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Bạn ơi hình như sai đề bài rồi. Phải là 5x chứ. Mk ngồi từ nãy giờ tính ko ra.
Ta có :
25 < 2^x < 3125
Suy ra : 5^2 < 2^x < 5^5
Không liên quan gì bạn ơi
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\(5^x.5^{x+1}.5^{x+2}=3125\)
\(\Rightarrow5^{x+x+1+x+2}=5^5\)
\(\Rightarrow5^{3x+3}=5^5\)
\(\Rightarrow3x+3=5\)
\(\Rightarrow3x=2\)
\(\Rightarrow x=\frac{2}{3}\)
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\(25\le5^n< 3125\)
\(\Rightarrow5^2\le5^n< 5^5\)
\(\Rightarrow2\le n< 5\)
Vậy \(n=\left\{2;3;4\right\}\)
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\(25< 5^x< 3125\)
\(\Leftrightarrow5^2\le5^x< 5^5\)
\(\Leftrightarrow2\le x< 5\)
\(\Leftrightarrow x\in\left\{2;3;4\right\}\)
Vậy \(x\in\left\{2;3;4\right\}\) là giá trị cần tìm
\(25\le5^x< 3125\)
\(\Leftrightarrow5^2\le5^x< 5^5\)
\(\Rightarrow2\le x< 5\)
\(\Rightarrow x=\left\{2;3;4\right\}\)
(3X-7)^5 =3125
= (3X-7)^5 = 5^5
= 3X-7 = 5
= 3X = 5 + 7
= 3X = 12
= X = 12 : 3
= X = 4
\(\Leftrightarrow\left(3x-7\right)^5=5^5\Rightarrow3x-7=5\Leftrightarrow x=4\)