√2x²+2√2 .x –1=0
(a=√2;b=2√2=>b'=√2;c=-1)
∆'=(√2)²–√2 .-1=2+√2>0
=> pt có 2 n° phân biệt
x1=$\frac{-√2 +√(2+√2)}{√2}$
x2=$\frac{-√2 -√(2+√2)}{√2}$
Vậy....
b,$\frac{3}{5}$x²—x+ $\frac{2}{3}$=0
(a=$\frac{3}{5}$; b=—1; c=$\frac{2}{3}$ )
∆=(-1)²—4.$\frac{3}{5}$. $\frac{2}{3}$ =$\frac{-3}{5}$ <0
=> pt vô n°
