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Đặt \(\left\{{}\begin{matrix}x-y=a\\xy=b\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a-b=7\\-ab=12\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=b+7\\ab+12=0\end{matrix}\right.\)
\(\Rightarrow\left(b+7\right)b+12=0\Leftrightarrow b^2+7b+12=0\Rightarrow\left[{}\begin{matrix}b=-3;a=4\\b=-4;a=3\end{matrix}\right.\)
TH1: \(\left\{{}\begin{matrix}a=4\\b=-3\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x-y=4\\xy=-3\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=y+4\\xy+3=0\end{matrix}\right.\)
\(\Rightarrow\left(y+4\right)y+3=0\Rightarrow y^2+4y+3=0\Rightarrow\left[{}\begin{matrix}y=-1;x=3\\y=-3;x=1\end{matrix}\right.\)
TH2: \(\left\{{}\begin{matrix}a=3\\b=-4\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x-y=3\\xy=-4\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=y+3\\xy+4=0\end{matrix}\right.\)
\(\Rightarrow\left(y+3\right)y+4=0\Rightarrow y^2+3y+4=0\) (vô nghiệm)
Vậy hệ đã cho có 2 cặp nghiệm \(\left(x;y\right)=\left(3;-1\right);\left(1;-3\right)\)
\(\left\{{}\begin{matrix}\left(x-y\right)^2+xy=3\left(x-y\right)\\\left(x-y\right)^2+3xy=7\left(x-y\right)^3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}3\left(x-y\right)^2+3xy=9\left(x-y\right)\\\left(x-y\right)^2+3xy=7\left(x-y\right)^3\end{matrix}\right.\)
\(\Rightarrow7\left(x-y\right)^3-9\left(x-y\right)=-2\left(x-y\right)^2\)
\(\Leftrightarrow7\left(x-y\right)^3+2\left(x-y\right)^2-9\left(x-y\right)=0\)
\(\Leftrightarrow\left(x-y\right)\left(7\left(x-y\right)^2+2\left(x-y\right)-9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-y=0\\x-y=1\\x-y=\dfrac{-9}{7}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}y=x\\y=x-1\\y=x+\dfrac{9}{7}\end{matrix}\right.\)
TH1: \(y=x\) thay vaò pt đầu:
\(x^2-x^2+x^2=3\left(x-x\right)\Rightarrow x^2=0\Rightarrow\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)
TH2: \(y=x-1\) thay vào pt đầu:
\(x^2-x\left(x-1\right)+\left(x-1\right)^2=3\Leftrightarrow x^2-x-2=0\)
\(\Rightarrow\left[{}\begin{matrix}x=2\Rightarrow y=1\\x=-1\Rightarrow y=-2\end{matrix}\right.\)
TH3: \(y=x+\dfrac{9}{7}\):
\(x^2-x\left(x+\dfrac{9}{7}\right)+\left(x+\dfrac{9}{7}\right)^2=\dfrac{-27}{7}\Leftrightarrow x^2+\dfrac{9}{7}x+\dfrac{270}{49}=0\) (vô nghiệm)
Vậy hệ đã cho có 3 cặp nghiệm:
\(\left(x;y\right)=\left(0;0\right);\left(2;1\right);\left(-1;-2\right)\)
Đặt \(\left\{{}\begin{matrix}x\left(x+1\right)=a\\y\left(y+1\right)=b\end{matrix}\right.\) thì ta có:
\(\left\{{}\begin{matrix}a+b=8\\ab=12\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=2\\b=6\end{matrix}\right.or\left\{{}\begin{matrix}a=6\\b=2\end{matrix}\right.\)
Tới đây thì đơn giải rồi nhé
a) x + y = 6 (1)
2x - 3y = 12 (2)
(1) ⇔ x = 6 - y (3)
Thế (3) vào (2) ta có:
2(6 - y) - 3y = 12
⇔ 12 - 2y - 3y = 12
⇔ -5y = 12 - 12
⇔ -5y = 0
⇔ y = 0
Thế y = 0 vào (3) ta có:
x = 6 - 0
⇔ x = 6
Vậy S = {6; 0}
b) x - y = 5 (4)
(x - 2)(y + 3) = 3 + xy (5)
(5) ⇔ xy + 3x - 2y - 6 = 3 + xy
⇔ 3x - 2y = 3 + 6
⇔ 3x - 2y = 9 (6)
(4) ⇔ x = y + 5 (7)
Thế x = y + 5 vào (6) ta có:
(6) ⇔ 3(y + 5) - 2y = 9
⇔ 3y + 15 - 2y = 9
⇔ y = 9 - 15
⇔ y = -6
Thế y = -6 vào (7) ta có:
x = -6 + 5
⇔ x = -1
Vậy S ={-1; -6}
Cộng vế với vế:
\(x^2+2xy+y^2+x+y=6\)
\(\Leftrightarrow\left(x+y\right)^2+\left(x+y\right)-6=0\)
\(\Rightarrow\left[{}\begin{matrix}x+y=-3\\x+y=2\end{matrix}\right.\)
TH1: \(\left\{{}\begin{matrix}x+y=-3\\xy=5\end{matrix}\right.\)
Theo Viet đảo, x và y là nghiệm của:
\(t^2+3t+5=0\) (vô nghiệm)
TH2: \(\left\{{}\begin{matrix}x+y=2\\xy=0\end{matrix}\right.\)
Theo Viet đảo, x và y là nghiệm:
\(t^2-2t=0\Rightarrow\left[{}\begin{matrix}t=0\\t=2\end{matrix}\right.\)
\(\Rightarrow\left(x;y\right)=\left(2;0\right);\left(0;2\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}xy-3x+2y-6=xy+1\\2x+2y=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2y-3x=7\\2x+2y=5\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{2}{5}\\y=\dfrac{29}{10}\end{matrix}\right.\)
c/
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+x+y^2+y=8\\\left(x^2+x\right)\left(y^2+y\right)=12\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x^2+x=a\\y^2+y=b\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=8\\ab=12\end{matrix}\right.\) theo Viet đảo, a và b là nghiệm:
\(t^2-8t+12=0\Rightarrow\left[{}\begin{matrix}t=6\\t=2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x^2+x=6\\y^2+y=2\end{matrix}\right.\\\left\{{}\begin{matrix}x^2+x=2\\y^2+y=6\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x^2+x-6=0\\y^2+y-2=0\end{matrix}\right.\\\left\{{}\begin{matrix}x^2+x-2=0\\y^2+y-6=0\end{matrix}\right.\end{matrix}\right.\)
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b/
\(\Leftrightarrow\left\{{}\begin{matrix}x+y+xy+1=0\\\left(x+y\right)^2-2xy-x-y=22\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2\left(x+y\right)+2xy+2=0\\\left(x+y\right)^2-2xy-x-y-22=0\end{matrix}\right.\)
\(\Rightarrow\left(x+y\right)^2+\left(x+y\right)-20=0\)
\(\Rightarrow\left[{}\begin{matrix}x+y=4\Rightarrow xy=-5\\x+y=-5\Rightarrow xy=4\end{matrix}\right.\)
TH1: \(\left\{{}\begin{matrix}x+y=4\\xy=-5\end{matrix}\right.\) thì x; y là nghiệm:
\(t^2-4t-5=0\Rightarrow\left[{}\begin{matrix}t=-1\\t=5\end{matrix}\right.\)
\(\Rightarrow\left(x;y\right)=\left(-1;5\right);\left(5;-1\right)\)
TH2: \(\left\{{}\begin{matrix}x+y=-5\\xy=4\end{matrix}\right.\) thì x; y là nghiệm:
\(t^2+5t+4=0\Rightarrow\left[{}\begin{matrix}t=-1\\t=-4\end{matrix}\right.\)
\(\Rightarrow\left(x;y\right)=\left(-1;-4\right);\left(-4;-1\right)\)
Đặt \(\left\{{}\begin{matrix}x-y=a\\xy=b\end{matrix}\right.\) : Hệ trở thành ;
\(\left\{{}\begin{matrix}a-b=7\\ab=-12\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=b+7\\b^2+7b+12=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=b+7\\\left(b+3\right)\left(b+4\right)=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a=4\\b=-3\end{matrix}\right.\\\left\{{}\begin{matrix}a=3\\b=-4\end{matrix}\right.\end{matrix}\right.\)
Với \(a=4;b=-3\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-y=4\\xy=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y+4\\y^2+4y+3=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=y+4\\\left(y+1\right)\left(y+3\right)=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=3\\y=-1\end{matrix}\right.\\\left\{{}\begin{matrix}x=1\\y=-3\end{matrix}\right.\end{matrix}\right.\)
Với \(a=3;b=-4\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-y=3\\xy=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y+3\\y^2+3y+4=0\end{matrix}\right.\) ( Vô nghiệm )
Vậy \(\left(x;y\right)=\left(3;-1\right)\) \(\left(x;y\right)=\left(1;-3\right)\)