a, \(\left(5x-10\right)\left(6x-\frac{1}{3}\right)=0\\ \Rightarrow\left[{}\begin{matrix}5x-10=0\\6x-\frac{1}{3}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}5x=10\\6x=\frac{1}{3}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=\frac{1}{18}\end{matrix}\right.\)
Vậy \(x\in\left\{2;\frac{1}{18}\right\}\)
b, \(\frac{-3}{4}-\left|\frac{4}{5}-x\right|=-10\\ \frac{-3}{4}+10=\left|\frac{4}{5}-x\right|\\ \left|\frac{4}{5}-x\right|=\frac{37}{4}\\ \Rightarrow\left[{}\begin{matrix}\frac{4}{5}-x=\frac{37}{4}\\\frac{4}{5}-x=\frac{-37}{4}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{4}{5}-\frac{37}{4}\\x=\frac{4}{5}-\frac{-37}{4}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{-169}{20}\\x=\frac{201}{20}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{-169}{20};\frac{201}{20}\right\}\)
c, \(\left|5+x\right|-\frac{-2}{3}=3\\ \left|5+x\right|=3+\frac{-2}{3}\\ \left|5+x\right|=\frac{7}{3}\\ \Rightarrow\left[{}\begin{matrix}5+x=\frac{7}{3}\\5+x=\frac{-7}{3}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{7}{3}-5\\x=\frac{-7}{3}-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{-8}{3}\\x=\frac{-22}{3}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{-8}{3};\frac{-22}{3}\right\}\)
d, Xem lại đề nhé vì không xuất hiện x thì đẳng thức sai.
, thứ 4 là kiểm tra 90'p rồi
