TÌM X BIẾT
\(0,5x-75\%x=\left(\frac{1}{3}-1\right)^2\)
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a) \(\left|0,5x-2\right|-\left|x+\frac{1}{3}\right|=0\)
=> \(\left|0,5x-2\right|=\left|x+\frac{1}{3}\right|\)
=> \(\orbr{\begin{cases}0,5x-2=x+\frac{1}{3}\\0,5x-2=-x-\frac{1}{3}\end{cases}}\)
=> \(\orbr{\begin{cases}-0,5x=\frac{7}{3}\\1,5x=\frac{5}{3}\end{cases}}\)
=> \(\orbr{\begin{cases}x=-\frac{14}{3}\\x=\frac{10}{9}\end{cases}}\)
b) \(2x-\left|x+1\right|=\frac{1}{2}\)
=> \(\left|x+1\right|=2x-\frac{1}{2}\) (Đk: \(2x-\frac{1}{2}\ge0\) <=> \(x\ge\frac{1}{4}\))
=> \(\orbr{\begin{cases}x+1=2x-\frac{1}{2}\\x+1=\frac{1}{2}-2x\end{cases}}\)
=> \(\orbr{\begin{cases}-x=-\frac{3}{2}\\3x=-\frac{1}{2}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{3}{2}\\x=-\frac{1}{6}\end{cases}}\)
\(\Leftrightarrow\left(\dfrac{2}{3}x-1\right)^3\cdot\dfrac{1}{50}=\dfrac{75}{2}-24=\dfrac{27}{2}\)
\(\Leftrightarrow\left(\dfrac{2}{3}x-1\right)^3=\dfrac{27}{2}:\dfrac{1}{50}=675\)
\(\Leftrightarrow x\cdot\dfrac{2}{3}-1=3\sqrt[3]{25}\)
\(\Leftrightarrow x\cdot\dfrac{2}{3}=3\sqrt[3]{25}+1\)
\(\Leftrightarrow x=\dfrac{3\left(3\sqrt[3]{25}+1\right)}{2}\)
b) \(\left(5x-1\right)\left(2x-\frac{1}{3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-1=0\\2x-\frac{1}{3}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}5x=1\\2x=\frac{1}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{5}\\x=\frac{1}{6}\end{matrix}\right.\)
e, \(-\frac{3}{4}-\left|\frac{4}{5}-x\right|=-1\)
\(\Leftrightarrow\left|\frac{4}{5}-x\right|=-\frac{3}{4}-\left(-1\right)\)
\(\Leftrightarrow\left|\frac{4}{5}-x\right|=\frac{1}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}\frac{4}{5}-x=\frac{1}{4}\\\frac{4}{5}-x=-\frac{1}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{7}{15}\\x=1,05\end{matrix}\right.\)
Vậy ....
\(\left(\frac{x}{-5}+1\frac{1}{2}\right):\frac{28}{75}-1,4.\frac{15}{49}=\left|-\frac{2}{3}\right|.\left(-\frac{3}{2}\right)^3\)
\(\left(\frac{x}{-5}+\frac{3}{2}\right).\frac{75}{28}-\frac{14}{10}.\frac{15}{49}=\frac{2}{3}.\frac{-27}{8}\)
\(\left(\frac{-x}{5}+\frac{3}{2}\right).\frac{75}{28}-\frac{3}{7}=\frac{-9}{4}\)
\(\left(\frac{-x}{5}+\frac{3}{2}\right).\frac{75}{28}=\frac{-9}{4}+\frac{3}{7}\)
\(\left(\frac{-x}{5}+\frac{3}{2}\right).\frac{75}{28}=\frac{-63}{28}+\frac{12}{28}\)
\(\left(\frac{-x}{5}+\frac{3}{2}\right).\frac{75}{28}=\frac{-51}{28}\)
\(\frac{-x}{5}+\frac{3}{2}=\frac{-51}{28}:\frac{75}{28}\)
\(\frac{-x}{5}+\frac{3}{2}=\frac{-51}{28}.\frac{28}{75}\)
\(\frac{-x}{5}+\frac{3}{2}=\frac{-17}{25}\)
\(\frac{-x}{5}=\frac{-17}{25}-\frac{3}{2}\)
\(\frac{-x}{5}=\frac{-34}{50}-\frac{75}{50}\)
\(\frac{-x}{5}=\frac{-109}{50}\)
\(\frac{-10x}{50}=\frac{-109}{50}\)
Hình như đề sai thì phải
=>1/2:(0,5x-1,5)=0,2-0,35=-0,15=-3/20
=>0,5x-1,5=1/2:(-3/20)=-1/2*20/3=-10/3
=>0,5x=-10/3+3/2=-11/6
=>x=-11/3
\(0,5x-75\%x=\left(\frac{1}{3}-1\right)^2\)
\(\Leftrightarrow\frac{1}{2}x-\frac{3}{4}x=\left(-\frac{2}{3}\right)^2\)
\(\Leftrightarrow-\frac{1}{4}x=\frac{4}{9}\)
\(\Leftrightarrow x=-\frac{16}{9}\)
\(0,5x-75\%x=\left(\frac{1}{3}-1\right)^2\)
\(< =>\frac{1}{2}x-\frac{3}{4}x=\left(-\frac{2}{3}\right)^2\)
\(< =>-\frac{1}{4}x=\frac{4}{9}\)
\(=>x=\frac{-16}{9}\)