Question

Giải hệ phương trình

a) \(\left\{{}\begin{matrix}x+y+xy=7\\x^2+y^2+xy=13\end{matrix}\right.\)

b) \(\left\{{}\begin{matrix}x+y+xy+1=0\\x^2+y^2-x-y=22\end{matrix}\right.\)

c) \(\left\{{}\begin{matrix}x+y+x^2+y^2=8\\xy\left(x+1\right)\left(y+1\right)=12\end{matrix}\right.\)

Answer 1

a/

\(\Leftrightarrow\left\{{}\begin{matrix}x+y+xy=7\\\left(x+y\right)^2-xy=13\end{matrix}\right.\)

\(\Rightarrow\left(x+y\right)^2+x+y=20\)

\(\Leftrightarrow\left(x+y\right)^2+x+y-20=0\)

\(\Rightarrow\left[{}\begin{matrix}x+y=4\Rightarrow xy=3\\x+y=-5\Rightarrow xy=12\end{matrix}\right.\)

TH1: \(\left\{{}\begin{matrix}x+y=4\\xy=3\end{matrix}\right.\) theo Viet đảo x; y là nghiệm:

\(t^2-4t+3=0\Rightarrow\left[{}\begin{matrix}t=1\\t=3\end{matrix}\right.\)

\(\Rightarrow\left(x;y\right)=\left(1;3\right);\left(3;1\right)\)

TH2: \(\left\{{}\begin{matrix}x+y=-5\\xy=12\end{matrix}\right.\) theo Viet đảo x; y là nghiệm:

\(t^2+5t+12=0\left(vn\right)\)

Answer 2

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:

\(t^2+5t+4=0\Rightarrow\left[{}\begin{matrix}t=-1\\t=-4\end{matrix}\right.\)

\(\Rightarrow\left(x;y\right)=\left(-1;-4\right);\left(-4;-1\right)\)

Answer 3

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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