Mình đặt là a, b, c cho dễ nhé
a) \(3x-\left|x\right|=2x\)
\(\Leftrightarrow\)\(\left|x\right|=3x-2x\)
+) Nếu \(x\ge0\) ta có :
\(x=3x-2x\)
\(\Leftrightarrow\)\(x=x\) ( thoã mãn )
+) Nếu \(x< 0\) ta có :
\(-x=3x-2x\)
\(\Leftrightarrow\)\(-x=x\) ( loại )
Vậy \(x\ge0\) là tập hợp các giá trị x thoã mãn đề bài
b) \(\left(3x-1\right)^2=25\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}\left(3x-1\right)^2=5^2\\\left(3x-1\right)^2=\left(-5\right)^2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x-1=5\\3x-1=-5\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}3x=5+1\\3x=-5+1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=6\\3x=-4\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=\dfrac{6}{3}\\x=\dfrac{-4}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{-4}{3}\end{matrix}\right.\)
Vậy \(x=2\) hoặc \(x=\dfrac{-4}{3}\)
c) \(25x^3-4x=0\)
\(\Leftrightarrow\)\(x\left(25x^2-4\right)=0\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=0\\25x^2-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\\left(5x\right)^2-2^2=0\end{matrix}\right.\)
Từ \(\left(5x\right)^2-2^2=0\) suy ra \(\left(5x-2\right)\left(5x+2\right)=0\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}5x-2=0\\5x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=0+2\\5x=0-2\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}5x=2\\5x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{5}\\x=\dfrac{-2}{5}\end{matrix}\right.\)
Vậy \(x=0\) ; \(x=\dfrac{2}{5}\) hoặc \(x=\dfrac{-2}{5}\)
Chúc bạn học tốt ~
