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a) \(\left|3x-\frac{1}{2}\right|+\left|\frac{1}{2}y+\frac{3}{5}\right|=0\)
\(\Rightarrow\left|3x-\frac{1}{2}\right|=0\) \(\Rightarrow\left|\frac{1}{2}y+\frac{3}{5}\right|=0\)
\(\Rightarrow3x-\frac{1}{2}=0\) \(\Rightarrow\frac{1}{2}y+\frac{3}{5}=0\)
\(3x=\frac{1}{2}\) \(\frac{1}{2}y=\frac{-3}{5}\)
\(x=\frac{1}{2}:3\) \(y=\left(\frac{-3}{5}\right):\frac{1}{2}\)
\(x=\frac{1}{6}\) \(y=\frac{-6}{5}\)
KL: x = 1/6; y = -6/5
b) \(\left|\frac{3}{2}x+\frac{1}{9}\right|+\left|\frac{1}{5}y-\frac{1}{2}\right|\le0\)
mà \(\left|\frac{3}{2}x+\frac{1}{9}\right|>0;\left|\frac{1}{5}y-\frac{1}{2}\right|>0\)
\(\Rightarrow\left|\frac{3}{2}x+\frac{1}{9}\right|+\left|\frac{1}{5}y-\frac{1}{2}\right|>0\)
=> trường hợp |3/2x +1/9| + |1/5y -1/2| < 0 không thế xảy ra
\(\Rightarrow\left|\frac{3}{2}x+\frac{1}{9}\right|+\left|\frac{1}{5}y-\frac{1}{2}\right|=0\)
rùi bn lm tương tự như phần a nhé!
Giải:
Vì:
\(\left\{{}\begin{matrix}\left|3x-\dfrac{1}{2}\right|\ge0\\\left|\dfrac{1}{2}y+\dfrac{3}{5}\right|\ge0\end{matrix}\right.\)
Nên dấu "=" xảy ra khi và chỉ khi:
\(\left\{{}\begin{matrix}\left|3x-\dfrac{1}{2}\right|=0\\\left|\dfrac{1}{2}y+\dfrac{3}{5}\right|=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x-\dfrac{1}{2}=0\\\dfrac{1}{2}y+\dfrac{3}{5}=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x=\dfrac{1}{2}\\\dfrac{1}{2}y=-\dfrac{3}{5}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{6}\\y=-\dfrac{6}{5}\end{matrix}\right.\)
Vậy ...
b) \(\left|\dfrac{3}{2}x+\dfrac{1}{9}\right|+\left|\dfrac{1}{5}y-\dfrac{1}{2}\right|\le0\)
Vì:
\(\left\{{}\begin{matrix}\left|\dfrac{3}{2}x+\dfrac{1}{9}\right|\ge0\\\left|\dfrac{1}{5}y-\dfrac{1}{2}\right|\ge0\end{matrix}\right.\)
Dấu "=" xảy ra, khi và chỉ khi:
\(\left\{{}\begin{matrix}\left|\dfrac{3}{2}x+\dfrac{1}{9}\right|=0\\\left|\dfrac{1}{5}y-\dfrac{1}{2}\right|=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{2}x+\dfrac{1}{9}=0\\\dfrac{1}{5}y-\dfrac{1}{2}=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{2}x=-\dfrac{1}{9}\\\dfrac{1}{5}y=\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{2}{27}\\y=\dfrac{5}{2}\end{matrix}\right.\)
Vậy ...
Bài 1 : \(a,\left|x-3,5\right|=7,5\)
\(\Rightarrow\orbr{\begin{cases}x-3,5=7,5\\x-3,5=-7,5\end{cases}}\Rightarrow\orbr{\begin{cases}x=11\\x=-4\end{cases}}\)
\(b,\left|x+\frac{3}{4}\right|-\frac{1}{2}=0\)
\(\Rightarrow\left|x+\frac{3}{4}\right|=\frac{1}{2}\)
\(\Rightarrow\orbr{\begin{cases}x+\frac{3}{4}=\frac{1}{2}\\x+\frac{3}{4}=-\frac{1}{2}\end{cases}}\Rightarrow\orbr{\begin{cases}x=-\frac{1}{4}\\x=-\frac{5}{4}\end{cases}}\)
\(c,3,6-\left|x-0,4\right|=0\)
\(\Rightarrow\left|x-0,4\right|=3,6\)
\(\Rightarrow\orbr{\begin{cases}x-0,4=3,6\\x-0,4=-3,6\end{cases}}\Rightarrow\orbr{\begin{cases}x=4\\x=-3,2\end{cases}}\)
\(d,\left|x-\frac{1}{2}\right|-\frac{1}{3}=1\)
\(\Rightarrow\left|x-\frac{1}{2}\right|=\frac{4}{3}\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{2}=\frac{4}{3}\\x-\frac{1}{2}=-\frac{4}{3}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{11}{6}\\x=-\frac{5}{6}\end{cases}}\)
a) \(\frac{3}{7}x-\frac{1}{35}=\frac{3}{5}\)
\(\frac{3}{7}x=\frac{3}{5}+\frac{1}{35}\)
\(\frac{3}{7}x=\frac{22}{35}\)
\(x=\frac{49}{35}=1,4\)
b) \(1,5-x:\frac{1}{2}=\frac{1}{4}\)
\(x:\frac{1}{2}=1,5-\frac{1}{4}\)
\(x:\frac{1}{2}=\frac{5}{4}\)
\(x=\frac{5}{4}.\frac{1}{2}\)
\(x=\frac{5}{8}\)
Vậy ..