\(25^{2x}:5^x=125^2\)

\(\Rightarrow5^{4x}:5^x=\left(5^3\right)^2\)

\(\Rightarrow5^{4x-x}=5^6\)

\(\Rightarrow5^{3x}=5^6\)

\(\Rightarrow3x=6\)

\(\Rightarrow x=2\)

x = 2

\(=\left(25^2\right)^x\div5x=125^2\)

\(=625^x\div5^x=125^2\)

\(=\left(625\div5\right)^x=125^2\)

\(=125^x=125^2\)

\(=>x=2\)

${\left({25}^2\right)}^x:5^x={125}^2$

${625}^x:5^x={125}^2$

${(625:5)}^x={125}^2$

${125}^x={125}^2$

$x=2$

25^2\(x\) : 5^\(x\)  =125²

5\(^{4x}\) : 5\(^x\) = 5⁶

5^\(3x\)       = 5⁶

3\(x\)      = 6

   \(x\)      = 2

Tham Khảo]

(5^2x.5^x+2) : 25 = 125²

5^2x.5^x+2 = 125².25 = 5⁶.5²

5^2x+x+2 = 5⁸

5^3x+2 = 5⁸

=> 3x + 2 = 8

=> 3x = 6

=> x = 2

Ta có: \(\left(5^{2x}\cdot5^{x+2}\right):25=125^2\)

\(\Leftrightarrow5^{3x+2}=5^6\cdot5^2=5^8\)

hay x=2

dẫu . là dấu nhân nha mn.

mong mn giải giúp mình mình đang cần gấp

\(C^n_n+C^{n-1}_n+C^{n-2}_n=37\)

\(\Leftrightarrow1+\dfrac{n!}{\left(n-1\right)!}+\dfrac{n!}{\left(n-2\right)!2!}=37\)

\(\Leftrightarrow1+n+\dfrac{n\left(n-1\right)}{2}=37\)

\(\Rightarrow n=8\)

\(P=\left(2+5x\right)\left(1-\dfrac{x}{2}\right)^8=\left(2+5x\right).\left(\sum\limits^8_{k=0}.C_8^k.\left(-\dfrac{x}{2}\right)^k\right)\)

\(=\left(2+5x\right).\left(\sum\limits^8_{k=0}.C_8^k.\left(-\dfrac{1}{2}\right)^k.x^k\right)\)

\(=2.\left(\sum\limits^8_{k=0}.C_8^k.\left(-\dfrac{1}{2}\right)^k.x^k\right)+5x\)\(\left(\sum\limits^8_{k=0}.C_8^k.\left(-\dfrac{1}{2}\right)^k.x^k\right)\)

\(=2.\left(\sum\limits^8_{k=0}.C_8^k.\left(-\dfrac{1}{2}\right)^k.x^k\right)+5\)\(\left(\sum\limits^8_{k=0}.C_8^k.\left(-\dfrac{1}{2}\right)^k.x^{k+1}\right)\)

Số hạng chứa \(x^3\) trong \(2.\left(\sum\limits^8_{k=0}.C_8^k.\left(-\dfrac{1}{2}\right)^k.x^k\right)\) là \(2C^3_8.\left(-\dfrac{1}{2}\right)^3x^3\)

Số hạng chứa \(x^3\) trong \(5\left(\sum\limits^8_{k=0}.C_8^k.\left(-\dfrac{1}{2}\right)^k.x^{k+1}\right)\) là \(5C^2_8.\left(-\dfrac{1}{2}\right)^2x^3\)

Vậy số hạng chứa x³ trong P là:\(\left[2.C^3_8\left(-\dfrac{1}{2}\right)^3+5C^2_8\left(-\dfrac{1}{2}\right)^2\right]x^3\)

`(5x-1)^2-(5x-4)(5x+4)=7`

`\Leftrightarrow 25x^2-10x+1-25x^2+16-7=0`

`\Leftrightarrow  -10x+10=0`

`\Leftrightarrow  x=1`

\(\left|5x-2\right|\le0\)

\(\Leftrightarrow\left[{}\begin{matrix}5x-2\le0\\5x-2\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x\le\dfrac{2}{5}\\x\ge-\dfrac{2}{5}\end{matrix}\right.\)

\(\text{Vì: }\)\(x\in Z\)

\(S=\left\{0\right\}\)

Vì \(\left|5x-2\right|\ge0\), kết hợp với đề bài 

\(\Rightarrow\left|5x-2\right|=0\) \(\Leftrightarrow5x-2=0\) \(\Leftrightarrow x=\dfrac{2}{5}\)  (Loại)

  Vậy bất phương trình vô nghiệm