Question

giải phương trình ;

\(\dfrac{\left(7x+1\right).\left(x-2\right)}{10}+\dfrac{2}{5}=\dfrac{\left(x-2\right)^2}{5}+\dfrac{\left(x-1\right).\left(x-3\right)}{2}\)

Answer 1

\(\dfrac{\left(7x+1\right)\left(x-2\right)}{10}+\dfrac{2}{5}=\dfrac{\left(x-2\right)^2}{5}+\dfrac{\left(x-1\right)\left(x-3\right)}{2}\)

\(\Leftrightarrow\dfrac{7x^2-14x+x-2+4}{10}=\dfrac{2\left(x-2\right)^2+5\left(x^2-3x-x+3\right)}{10}\) \(\Leftrightarrow7x^2-13x+2=2x^2-8x+8+5x^2-15x-5x+15\)

\(\Leftrightarrow7x^2-13x+2=7x^2-28x+23\)

\(\Leftrightarrow7x^2-13x+2-7x^2+28x-23=0\)

\(\Leftrightarrow15x-21=0\)

\(\Leftrightarrow15x=21\)

\(\Leftrightarrow x=\dfrac{7}{5}\)

Vậy \(x=\dfrac{7}{5}\).