) - 25) = 0

(\(x^3\) + 5³) + (\(x\) + 5)(\(x\) - 25) = 0

(\(x\) + 5)(\(x^2\) - 5\(x\) + 25) + (\(x\) + 5)(\(x\) - 25) =0

(\(x\) + 5)(\(x^2\) - 5\(x\) + 25 + \(x\) - 25) = 0

(\(x\) + 5)(\(x^2\) - 4\(x\)) = 0

\(x\)(\(x\) + 5)(\(x\) - 4) = 0

\(\left[{}\begin{matrix}x=0\\x+5=0\\x-4=0\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=0\\x=-5\\x=4\end{matrix}\right.\)

`x^3 + 125 + (x + 5)(x - 25) = 0`

`<=>(x + 5)(x^2 + 5x + 25) + (x + 5)(x - 25) = 0`

`<=>(x + 5)(x^2 + 6x) = 0`

`<=>x(x + 5)(x + 6) = 0`

`<=>x = 0` hoặc `x = -5` hoặc `x=-6`

Sửa:

`x^3 + 125 + (x + 5)(x - 25) = 0`

`<=>(x + 5)(x^2 - 5x + 25) + (x + 5)(x - 25) = 0`

`<=>(x+5)(x^2-4x)=0`

`<=>x(x+5)(x-4)=0`

`<=>x=0` hoặc `x=-5` hoặc `x=4`

a) 5^x=125

=> 5^x=5^2

=>x=2

b) 3^2*x=81

=>3^2*x=9^2

=>3^2*x=(3^2)^2

=>x=2

câu c chưa hiểu chỗ 5^2x3 hay 5^2*3 vậy ?

a) \(5^x=125\)

\(5^x=5^3\)

\(\Rightarrow x=3\)

b)\(3^{2x}=81\)

\(3^{2x}=3^4\)

\(\Rightarrow2x=4\)

\(x=4:2\)

\(x=2\)

c)\(5^{2x-3}-2.5^2=5^2.3\)

\(5^{2x-3}=5^2.3+5^2.2\)

\(5^{2x-3}=5^2.\left(3+2\right)\)

\(5^{2x-3}=5^3\)

\(\Rightarrow2x-3=3\)

\(2x=3+3\)

\(2x=6\)

\(x=6:2\)

\(x=3\)

5^2x3 nha

a=600

b=24

k anh nhe

1/\(\left(x-6\right)^2=9\Rightarrow\orbr{\begin{cases}x-6=-3\\x-6=3\end{cases}\Rightarrow\orbr{\begin{cases}x=3\\x=9\end{cases}}}\)
2/\(5^{x+1}=125\Leftrightarrow5^{x+1}=5^3\Rightarrow x+1=3\Rightarrow x=2\)
3/ đề thiếu
 

x = 9

x = 2

câu 3 mình ko hỉu

à xin lỗi mình nghi thiếu

5^2x-3-2x5²=5²x3

Bài 1: 

a: \(\Leftrightarrow3^x\cdot10=810\)

\(\Leftrightarrow3^x=81\)

hay x=4

c: \(\Leftrightarrow5^x\cdot5+5^x\cdot\dfrac{1}{25}=126\)

\(\Leftrightarrow5^x\cdot\dfrac{126}{25}=126\)

\(\Leftrightarrow5^x=25\)

hay x=2

Bài 2: 

a: \(27^{11}=3^{33}\)

\(81^8=3^{32}\)

mà 33>32

nên \(27^{11}>81^8\)

c: \(625^5=\left(5^4\right)^5=5^{20}\)

\(125^7=\left(5^3\right)^7=5^{21}\)

mà 20<21

nên \(625^5< 125^7\)

2: \(=\dfrac{\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)}{-\left(x-y\right)\left(x^2+xy+y^2\right)}=\dfrac{-\left(x+y\right)\left(x^2+y^2\right)}{x^2+xy+y^2}\)