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Sửa đề: x<-2/3
=>x+2/3<0
=>6x+4<0
|6x+4|+3x=-6x-4+3x=-3x-4
a) \(x\ge\frac{2}{3}\Rightarrow3x-2\ge0\Rightarrow\left|3x-2\right|=3x-2\)
\(\Rightarrow P=\frac{1}{2}-\frac{1}{2}\left(x:\frac{1}{6}-\frac{1}{4}\right)-2\left(3x-2\right)\)
\(\Rightarrow P=\frac{1}{2}-\frac{1}{2}\left(6x-\frac{1}{4}\right)-6x+4\)
\(\Rightarrow P=\frac{4}{8}-3x+\frac{1}{8}-6x+\frac{32}{8}\)
\(\Rightarrow P=\frac{37}{8}-9x\)
b) \(x< \frac{2}{3}\Rightarrow3x-2< 0\Rightarrow\left|3x-2\right|=2-3x\)
\(\Rightarrow P=\frac{1}{2}-\frac{1}{2}\left(x:\frac{1}{6}-\frac{1}{4}\right)-2\left(2-3x\right)\)
\(\Rightarrow P=\frac{1}{2}-\frac{1}{2}\left(6x-\frac{1}{4}\right)-4+6x\)
\(\Rightarrow P=\frac{4}{8}-3x+\frac{1}{8}-\frac{32}{8}+6x\)
\(\Rightarrow P=\frac{-27}{8}+3x\)
\(M=\frac{-2x}{3}+3x\left(\frac{x}{6}-\frac{-2}{9}-\frac{7}{5}\right)-\frac{5x}{2}\left(\frac{x}{5}-\frac{4}{5}\right)\)
\(M=\frac{-2x}{3}+3x\left[\frac{x}{6}-\left(-\frac{2}{9}\right)-\frac{7}{5}\right]-\frac{5x}{4}\left(\frac{x}{5}-\frac{4}{5}\right)\)
\(M=\frac{-2x}{3}+3x\left(\frac{x}{6}+\frac{2}{9}-\frac{7}{5}\right)-\frac{5x}{2}\left(\frac{x}{5}-\frac{4}{5}\right)\)
\(M=-\frac{2x}{3}+3x\left(\frac{x}{6}-\frac{53}{45}\right)-\frac{5x}{2}.\frac{x-4}{5}\)
\(M=-\frac{2x}{3}+3x\left(\frac{x}{6}-\frac{53}{45}\right)-\frac{5x\left(x-4\right)}{10}\)
\(M=-\frac{2x}{3}+3x\left(\frac{x}{6}-\frac{53}{45}\right)-\frac{x\left(x-4\right)}{2}\)
\(M=-\frac{2x}{3}+\frac{x^2}{2}-\frac{53x}{15}-\frac{x\left(x-4\right)}{2}\)
\(M=\left(-\frac{2x}{3}-\frac{53x}{15}\right)+\frac{x^2}{2}-\frac{x\left(x-4\right)}{2}\)
\(M=-\frac{21x}{5}+\frac{x^2}{2}-\frac{x\left(x-4\right)}{2}\)
\(M=\frac{-2.21x+5x^2-5x\left(x-4\right)}{10}\)
\(M=\frac{-42x+5x^2-5x\left(x-4\right)}{10}\)
\(M=\frac{-x\left[42-5x+5\left(x-4\right)\right]}{10}\)
\(M=\frac{-x\left(42-5x+5x-20\right)}{10}\)
\(M=\frac{-x\left(42-20\right)}{10}\)
\(M=\frac{-x.22}{10}\)
\(M=-\frac{22x}{10}\)
\(M=-\frac{11x}{5}\)