a. \(\Delta=5^2-4.3.2=1>0\) => P/t luôn có no (đpcm)
b. \(Theo\) Viet ta có : \(S=\dfrac{5}{3};P=\dfrac{2}{3}\)
c. Ta có : \(\dfrac{x_1-2}{x_2+2}+\dfrac{x_2-2}{x_1+2}=\dfrac{x_1^2-4+x_2^2-4}{\left(x_1+2\right)\left(x_2+2\right)}=\dfrac{\left(x_1+x_2\right)^2-2x_1x_2-8}{x_1x_2+2\left(x_1+x_2\right)+4}\)
\(=\dfrac{\left(\dfrac{5}{3}\right)^2-\dfrac{4}{3}-8}{\dfrac{2}{3}+2.\dfrac{5}{3}+4}=-\dfrac{59}{72}\)
\(a,\Delta=b^2-4ac=\left(-5\right)^2-4.3.2=1>0\)
\(\Rightarrow\) Pt luôn có 2 nghiệm phân biệt
\(\left\{{}\begin{matrix}x_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{5+1}{6}=1\\x_2=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{5-1}{6}=\dfrac{2}{3}\end{matrix}\right.\)
\(b,\left\{{}\begin{matrix}S=x_1+x_2=-\dfrac{b}{a}=\dfrac{5}{3}\\P=x_1x_2=\dfrac{c}{a}=\dfrac{2}{3}\end{matrix}\right.\)
\(c,\dfrac{x_1-2}{x_2+2}+\dfrac{x_2-2}{x_1+2}\)
\(=\dfrac{x_1^2-4+x_2^2-4}{\left(x_2+2\right)\left(x_1+2\right)}\)
\(=\dfrac{x_1^2+x_2^2-8}{x_1x_2+2x_2+2x_1+4}\)
\(=\dfrac{x_1^2+x_2^2-8}{x_1x_2+2\left(x_1+x_2\right)+4}\)
\(=\dfrac{S^2-2P-8}{P+2S+4}\)
\(=\dfrac{\dfrac{5}{3}^2-2.\dfrac{2}{3}-8}{\dfrac{2}{3}+2.\dfrac{5}{3}+4}\)
\(=-\dfrac{59}{72}\)
