x.(x-2)= 16; 2 mũ x+3.2 mũ x=144
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\(A=\dfrac{\left(x-4\right)\left(x+4\right)}{x}\cdot\dfrac{x}{\left(x-4\right)^2}=\dfrac{x+4}{x-4}\)
Để A=2 thì 2x-8=x+4
=>x=12
![](https://rs.olm.vn/images/avt/0.png?1311)
Đề bài sai, pt này ko giải được
Đề đúng: \(\dfrac{4x^2+16}{x^2+6}=...\)
Mẫu số bên trái thừa mất số 1
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1. \(\text{6 + 2.(x - 19) = 16}.\)
\(\Leftrightarrow2.\left(x-19\right)=10.\)
\(\Leftrightarrow x-19=5.\)
\(\Leftrightarrow x=24.\)
Vậy \(x=24.\)
2. \(\text{(-240) : x – 16 = 64}.\)
\(\Leftrightarrow\left(-240\right):x=80.\)
\(\Leftrightarrow x=-3.\)
Vậy \(x=-3.\)
3. \(2x^3=16.\)
\(\Leftrightarrow x^3=8.\)
\(\Leftrightarrow x=2.\)
Vậy \(x=2.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=\dfrac{\left(x-4\right)\cdot\left(x+4\right)}{x}\cdot\dfrac{x}{\left(x-4\right)^2}=\dfrac{x+4}{x-4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) 2^x . 16^2 = 1024 b) 64 . 4^x = 16^8 c) 2^x = 16
=> 2^x . 256 = 1024 => 64 . 4^x = (4^2) ^ 8 => 2^x = 2^4
=> 2^x = 1024 : 256 => 4^3 . 4^x = 4^16 => x = 4
=> 2^x = 4 => 4^x = 4^16 : 4^3
=> 2^x = 2^2 => 4^x = 4^13
=> x = 13
=> x = 2
a) \(2^x.16^2=1024\Rightarrow2^x=1024:16^2=2^{10}:\left(2^4\right)^2=2^{10}:2^8=2^2\)\(\Rightarrow x=2\)
b) \(64.4^x=16^8\Rightarrow4^x=16^8:64=\left(4^2\right)^8:4^3=4^{16}:4^3=4^{13}\Rightarrow x=13\)
c)\(2^x=16\Rightarrow2^x=2^4\Rightarrow x=4\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x\)(\(x\) - 2) = 16
\(x^2\) - 2\(x\) - 16 = 0
(\(x^2\) - \(x\)) - (\(x\) - 1) - 17 = 0
\(x\)(\(x\) - 1) - (\(x-1\)) = 17
(\(x\) - 1)(\(x\) - 1) = 17
(\(x-1\))2 = 17
\(\left[{}\begin{matrix}x-1=\sqrt{17}\\x-1=-\sqrt{17}\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=1+\sqrt{17}\\x=1-\sqrt{17}\end{matrix}\right.\)
Vậy: \(\in\) {1 - \(\sqrt{17}\); 1 + \(\sqrt{17}\)}
Bài 2:
2\(x\) + 3.2\(x\) = 144
2\(x\).(1 + 3) = 144
2\(x\).4 = 144
2\(x\) = 144 : 4
2\(x\) = 36
2\(^x\) = 36
Nếu \(x\) = 6 ⇒ 2\(^x\) = 64 > 36 (loại)
Nếu \(x\) ≤ 5 ⇒2\(^x\) ≤ 25 = 32 < 36 (loại)
Vậy \(x\in\) \(\varnothing\)