+) \(5\frac{2}{3}x+1\frac{2}{3}=4\frac{1}{2}\Leftrightarrow\frac{17}{3}x+\frac{5}{3}=\frac{9}{2}\Leftrightarrow\frac{17}{3}x=\frac{17}{6}\Leftrightarrow x=\frac{1}{2}\)
+) \(\frac{x}{27}=\frac{-2}{9}\Leftrightarrow x=\frac{-2}{9}.27=-6\)
+) \(\left|x+1,5\right|=2\Leftrightarrow\orbr{\begin{cases}x+1,5=2\\x+1,5=-2\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0,5\\x=-3,5\end{cases}}}\)
+) \(A=\left|x-1004\right|-\left|x+1003\right|\)
Ta có BĐT \(\left|x\right|-\left|y\right|\le\left|x-y\right|,\)dấu "=" xảy ra khi và chỉ khi x,y cùng dấu hay \(xy\ge0\)
Áp dụng: \(A=\left|x-1004\right|-\left|x+1003\right|\le\left|x-1004-x-1003\right|=\left|-2007\right|=2007\)
Vậy \(maxA=2007\Leftrightarrow\left(x-1004\right)\left(x+1003\right)\ge0\Leftrightarrow\orbr{\begin{cases}x\ge1004\\x\le-1003\end{cases}}\)
