Question

[Muối Họ Trần]

\(\frac{12}{16}=\frac{-x}{4}=\frac{21}{y}=\frac{z}{80}\) \((-0,6x-\frac{1}{2}).\frac{3}{4}-\left(-1\right)=\frac{1}{3}\)

\(\frac{1}{3}x+\frac{2}{5}\left(x-1\right)=0\)

\(\left(2x-3\right).\left(6-2x\right)=0\)

\(\frac{-2}{3}-\frac{1}{3}\left(2x-5\right)=\frac{3}{2}\)

\(2|\frac{1}{2}x-\frac{1}{3}|-\frac{3}{2}=\frac{1}{4}\)

\(\frac{3}{4}-2|2x-\frac{2}{3}|=2\)

Answer 1

[Phùng Tuệ Minh]

c) \(\left(2x-3\right).\left(6-2x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}2x-3=0\\6-2x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=3\\2x=6\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{3}{2}\\x=3\end{matrix}\right.\)

Vậy \(x\in\left\{\frac{3}{2};3\right\}\)

e) \(2\left|\frac{1}{2}x-\frac{1}{3}\right|-\frac{3}{2}=\frac{1}{4}\)

\(\Leftrightarrow2\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{1}{4}+\frac{3}{2}=\frac{7}{4}\)

\(\Leftrightarrow\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{7}{4}:2=\frac{7}{4}.\frac{1}{2}=\frac{7}{8}\)

\(\Rightarrow\left[{}\begin{matrix}\frac{1}{2}x-\frac{1}{3}=\frac{7}{8}\\\frac{1}{2}x-\frac{1}{3}=\left(-\frac{7}{8}\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{29}{12}\\x=\frac{-13}{12}\end{matrix}\right.\)

Vậy \(x\in\left\{\frac{29}{12};\frac{-13}{12}\right\}\)

Answer 2

[Trần Minh Hoàng]

Mấy bài này ko quá khó, tải MathPhoto trong đt về nó tự lm