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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\(x\in\left\{3;1\right\}\)
\(\left(2x-1\right)^3=-8\)
\(\left(2x-1\right)^3=\left(-2\right)^3\)
\(2x-1=-2\)
\(2x=\left(-2\right)+1\)
\(2x=-1\)
\(x=-1\times2\)
\(x=-2\)
\(x\left(\frac{1}{2}\right)^2=\frac{1}{16}\)
\(x\left(\frac{1}{2}\right)^2=\left(\frac{1}{4}\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}x\frac{1}{2}=\frac{1}{4}\\x\frac{1}{2}=-\frac{1}{4}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{4}:\frac{1}{2}\\x=-\frac{1}{4}:\frac{1}{2}\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{1}{2}\end{cases}}}\)
a/ \(\dfrac{x-7}{2}< 0\Rightarrow x-7< 0\Rightarrow x< 7\)
Vậy........
b/ \(\dfrac{x+3}{x+5}< 0\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+3< 0\\x+5>0\end{matrix}\right.\\\left\{{}\begin{matrix}x+3>0\\x+5< 0\end{matrix}\right.\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< -3\\x>-5\end{matrix}\right.\\\left\{{}\begin{matrix}x>-3\\x< -5\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow-5< x< -3\)
Vậy........
Bài 1:
a: \(\left|x\right|=\dfrac{1}{7}\)
b: |x|=1/7
c: \(\left|x\right|=\left|-3\dfrac{1}{5}\right|=\dfrac{16}{5}\)
d: |x|=0
Bài 2
a: \(=-2.853\)
b: \(=7.992\)
\(2x\left(x-\frac{1}{7}\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-\frac{1}{7}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{7}\end{cases}}\)
Vậy \(x=0\)hoặc \(x=\frac{1}{7}\)
\(2x\left(x-\frac{1}{7}\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}2x=0\\x-\frac{1}{7}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=0\\x=\frac{1}{7}\end{cases}}\)
Vậy ...