<=>
\(\left(x+\dfrac{1}{5}\right)^2=\dfrac{26}{25}-\left(17.25\right)=-423,96\)
<=>
\(x^2+2.x.\dfrac{1}{5}+\left(\dfrac{1}{5}\right)^2=-423,96\)
<=>
\(x^2+\dfrac{2}{5}x+\dfrac{1}{25}+423,96=0\)
<=>
\(x^2+\dfrac{2}{5}x+424=0\)
xin lũi e coi trc vụ denta l9 nhe^^ :>>
\(\Delta=b^2-4ac=\left(\dfrac{2}{5}\right)^2-4.1.424=-1695,84< 0\)
=> pt vô nghiệm