Bài 2:
\(a,-2x^2+11x-9=0\)
\(\Delta=b^2-4ac=11^2-4.\left(-2\right).\left(-9\right)=49>0\Rightarrow\sqrt{\Delta}=7\)
\(\left\{{}\begin{matrix}x_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{-11+7}{2.\left(-2\right)}=1\\x_2=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{-11-7}{2.\left(-2\right)}=\dfrac{9}{2}\end{matrix}\right.\)
Vậy \(S=\left\{1;\dfrac{9}{2}\right\}\)
\(b,2x^2-\sqrt{3}x+\dfrac{1}{4}=0\)
\(\Delta=b^2-4ac=\left(-\sqrt{3}\right)^2-4.2.\dfrac{1}{4}=1>0\Rightarrow\sqrt{\Delta}=1\)
\(\left\{{}\begin{matrix}x_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{\sqrt{3}+1}{2.2}=\dfrac{1+\sqrt{3}}{4}\\x_2=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{\sqrt{3}-1}{2.2}=\dfrac{-1+\sqrt{3}}{4}\end{matrix}\right.\)
Vậy \(S=\left\{\dfrac{1+\sqrt{3}}{4};\dfrac{-1+\sqrt{3}}{4}\right\}\)
\(c,4x^2-4x=x^2-\dfrac{4}{3}\)
\(\Leftrightarrow4x^2-4x-x^2+\dfrac{4}{3}=0\)
\(\Leftrightarrow3x^2-4x+\dfrac{4}{3}=0\)
\(\Leftrightarrow x=\dfrac{2}{3}\)
Vậy \(S=\left\{\dfrac{2}{3}\right\}\)
Bài 3 : \(x^2+5x-15=0\)
\(+\)Do pt có 2 nghiệm pb \(x_1,x_2\)
nên theo đ/l Vi-ét , ta có :
\(\left\{{}\begin{matrix}S=x_1+x_2=-\dfrac{b}{a}=-5\\P=x_1x_2=\dfrac{c}{a}=-15\end{matrix}\right.\)
Ta có :
\(+\) \(x_1+x_2=S=-5\)
\(+\)\(x_1x_2=P=-15\)
\(+\dfrac{3}{x_1}+\dfrac{3}{x_2}=\dfrac{3x_2+3x_1}{x_1x_2}=\dfrac{3\left(x_1+x_2\right)}{x_1x_2}=\dfrac{3S}{P}=\dfrac{3.\left(-5\right)}{-15}=1\)
