tìm số hạng đầu và công sai của dãy số sau u2-3u8=-20 và u3*u4=24 14/07/2021 Bởi Eliza tìm số hạng đầu và công sai của dãy số sau u2-3u8=-20 và u3*u4=24
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: $\left \{ {{U_2-3U_8=-20} \atop {U_3.U_4=24}} \right.$ <=>$\left \{ {{U_1+d-3.(U_1+7d)=-20} \atop {(U_1+2d).(U_1+3d)=24}} \right.$ <=>$\left \{ {{U_1+d-3.U_1-21d=-20} \atop {U_1^2+5U1d+6d^2=24}} \right.$ <=>$\left \{ {{-U_1-10d=-10} \atop {U_1^2+5U1d+6d^2=24}} \right.$ <=>$\left \{ {{U_1=10-10d} \atop {(10-10d)^2+5.(10-10d)d+6d^2=24}} \right.$ <=>$\left \{ {{U_1=10-10d} \atop {100-200d+100d^2+50d-50d^2+6d^2=24}} \right.$ <=>$\left \{ {{U_1=10-10d} \atop {56d^2-150d+76=0}} \right.$ <=>$\left \{ {{U_1=10-10d} \atop {\left[ \begin{array}{l}d=2(n)\\d=\frac{19}{28}(l)\end{array} \right.}} \right.$ <=>$\left \{ {{U_1=-10} \atop {d=2}} \right.$ Bình luận
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: \(\begin{array}{l}\left\{ \begin{array}{l}{u_2} – 3{u_8} = – 20\\{u_3}.{u_4} = 24\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}{u_1} + d – 3\left( {{u_1} + 7d} \right) = – 20\\\left( {{u_1} + 2d} \right)\left( {{u_1} + 3d} \right) = 24\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} – 2{u_1} – 20d = – 20\\\left( {{u_1} + 2d} \right)\left( {{u_1} + 3d} \right) = 24\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}{u_1} = 10 – 10d\\\left( {10 – 8d} \right)\left( {10 – 7d} \right) = 24\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}{u_1} = 10 – 10d\\100 – 150d + 56{d^2} = 24\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}{u_1} = 10 – 10d\\56{d^2} – 150d + 76 = 0\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}d = 2\\{u_1} = – 10\end{array} \right.\\\left\{ \begin{array}{l}d = \frac{{19}}{{28}}\\{u_1} = \frac{{45}}{{14}}\end{array} \right.\end{array} \right.\end{array}\) Bình luận
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:
$\left \{ {{U_2-3U_8=-20} \atop {U_3.U_4=24}} \right.$
<=>$\left \{ {{U_1+d-3.(U_1+7d)=-20} \atop {(U_1+2d).(U_1+3d)=24}} \right.$
<=>$\left \{ {{U_1+d-3.U_1-21d=-20} \atop {U_1^2+5U1d+6d^2=24}} \right.$
<=>$\left \{ {{-U_1-10d=-10} \atop {U_1^2+5U1d+6d^2=24}} \right.$
<=>$\left \{ {{U_1=10-10d} \atop {(10-10d)^2+5.(10-10d)d+6d^2=24}} \right.$
<=>$\left \{ {{U_1=10-10d} \atop {100-200d+100d^2+50d-50d^2+6d^2=24}} \right.$
<=>$\left \{ {{U_1=10-10d} \atop {56d^2-150d+76=0}} \right.$
<=>$\left \{ {{U_1=10-10d} \atop {\left[ \begin{array}{l}d=2(n)\\d=\frac{19}{28}(l)\end{array} \right.}} \right.$
<=>$\left \{ {{U_1=-10} \atop {d=2}} \right.$
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:
\(\begin{array}{l}
\left\{ \begin{array}{l}
{u_2} – 3{u_8} = – 20\\
{u_3}.{u_4} = 24
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{u_1} + d – 3\left( {{u_1} + 7d} \right) = – 20\\
\left( {{u_1} + 2d} \right)\left( {{u_1} + 3d} \right) = 24
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
– 2{u_1} – 20d = – 20\\
\left( {{u_1} + 2d} \right)\left( {{u_1} + 3d} \right) = 24
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{u_1} = 10 – 10d\\
\left( {10 – 8d} \right)\left( {10 – 7d} \right) = 24
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{u_1} = 10 – 10d\\
100 – 150d + 56{d^2} = 24
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{u_1} = 10 – 10d\\
56{d^2} – 150d + 76 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
d = 2\\
{u_1} = – 10
\end{array} \right.\\
\left\{ \begin{array}{l}
d = \frac{{19}}{{28}}\\
{u_1} = \frac{{45}}{{14}}
\end{array} \right.
\end{array} \right.
\end{array}\)