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Giải các hệ phương trình sau
\(\left\{ \matrix{
{x^2} – xy = 28 \hfill \cr
{y^2} – xy = – 12; \hfill \cr} \right.\)
\(\left\{ \matrix{
5(x + y) + 2xy = – 19 \hfill \cr
15xy + 5(x + y) = – 175. \hfill \cr} \right.\)
Gợi ý làm bài
a)
\(\eqalign{
& \left\{ \matrix{
x_{}^2 – xy = 28 \hfill \cr
y_{}^2 – xy = – 12 \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{
x_{}^2 – 2xy + y_{}^2 = 16 \hfill \cr
x_{}^2 – xy = 28 \hfill \cr} \right. \cr
& \Leftrightarrow \left\{ \matrix{
(x – y)_{}^2 = 16 \hfill \cr
x(x – y) = 28 \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{
\left[ \matrix{
x – y = 4 \hfill \cr
x – y = – 4 \hfill \cr} \right. \hfill \cr
x(x – y) = 28 \hfill \cr} \right. \cr} \)
\( \Leftrightarrow \left[ \matrix{
\left\{ \matrix{
x – y = 4 \hfill \cr
x(x – y) = 28 \hfill \cr} \right. \hfill \cr
\left\{ \matrix{
x – y = – 4 \hfill \cr
x(x – y) = 28 \hfill \cr} \right. \hfill \cr} \right. \Leftrightarrow \left[ \matrix{
\left\{ \matrix{
x = 7 \hfill \cr
y = 3 \hfill \cr} \right. \hfill \cr
\left\{ \matrix{
x = – 7 \hfill \cr
y = – 3 \hfill \cr} \right. \hfill \cr} \right.\)
b)
\(\left\{ \matrix{
5(x + y) + 2xy = – 19 \hfill \cr
15xy + 5(x + y) = – 175 \hfill \cr} \right.\)
Đặt \(\left\{ \matrix{
x + y = a \hfill \cr
xy = b \hfill \cr} \right.\)
ta có hệ phương trình đã cho tương đương với hệ phương trình:
\(\left\{ \matrix{
5a + 2b = – 19 \hfill \cr
5a + 15b = – 175 \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{
13b = – 156 \hfill \cr
5a + 2b = – 19 \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{
b = – 12 \hfill \cr
a = 1 \hfill \cr} \right.\)
Vậy \(\left\{ \matrix{
x + y = 1 \hfill \cr
xy = – 12 \hfill \cr} \right.\) \( \Rightarrow \,x,y\) là 2 nghiệm của phương trình
\(X_{}^2 – X – 12 = 0 \Leftrightarrow \left\{ \matrix{
X_1^{} = – 3 \hfill \cr
X_2^{} = 4 \hfill \cr} \right.\)
Vậy hệ phương trình có 2 nghiệm
\(\left\{ \matrix{
x = – 3 \hfill \cr
y = 4 \hfill \cr} \right.\)
và
\(\left\{ \matrix{
x = 4 \hfill \cr
y = – 3 \hfill \cr} \right.\)