\(u_3=u_1+2d;u_2=u_1+d;u_7=u_1+6d\\ Có:\left\{{}\begin{matrix}u_3-u_7=-8\\u_2.u_7=75\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\left(u_1+2d\right)-\left(u_1+6d\right)=-8\\\left(u_1+d\right).\left(u_1+6d\right)=75\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}-4d=-8\\u_1^2+7u_1d+6d^2=75\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}d=2\\u^2_1+14u_1+24=75\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}d=2\\u_1^2+14u_1-51=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}d=2\\u_1^2-3u_1+17u_1-51=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}d=2\\u_1\left(u_1-3\right)+17\left(u_1-3\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}d=2\\\left(u_1+17\right)\left(u_1-3\right)=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}d=2\\\left[{}\begin{matrix}u_1+17=0\\u_1-3=0\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}d=2\\\left[{}\begin{matrix}u_1=-17\\u_1=3\end{matrix}\right.\end{matrix}\right.\)
Bài 1:
\(a\in\left(-\Omega;-\dfrac{\Omega}{2}\right)\)
=>\(cosa< 0\)
\(sin^2a+cos^2a=1\)
=>\(cos^2a=1-\left(-\dfrac{2}{3}\right)^2=\dfrac{5}{9}\)
mà cosa<0
nên \(cosa=-\dfrac{\sqrt{5}}{3}\)
\(sin2a=2\cdot sina\cdot cosa=2\cdot\dfrac{-2}{3}\cdot\dfrac{-\sqrt{5}}{3}=\dfrac{4\sqrt{5}}{9}\)
