a) \(5^{x+2}-5^{x-1}=3100\) \(\Leftrightarrow5^x.5^2-5^x:5=3100\)

\(\Leftrightarrow5^x.25-5^x.\frac{1}{5}=3100\)\(\Leftrightarrow5^x.\left(25-\frac{1}{5}\right)=3100\)

\(\Leftrightarrow5^x.\frac{124}{5}=3100\)\(\Leftrightarrow5^x=125=5^3\)\(\Leftrightarrow x=3\)

Vậy \(x=3\)

b) \(\left(x-4\right)\left(2x+3\right)< 0\)

TH1: \(\hept{\begin{cases}x-4>0\\2x+3< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}x>4\\2x< -3\end{cases}}\Leftrightarrow\hept{\begin{cases}x>4\\x< \frac{-3}{2}\end{cases}}\)( vô lý )

TH2: \(\hept{\begin{cases}x-4< 0\\2x+3>0\end{cases}}\Leftrightarrow\hept{\begin{cases}x< 4\\2x>-3\end{cases}}\Leftrightarrow\hept{\begin{cases}x< 4\\x>\frac{-3}{2}\end{cases}}\Leftrightarrow\frac{-3}{2}< x< 4\)

mà x là số nguyên \(\Rightarrow-1< x< 4\)

Vậy \(-1< x< 4\)

Bạn xem lại đề

\(53.x+2-25=3100\)

\(53.x+2=3100+25\)

\(53.x+2=3125\)

\(53.x=3125-2\)

\(53.x=3123\)

\(x=3123:53\)

\(x=\dfrac{3123}{53}\)

a) ( x - 1 )( x + 4 ) < 0

Xét hai trường hợp :

1. \(\hept{\begin{cases}x-1< 0\\x+4>0\end{cases}}\Leftrightarrow\hept{\begin{cases}x< 1\\x>-4\end{cases}}\Rightarrow-4< x< 1\)

2. \(\hept{\begin{cases}x-1>0\\x+4< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}x>1\\x< -4\end{cases}}\)( loại )

Vậy với -4 < x < 1 thì ( x - 1 )( x + 4 ) < 0

b) 5^x+2 - 5^x-1 = 3100

<=> 5^x( 5² - 5^-1 ) = 3100

<=> 5^x( 25 - 1/5 ) = 3100

<=> 5^x.124/5 = 3100

<=> 5^x = 125

<=> 5^x = 5³

<=> x = 3

c) 3^x+1 - 3^x-2 = 702

<=> 3^x( 3 - 3^-2 ) = 702

<=> 3^x( 3 - 1/9 ) = 702

<=> 3^x.26/9 = 702

<=> 3^x = 243

<=> 3^x = 3⁵

<=> x = 5

a) (x - 1)(x + 4) < 0

Xét các trường hợp

TH1\(\hept{\begin{cases}x-1>0\\x+4< 0\end{cases}}\Rightarrow\hept{\begin{cases}x>1\\x< -4\end{cases}}\left(\text{loại}\right)\)

TH2\(\hept{\begin{cases}x-1< 0\\x+4>0\end{cases}}\Rightarrow\hept{\begin{cases}x< 1\\x>-4\end{cases}}\Rightarrow-4< x< 1\left(tm\right)\)

Vậy -4 < x < 1

b) 5^{x + 2} - 5^{x - 1} = 3100

=> 5^x(5² - 1/5) = 3100

=> 5^x.124/5 = 3100

=> 5^x = 125

=> 5^x = 5³

=> x = 3

c) 3^{x + 1} - 3^{x - 2} = 702

=> \(3^x.3-3^x.\frac{1}{3^2}=702\)

=> 3^x(3 - 1/9) = 702

=> 3^x.26/9 = 702

=> 3^x = 243

=> 3^x = 3⁵

=> x = 5

Vậy x = 5

\(\left(x-1\right)\left(x+4\right)< 0\)

\(th1\orbr{\begin{cases}x-1>0\\x+4< 0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x>1\\x< -4\end{cases}\Leftrightarrow1< x< -4\left(1\right)}\)

\(th2\orbr{\begin{cases}x-1< 0\\x+4>0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x< 1\\x>-4\end{cases}\Leftrightarrow-4< x< 1\left(tm\right)}\)

vậy ....

b) \(5^{x+2}-5^{x-1}=3100\)

\(\Leftrightarrow5^x.\left(5^2-5^{-1}\right)=3100\)

\(\Leftrightarrow5^x.\frac{124}{5}=3100\)

\(\Leftrightarrow5^x=125\Leftrightarrow5^x=5^3\Leftrightarrow x=3\)

c)\(3^{x+1}-3^{x-2}=702\)

\(\Leftrightarrow3^x\left(3-3^{-2}\right)=702\)

\(\Leftrightarrow3^x.\frac{26}{9}=702\)

\(\Leftrightarrow3^x=243\Leftrightarrow3^x=3^5\Leftrightarrow x=5\)

a. 3100 : (5 x 2)

= 3100 : 10

= 310

b.  4 x 21 x 25

= ( 4 x 25) x 21

= 100 x 21

= 2100

\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{99}}\)

\(\Rightarrow\dfrac{A}{3}=\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\)

\(\Rightarrow A-\dfrac{A}{3}=\dfrac{2A}{3}=\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)-\left(\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\right)\)

\(\Rightarrow\dfrac{2A}{3}=\left(\dfrac{1}{3^2}-\dfrac{1}{3^2}\right)+\left(\dfrac{1}{3^3}-\dfrac{1}{3^3}\right)+...+\left(\dfrac{1}{3^{99}}-\dfrac{1}{3^{99}}\right)+\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)=\dfrac{1}{3}-\dfrac{1}{3^{100}}\)

\(\Rightarrow2A=3\cdot\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)\)

\(\Rightarrow\text{A}=\dfrac{1-\dfrac{1}{3^{99}}}{2}\)

\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2.3^{99}}< \dfrac{1}{2}\)

a: \(A=3^{100}-3^{99}+3^{98}-...+3^2-3\)

=>\(3A=3^{101}-3^{100}+3^{99}-...+3^3-3^2\)

=>\(4A=3^{101}-3\)

=>\(A=\dfrac{3^{101}-3}{4}\)

b: \(B=\left(-2\right)^0+\left(-2\right)^1+...+\left(-2\right)^{2024}\)

=>\(B\cdot\left(-2\right)=\left(-2\right)^1+\left(-2\right)^2+...+\left(-2\right)^{2025}\)

=>\(-2B-B=\left(-2\right)^1+\left(-2\right)^2+...+\left(-2\right)^{2025}-\left(-2\right)^0-\left(-2\right)^1-...-\left(-2\right)^{2024}\)

=>\(-3B=-2^{2025}-1\)

=>\(B=\dfrac{2^{2025}+1}{3}\)

c: \(C=\left(-\dfrac{1}{5}\right)^0+\left(-\dfrac{1}{5}\right)^1+...+\left(-\dfrac{1}{5}\right)^{2023}\)

=>\(\left(-\dfrac{1}{5}\right)\cdot C=\left(-\dfrac{1}{5}\right)^1+\left(-\dfrac{1}{5}\right)^2+...+\left(-\dfrac{1}{5}\right)^{2024}\)

=>\(\left(-\dfrac{6}{5}\right)\cdot C=\left(-\dfrac{1}{5}\right)^{2024}-\left(-\dfrac{1}{5}\right)^0\)

=>\(C\cdot\dfrac{-6}{5}=\dfrac{1}{5^{2024}}-1=\dfrac{1-5^{2024}}{5^{2024}}\)

=>\(C\cdot\dfrac{6}{5}=\dfrac{5^{2024}-1}{5^{2024}}\)

=>\(C=\dfrac{5^{2024}-1}{5^{2024}}:\dfrac{6}{5}=\dfrac{5^{2024}-1}{6\cdot5^{2023}}\)