1. Cho (x, y) là nghiệm của hệ phương trình \(\left\{ \begin{array}{l}x - y = m + 1\\2x - 3y = m + 3\end{array} \right.

* Hướng dẫn giải

1)

\(\begin{array}{l}
\left\{ \begin{array}{l}
x - y = m + 1\\
2x - 3y = m + 3
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
3x - 3y = 3m + 3\\
2x - 3y = m + 3
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = 2m\\
y = x - m - 1
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x = 2m\\
y = m - 1
\end{array} \right.{\rm{    (}}\forall {\rm{m}} \in {\rm{R)}}
\end{array}\)

Ta có:

\(\begin{array}{l}
P = {x^2} + 8y = 4{m^2} + 8(m - 1) = 4{m^2} + 8m - 8\\
 = {\left( {2m + 2} \right)^2} - 12 \ge  - 12
\end{array}\)

Dấu “=” xẩy ra khi 2m + 2 = 0\( \Leftrightarrow m =  - 1\)

Giá trị nhỏ nhất của P là - 12 khi m = - 1

2) \(\left\{ \begin{array}{l}
{x^2} + {y^2} = 1\\
{x^3} - {y^3} =  - 1
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
{\left( {x - y} \right)^2} + 2xy = 1\\
{(x - y)^3} - 3xy\left( {x - y} \right) =  - 1
\end{array} \right.\)

Đặt \(\left\{ \begin{array}{l}
x - y = S\\
xy = P
\end{array} \right.\)

Ta có: \(\left\{ \begin{array}{l}
{S^2} + 2P = 1\\
{S^3} - 3SP =  - 1
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
P = \frac{{1 - {S^2}}}{2}\\
{S^3} - 3S.\frac{{1 - {S^2}}}{2} =  - 1
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
P = \frac{{1 - {S^2}}}{2}\\
2{S^3} + 3{S^3} - 3S + 2 = 0
\end{array} \right.\)

\(\begin{array}{l}
 \Leftrightarrow \left\{ \begin{array}{l}
P = \frac{{1 - {S^2}}}{2}\\
5{S^3} - 3S + 2 = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
P = \frac{{1 - {S^2}}}{2}\\
\left( {S + 1} \right)\left( {5{S^2} - 5S + 2} \right) = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
P = \frac{{1 - {S^2}}}{2}\\
\left( {S + 1} \right)\left( {5{S^2} - 5S + 2} \right) = 0
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
P = \frac{{1 - {S^2}}}{2}\\
\left[ \begin{array}{l}
\left( {S + 1} \right) = 0\\
5{S^2} - 5S + 2 = 0{\rm{  (vn)}}
\end{array} \right.
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
P = 0\\
S =  - 1
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x - y =  - 1\\
xy = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x = 0\\
y = 1
\end{array} \right.\\
\left\{ \begin{array}{l}
y = 0\\
x =  - 1
\end{array} \right.
\end{array} \right.
\end{array}\)