Đáp án:

\[\left[ \begin{array}{l}
\left\{ \begin{array}{l}
d = 2\\
{u_1} =  - 10
\end{array} \right.\\
\left\{ \begin{array}{l}
d = \frac{{19}}{{28}}\\
{u_1} = \frac{{45}}{{14}}
\end{array} \right.
\end{array} \right.\]

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
\left\{ \begin{array}{l}
{u_2} - 3{u_8} =  - 20\\
{u_3}.{u_4} = 24
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
{u_1} + d - 3\left( {{u_1} + 7d} \right) =  - 20\\
\left( {{u_1} + 2d} \right)\left( {{u_1} + 3d} \right) = 24
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
 - 2{u_1} - 20d =  - 20\\
\left( {{u_1} + 2d} \right)\left( {{u_1} + 3d} \right) = 24
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
{u_1} = 10 - 10d\\
\left( {10 - 8d} \right)\left( {10 - 7d} \right) = 24
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
{u_1} = 10 - 10d\\
100 - 150d + 56{d^2} = 24
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
{u_1} = 10 - 10d\\
56{d^2} - 150d + 76 = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
d = 2\\
{u_1} =  - 10
\end{array} \right.\\
\left\{ \begin{array}{l}
d = \frac{{19}}{{28}}\\
{u_1} = \frac{{45}}{{14}}
\end{array} \right.
\end{array} \right.
\end{array}\)

Đáp án:

$\left \{ {{U_1=-10} \atop {d=2}} \right.$  

Giải thích các bước giải:

 Áp dụng công thức:$U_n=U_1+(n-1).d$

*$U_2=U_1+(2-1).d=U_1+d$

*$U_8=U_1+(8-1).d=U_1+7.d$

*$U_3=U_1+(3-1).d=U_1+2.d$

*$U_4=U_1+(4-1).d=U_1+3.d$

Ta có:

$\left \{ {{U_2-3U_8=-20} \atop {U_3.U_4=24}} \right.$ 

<=>$\left \{ {{U_1+d-3.(U_1+7d)=-20} \atop {(U_1+2d).(U_1+3d)=24}} \right.$ 

<=>$\left \{ {{U_1+d-3.U_1-21d=-20} \atop {U_1^2+5U1d+6d^2=24}} \right.$ 

<=>$\left \{ {{-U_1-10d=-10} \atop {U_1^2+5U1d+6d^2=24}} \right.$ 

<=>$\left \{ {{U_1=10-10d} \atop {(10-10d)^2+5.(10-10d)d+6d^2=24}} \right.$ 

<=>$\left \{ {{U_1=10-10d} \atop {100-200d+100d^2+50d-50d^2+6d^2=24}} \right.$ 

<=>$\left \{ {{U_1=10-10d} \atop {56d^2-150d+76=0}} \right.$ 

<=>$\left \{ {{U_1=10-10d} \atop {\left[ \begin{array}{l}d=2(n)\\d=\frac{19}{28}(l)\end{array} \right.}} \right.$ 

<=>$\left \{ {{U_1=-10} \atop {d=2}} \right.$