Question

giả phương tình

(\(\dfrac{2}{5}\)x-1).(x-1)+5=2x

 

Answer 1

\(\left(\dfrac{2}{5}x-1\right)\left(x-1\right)+5=2x\\ \Leftrightarrow\dfrac{2}{5}x^2-\dfrac{2}{5}x-x+1=2x-5\\ \Leftrightarrow\dfrac{2}{5}x^2-\dfrac{7}{5}x+1=2x-5\\ \Leftrightarrow\dfrac{2}{5}x^2-\dfrac{7}{5}x-2x+1+5=0\\ \Leftrightarrow\dfrac{2}{5}x^2-\dfrac{17}{5}x+6=0\\ \Leftrightarrow\dfrac{2}{5}x\left(x-6\right)-\left(x-6\right)=0\\ \Leftrightarrow\left(x-6\right)\left(\dfrac{2}{5}x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\\dfrac{2}{5}x-1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=6\\x=\dfrac{5}{2}\end{matrix}\right.\)

Vậy Phương trình có tập nghiệm `S={6;5/2}`