Bài 2:
\(\left\{{}\begin{matrix}\left(2x-\dfrac{1}{2}\right)^2\ge0\\\left(y+\dfrac{1}{2}\right)^2\ge0\\\left(z-\dfrac{1}{3}\right)^2\ge0\end{matrix}\right.\Rightarrow\left(2x-\dfrac{1}{2}\right)^2+\left(y+\dfrac{1}{2}\right)^2+\left(z-\dfrac{1}{3}\right)^2\ge0\)Mà \(\left(2x-\dfrac{1}{2}\right)^2+\left(y+\dfrac{1}{2}\right)^2+\left(z-\dfrac{1}{3}\right)^2=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left(2x-\dfrac{1}{2}\right)^2=0\\\left(y+\dfrac{1}{2}\right)^2=0\\\left(z-\dfrac{1}{3}\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{4}\\y=\dfrac{-1}{2}\\z=\dfrac{1}{3}\end{matrix}\right.\)
Vậy \(x=\dfrac{1}{4},y=\dfrac{-1}{2},z=\dfrac{1}{3}\)
1)
a) \(2x+\dfrac{5}{2}=\dfrac{7}{2}\)
\(\Leftrightarrow2x=\dfrac{7}{2}-\dfrac{5}{2}\)
\(\Leftrightarrow2x=1\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
Vậy \(x=\dfrac{1}{2}\)
b) \(\left|5-\dfrac{1}{2}x\right|=\left|-\dfrac{1}{5}\right|\)
\(\Leftrightarrow\left|5-\dfrac{1}{2}x\right|=\dfrac{1}{5}\)
\(\Leftrightarrow\left[{}\begin{matrix}5-\dfrac{1}{2}x=\dfrac{1}{5}\\5-\dfrac{1}{2}x=-\dfrac{1}{5}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{48}{5}\\x=\dfrac{52}{5}\end{matrix}\right.\)
Vậy \(x_1=\dfrac{48}{5};x_2=\dfrac{52}{5}\)
a/ \(2x+\dfrac{5}{2}=\dfrac{7}{2}\)
\(\Rightarrow2x=1\Rightarrow x=\dfrac{1}{2}\)
b/ \(\left|5-\dfrac{1}{2}x\right|=\left|-\dfrac{1}{5}\right|\)
\(\Rightarrow\left|5-\dfrac{1}{2}x\right|=\dfrac{1}{5}\)
\(\Rightarrow\left[{}\begin{matrix}5-\dfrac{1}{2}x=\dfrac{1}{5}\\5-\dfrac{1}{2}x=\dfrac{1}{5}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{12}{5}\\x=\dfrac{13}{5}\end{matrix}\right.\)
Bài 2:
Vì \(\left\{{}\begin{matrix}\left(2x-\dfrac{1}{2}\right)^2\ge0\forall x\\\left(y+\dfrac{1}{2}\right)^2\ge0\forall y\\\left(x-\dfrac{1}{3}\right)^2\ge0\forall z\end{matrix}\right.\)
=> Để bt = 0 thì\(\left\{{}\begin{matrix}\left(2x-\dfrac{1}{2}\right)^2=0\\\left(y+\dfrac{1}{2}\right)^2=0\\\left(z-\dfrac{1}{3}\right)^2=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2x-\dfrac{1}{2}=0\\y+\dfrac{1}{2}=0\\z-\dfrac{1}{3}=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{4}\\y=-\dfrac{1}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy.....................
1) a) \(2x+\dfrac{5}{2}=\dfrac{7}{2}\) \(\Leftrightarrow\) \(2x=\dfrac{7}{2}-\dfrac{5}{2}\) \(\Leftrightarrow\) \(2x=1\) \(\Leftrightarrow\) \(x=\dfrac{1}{2}\)
b) \(\left|5-\dfrac{1}{2}x\right|=\left|\dfrac{-1}{5}\right|\) \(\Leftrightarrow\) \(\left(\left|5-\dfrac{1}{2}x\right|\right)^2=\left(\left|\dfrac{-1}{5}\right|\right)^2\)
\(\Leftrightarrow\) \(\left(5-\dfrac{1}{2}x\right)^2=\dfrac{1}{25}\) \(\Leftrightarrow\) \(5-\dfrac{1}{2}x=\pm\dfrac{1}{5}\)
th1 : \(5-\dfrac{1}{2}x=\dfrac{1}{5}\) \(\Leftrightarrow\) \(-\dfrac{1}{2}x=\dfrac{1}{5}-5=-\dfrac{24}{5}\)
\(\Leftrightarrow\) \(x=-\dfrac{24}{5}:-\dfrac{1}{2}=\dfrac{48}{5}\)
th2 : \(5-\dfrac{1}{2}x=-\dfrac{1}{5}\) \(\Leftrightarrow\) \(-\dfrac{1}{2}x=-\dfrac{1}{5}-5=-\dfrac{26}{5}\)
\(x=-\dfrac{26}{5}:-\dfrac{1}{2}=\dfrac{52}{5}\)
vậy \(x=\dfrac{48}{5};x=\dfrac{52}{5}\)
2) ta có : \(\left\{{}\begin{matrix}\left(2x-\dfrac{1}{2}\right)^2\ge0\\\left(y+\dfrac{1}{2}\right)^2\ge0\\\left(z-\dfrac{1}{3}\right)^2\ge0\end{matrix}\right.\) \(\Rightarrow\) \(\left(2x-\dfrac{1}{2}\right)^2+\left(y+\dfrac{1}{2}\right)^2+\left(z-\dfrac{1}{3}\right)^2\ge0\)
vậy \(\left(2x-\dfrac{1}{2}\right)^2+\left(y+\dfrac{1}{2}\right)^2+\left(z-\dfrac{1}{3}\right)^2=0\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}\left(2x-\dfrac{1}{2}\right)^2=0\\\left(y+\dfrac{1}{2}\right)^2=0\\\left(z-\dfrac{1}{3}\right)^2=0\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}2x-\dfrac{1}{2}=0\\y+\dfrac{1}{2}=0\\z-\dfrac{1}{3}=0\end{matrix}\right.\) \(\Rightarrow\) \(\left\{{}\begin{matrix}x=\dfrac{1}{4}\\y=-\dfrac{1}{2}\\z=\dfrac{1}{3}\end{matrix}\right.\)
vậy \(x=\dfrac{1}{4};y=-\dfrac{1}{2};z=\dfrac{1}{3}\)
