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(x - 5)2 = 16
=> (x - 5)2 = 42
=> \(\orbr{\begin{cases}x-5=4\\x-5=-4\end{cases}}\)
=> \(\orbr{\begin{cases}x=9\\x=1\end{cases}}\)
(2x - 1)3 = -64
=> (2x - 1)3 = -43
=> 2x - 1 = -4
=> 2x = -4 + 1
=> 2x = -3
=> x = -3/2
( x - 5)2 = 16
=> (x - 5)2 = 42
=> \(\orbr{\begin{cases}x-5=4\\x-5=-4\end{cases}}\)
=> \(\orbr{\begin{cases}x=9\\x=1\end{cases}}\)
a) \(2^x+2^{x+3}=144\)
\(2^x\left(1+2^3\right)=144\)
\(2^x.9=144\)
\(2^x=144:9=16\)
=> \(2^x=2^4\Rightarrow x=4\)
b) \(2^{x-1}+5.2^{x-2}=224\)
\(2^{x-2}\left(2+5\right)=224\)
\(2^{x-2}.7=224\Rightarrow2^{x-2}=32\Rightarrow2^{x-2}=2^5\)
=> x - 2 = 5 => x = 7
b) (5/2-3x)=25/9
3x = 5/2-25/9
3x =-5/18
x =-5/18:3
x=-5/54
\(e.\left(x-1\right)^5=-32\)
\(\left(x-1\right)^5=\left(-2\right)^5\)
\(x-1=-2\)
\(x\) \(=-2+1\)
\(x\) \(=-1\)
Vậy \(x=-1\)
\(\left(x-\frac{1}{2}\right)^2=0\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right)^2=0^2\)
\(\Leftrightarrow x-\frac{1}{2}=0\)
\(\Leftrightarrow x=\frac{1}{2}\)
Vậy x = 1/2
\(\left(x-2\right)^2=1\)
\(\Leftrightarrow\left(x-2\right)^2=1^2\)
\(\Leftrightarrow x-2=1\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=1\\x-2=-1\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}}\)
Vậy x = 3 hoặc x = 1
\(\left(2x-1\right)^3=-8\)
\(\Leftrightarrow\left(2x-1\right)^3=\left(-2\right)^3\)
\(\Leftrightarrow2x-1=-2\)
<=> 2x = -1
<=> x = -0,5
Vậy x = -0,5
\(\left(x-\frac{1}{2}\right)^2=0\)
\(x-\frac{1}{2}=0\)
\(x=\frac{1}{2}\)
\(\left(x-2\right)^2=1\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=1\\x-2=-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1+2\\x=-1+2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}\)
Vậy\(x\in\left\{3;1\right\}\)
\(\left(2x-1\right)^3=-8\)
\(\left(2x-1\right)^3=\left(-2\right)^3\)
\(2x-1=-2\)
\(2x=\left(-2\right)+1\)
\(2x=-1\)
\(x=-1\times2\)
\(x=-2\)
\(x\left(\frac{1}{2}\right)^2=\frac{1}{16}\)
\(x\left(\frac{1}{2}\right)^2=\left(\frac{1}{4}\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}x\frac{1}{2}=\frac{1}{4}\\x\frac{1}{2}=-\frac{1}{4}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{4}:\frac{1}{2}\\x=-\frac{1}{4}:\frac{1}{2}\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{1}{2}\end{cases}}}\)
\(x\)(\(x\) - 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 \(x\in\) \(\varnothing\)