đặt t=x2 (t>0)
ta có pt: 3t2 -11t +10=0
Δ=(-11)2 -4.3.10=121 -120=1
t1=\(\dfrac{-b+\sqrt{\Delta}}{2a}\)=\(\dfrac{11+1}{6}=\)2
t2=\(\dfrac{-b-\sqrt{\Delta}}{2a}\)=\(\dfrac{11-1}{6}\)=\(\dfrac{5}{3}\)
khi t1=2⇒x=\(^{^+_-}\sqrt{2}\)
khi t2=\(\dfrac{5}{3}\)⇒x=\(^{^+_-\sqrt{\dfrac{5}{3}}}\)
vậy...
1) 3x4-11x2+10=0
Đặt x2=t \(\left(t\ge0\right)\rightarrow x^4=t\)
Ta có PT: 3t2-11t+10=0 (a=3;b=-11;c=10)
\(\Delta=b^2-4ac=\left(-11\right)^2-4.3.10=1\\ \rightarrow t_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{-\left(-11\right)+\sqrt{1}}{2.3}=2\left(t/m\right)\rightarrow x_1=\pm\sqrt{2}\\ \rightarrow t_2=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{-\left(-11\right)-\sqrt{1}}{2.3}=\dfrac{5}{3}\left(t/m\right)\rightarrow x_2=\pm\dfrac{\sqrt{15}}{3}\\ \rightarrow S=\left\{\sqrt{2};-\sqrt{2};\dfrac{\sqrt{15}}{3};-\dfrac{\sqrt{15}}{3}\right\}\)
