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\(\left(x-\frac{1}{2}\right)^2=0\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right)^2=0^2\)
\(\Leftrightarrow x-\frac{1}{2}=0\)
\(\Leftrightarrow x=\frac{1}{2}\)
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\(x\in\left\{3;1\right\}\)
\(\left(2x-1\right)^3=-8\)
\(\left(2x-1\right)^3=\left(-2\right)^3\)
\(2x-1=-2\)
\(2x=\left(-2\right)+1\)
\(2x=-1\)
\(x=-1\times2\)
\(x=-2\)
\(x\left(\frac{1}{2}\right)^2=\frac{1}{16}\)
\(x\left(\frac{1}{2}\right)^2=\left(\frac{1}{4}\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}x\frac{1}{2}=\frac{1}{4}\\x\frac{1}{2}=-\frac{1}{4}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{4}:\frac{1}{2}\\x=-\frac{1}{4}:\frac{1}{2}\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{1}{2}\end{cases}}}\)
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\(\left(\frac{1}{2}\right)^{2x-1}=\frac{1}{8}\)
\(\left(\frac{1}{2}\right)^{2x-1}=\left(\frac{1}{2}\right)^3\)
\(\Rightarrow2.x-1=3\)
\(2.x=3+1\)
\(2.x=4\)
\(x=4:2\)
\(x=2\)
Vậy \(x=2\)
Chúc bạn học tốt !!!
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(1/2)^(2x-1) = 1/8
(1/2)^(2x-1) = (1/2)^3
2x - 1 = 3
2x = 3 + 1
2x = 4
x = 4 : 2
x = 2
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a) m + 2 + 8 = 0 \(\Leftrightarrow\)m = ( - 10)
b) f(x) = x2 + 3x + 2
c) 1 + ( -3) + m = 0 \(\Leftrightarrow\)m = 2
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x13 = 27.x16
=> x13 - 27x16 = 0
=> x13(1 - 27x3) = 0
=> \(\orbr{\begin{cases}x^{13}=0\\1-27x^3=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=0\\27x^3=1\end{cases}}\)
=> \(\orbr{\begin{cases}x=0\\x^3=\frac{1}{27}\end{cases}}\)
=> \(\orbr{\begin{cases}x=0\\x=\frac{1}{3}\end{cases}}\)
c) \(\left(\frac{1}{2}\right)^{2x-1}=\frac{1}{8}\)
=> \(\left(\frac{1}{2}\right)^{2x-1}=\left(\frac{1}{2}\right)^3\)
=> \(2x-1=3\)
=> \(2x=3+1\)
=> \(2x=4\)
=> \(x=4:2=2\)
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\(\left(2x-1\right)^6=\left(2x-1\right)^8\\ \Rightarrow\left(2x-1\right)^6-\left(2x-1\right)^8=0\\ \Rightarrow\left(2x-1\right)^6\left[1-\left(2x-1\right)^2\right]=0\\ \Rightarrow\left[{}\begin{matrix}2x-1=0\\\left(2x-1\right)^2=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\2x-1=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=1\end{matrix}\right.\)
\(\left(2x-1\right)^6=\left(2x-1\right)^8\)
\(\Rightarrow\left[{}\begin{matrix}2x-1=-1\\2x-1=1\\2x-1=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\\x=\dfrac{1}{2}\end{matrix}\right.\)
#) TL :
\(\left(\frac{1}{2}\right)^{2x-1}=\frac{1}{8}\)
\(\left(\frac{1}{2}\right)^{2x-1}=\left(\frac{1}{2}\right)^3\)
\(\Rightarrow2x-1=3\)
\(\Rightarrow2x=4\)
\(\Rightarrow x=2\)
Chúc bn hok tốt ạ :33