Mới sửa lại phần tương đương nhé :(
\(-5\left(4x-1\right)\left(x-2\right)=2\left(4x-1\right)^2\)
\(\Leftrightarrow-5\left(4x-1\right)\left(x-2\right)-2\left(4x-1\right)^2=0\)
\(\Leftrightarrow\left(4x-1\right)\left[-5\left(x-2\right)-2\left(4x-1\right)\right]=0\)
\(\Leftrightarrow\left(4x-1\right)\left(-5x+10-8x+2\right)=0\)
\(\Leftrightarrow\left(4x-1\right)\left(-13x+12\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}4x-1=0\\-13x+12=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=\dfrac{12}{13}\end{matrix}\right.\)
Vậy: Phương trình có tập nghiệm \(S=\left\{\dfrac{1}{4};\dfrac{12}{13}\right\}\)
\(-5\left(4x-1\right)\left(x-2\right)=2\left(4x-1\right)^2\)
\(\Leftrightarrow-5\left(4x-1\right)\left(x-2\right)-2\left(4x-1\right)^2=0\)
\(\Leftrightarrow\left(4x-1\right)\left[-5\left(x-2\right)-2\left(4x-1\right)\right]=0\)
\(\Leftrightarrow\left(4x-1\right)\left(-5x+10-8x+2\right)=0\)
\(\Leftrightarrow\left(4x-1\right)\left(-13x+12\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}4x-1=0\\-13x+12=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{4}\\x=\dfrac{12}{13}\end{matrix}\right.\)
Vậy: Phương trình có tập nghiệm \(S=\left\{\dfrac{1}{4};\dfrac{12}{13}\right\}\)
\(-5\left(4x-1\right)\left(x-2\right)=2\left(4x-1\right)^2\)
\(\Leftrightarrow-5\left(4x-1\right)\left(x-2\right)-2\left(4x-1\right)\left(4x-1\right)=0\)
\(\Leftrightarrow\left(4x-1\right)\left(3x+8\right)=0\)
\(\left\{{}\begin{matrix}4x-1=0\\3x+8=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{4}\\x=-\dfrac{8}{3}\end{matrix}\right.\)
Vậy pt có tập nghiệm \(S\in\left\{\dfrac{1}{4};-\dfrac{8}{3}\right\}\)
