giup minh may cau nay voi? toan hoc 10 nha

Question

a/ $\left\{ \begin{array}{l} (x - y)( x^{2} - y^{2}) = 3\\ (x + y)(x^{2} + y^{2}= 15\end{array} \right.$
b/ $\left\{ \begin{array}{l} x + y +x^{2} + y^{2}= 8\\ xy( x+ 1)(y + 1) = 12\end{array} \right.$
c/ $\left\{ \begin{array}{l} x^{2} + y^{2} = 4\\ (x + y)^2 = 4 \end{array} \right.$

score 3 · Answer 1

Bình chọn tăng 3 Bình chọn giảm

c)
$\left\{ \begin{array}{l} x^{2} + y^{2} = 4\\ (x + y)^2 = 4 \end{array} \right.$
$\Leftrightarrow \left\{ \begin{array}{l} 2xy=0\\ x^2+y^2=4 \end{array} \right.$
$\Leftrightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} x=0\\ y=\pm2 \end{array} \right.\\\left\{ \begin{array}{l} x=\pm2\\ y=0 \end{array} \right. \end{array} \right.$

score 3 · Answer 2

Bình chọn tăng 3 Bình chọn giảm

b)
$\left\{ \begin{array}{l} x + y +x^{2} + y^{2}= 8\\ xy( x+ 1)(y + 1) = 12\end{array} \right.$
$\Leftrightarrow \left\{ \begin{array}{l} x(x+1)+y(y+1)=8\\ x(x+1)y(y+1)=12 \end{array} \right.$
$\Leftrightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} x(x+1)=2\\ y(y+1)=6 \end{array} \right.\\ \left\{ \begin{array}{l} x(x+1)=6\\ y(y+1)=2 \end{array} \right. \end{array} \right.$
Từ đó suy ra nghiệm:
$(x;y)\in\{(1;2),(1;-3),(-2;2),(-2;-3),(2;1),(-3;1),(2;-2),(-3;-2)\}$

score 3 · Answer 3

Bình chọn tăng 3 Bình chọn giảm

a)
$\left\{ \begin{array}{l} (x - y)( x^{2} - y^{2}) = 3\\ (x + y)(x^{2} + y^{2})= 15\end{array} \right.$
$\Leftrightarrow \left\{ \begin{array}{l} (x^3+y^3)-(xy^2+x^2y)=3\\ (x^3+y^3)+(xy^2+x^2y)=15 \end{array} \right.$
$\Leftrightarrow \left\{ \begin{array}{l} x^3+y^3=9\\ xy^2+yx^2=6 \end{array} \right.$
$\Leftrightarrow \left\{ \begin{array}{l} xy(x+y)=6\\ x^3+y^3+3xy(x+y)=27 \end{array} \right.$
$\Leftrightarrow \left\{ \begin{array}{l} xy(x+y)=6\\ (x+y)^3=27 \end{array} \right.$
$\Leftrightarrow \left\{ \begin{array}{l} x+y=3\\ xy=2 \end{array} \right.\Leftrightarrow \left[ \begin{array}{l} x=2;y=1\\ x=1;y=2 \end{array} \right.$