Theo Vi-et, ta có:
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=\dfrac{9}{4}\\x_1x_2=\dfrac{c}{a}=\dfrac{3}{4}\end{matrix}\right.\)
\(P=x_1^2+x_1x_2+x_2^2\)
\(=\left(x_1+x_2\right)^2-x_1x_2\)
\(=\left(\dfrac{9}{4}\right)^2-\dfrac{3}{4}=\dfrac{81}{16}-\dfrac{3}{4}=\dfrac{81}{16}-\dfrac{12}{16}=\dfrac{69}{16}\)
\(Q=x_1^2-2x_1x_2+x_2^2\)
\(=\left(x_1-x_2\right)^2=\left(x_1+x_2\right)^2-4x_1x_2\)
\(=\left(\dfrac{9}{4}\right)^2-4\cdot\dfrac{3}{4}=\dfrac{81}{16}-3=\dfrac{81}{16}-\dfrac{48}{16}=\dfrac{33}{16}\)
Đúng 1
Bình luận (0)
