Câu trong ảnh làm như nào các bạn ơi...

a) Ta có $u_5=u_1+4d$ và $S_4=\frac{(u_1+u_4)\times 4}{2}=2(u_1+u_4)$ Thay vào hệ phương trình ta có: $\left\{\begin{array}{l}5u_1+10(u_1+4d)=0\\2(u_1+u_1+3d)=14\end{array}\right.$ $\Rightarrow \left\{\begin{array}{l}3u_1+8d=0\\2u_1+3d=7\end{array}\right.$ $\Rightarrow \left\{\begin{array}{l}6u_1+16d=0\\6u_1+9d=21\end{array}\right.$ $\Rightarrow 7d=-21$ $\Rightarrow d=-3$ $\Rightarrow u_1=8$ b) Ta có $u_7=u_1+6d; u_{15}=u_1+14d; u_4=u_1+3d; u_{12}=u_1+11d$ Thay vào hệ phương trình ta có: $\left\{\begin{array}{l}(u_1+6d)+(u_1+14d)=60\\(u_1+3d)^2+(u_1+11d)^2=1170\end{array}\right.$ $\Rightarrow \left\{\begin{array}{l}u_1+10d=30\\(u_1+3d)^2+(u_1+11d)^2=1170\end{array}\right.$ $\Rightarrow \left\{\begin{array}{l}u_1=30-10d\\(30-10d+3d)^2+(30-10d+11d)^2=1170\end{array}\right.$ $\Rightarrow \left\{\begin{array}{l}u_1=30-10d\\(30-7d)^2+(30+d)^2=1170\end{array}\right.$ $\Rightarrow \left\{\begin{array}{l}u_1=30-10d\\900-420d+49d^2+900+60d+d^2=1170\end{array}\right.$ $\Rightarrow \left\{\begin{array}{l}u_1=30-10d\\50d^2-360d+630=0\end{array}\right.$ $\Rightarrow \left\{\begin{array}{l}u_1=30-10d\\d^2-7,2d+12,6=0\end{array}\right.$ $\Rightarrow \left\{\begin{array}{l}u_1=30-10d\\(d-3)(d-4,2)=0\end{array}\right.$ $\Rightarrow \left[\begin{array}{l}\left\{\begin{array}{l}d=3\\u_1=0\end{array}\right.\\\left\{\begin{array}{l}d=4,2\\u_1=-12\end{array}\right.\end{array}\right.$