Giải bài 8 trang 107 – SGK môn Đại số và Giải tích 11

a) 

$\begin{aligned} & \left\{ \begin{aligned} & 5{{u}_{1}}+10{{u}_{5}}=0 \\ & {{S}_{4}}=14 \\ \end{aligned} \right.\\&\Leftrightarrow \left\{ \begin{aligned} & {{u}_{1}}+2({{u}_{1}}+4d)=0 \\ & \dfrac{4\left[ {{u}_{1}}+\left( {{u}_{1}}+3d \right) \right]}{2}=14 \\ \end{aligned} \right. \\ & \Leftrightarrow \left\{ \begin{aligned} & 3{{u}_{1}}+8d=0 \\ & 4{{u}_{1}}+6d=14 \\ \end{aligned} \right.\\&\Leftrightarrow \left\{ \begin{aligned} & {{u}_{1}}=8 \\ & d=-3 \\ \end{aligned} \right. \\ \end{aligned}$

b) 

$\begin{aligned} & \left\{ \begin{aligned} & {{u}_{7}}+{{u}_{15}}=60 \\ & u_{4}^{2}+u_{12}^{2}=1170 \\ \end{aligned} \right.\\&\Leftrightarrow \left\{ \begin{aligned} & {{u}_{1}}+6d+{{u}_{1}}+14d=60 \\ & {{\left( {{u}_{1}}+3d \right)}^{2}}+{{\left( {{u}_{1}}+11d \right)}^{2}}=1170 \\ \end{aligned} \right. \\ & \Leftrightarrow \left\{ \begin{aligned} & 2{{u}_{1}}+20d=60 \\ & u_{1}^{2}+14{{u}_{1}}d+65{{d}^{2}}=585 \\ \end{aligned} \right. \\ & \Leftrightarrow \left\{ \begin{aligned} & {{u}_{1}}=30-10d \\ & {{\left( 30-10d \right)}^{2}}+14\left( 30-10d \right)d+65{{d}^{2}}=585 \\ \end{aligned} \right. \\ & \Leftrightarrow \left\{ \begin{aligned} & {{u}_{1}}=30-10d \\ & 25{{d}^{2}}-180d+315=0 \\ \end{aligned} \right. \\ & \Leftrightarrow \left[ \begin{aligned} & \left\{ \begin{aligned} & {{u}_{1}}=0 \\ & d=3 \\ \end{aligned} \right. \\ & \left\{ \begin{aligned} & {{u}_{1}}=-12 \\ & d=\dfrac{21}{5} \\ \end{aligned} \right. \\ \end{aligned} \right. \\ \end{aligned}$