Hệ phương trình đối xứng

Lời giải:

Lấy PT thứ nhất cộng phương trình thứ 2:

\(\Rightarrow 4(x+y)=\frac{1}{x^2}+\frac{1}{y^2}>0\Rightarrow x+y>0\)

Lấy PT thứ nhất trừ đi phương trình thứ 2:

\((3x+y)-(3y+x)=\frac{1}{x^2}-\frac{1}{y^2}\)

\(\Leftrightarrow 2(x-y)=\frac{y^2-x^2}{x^2y^2}\)

\(\Leftrightarrow (x-y)\left(2+\frac{x+y}{x^2y^2}\right)=0\)

Vì \(x+y>0\Rightarrow 2+\frac{x+y}{x^2y^2}>0\)

Do đó: \(x-y=0\Rightarrow x=y\). Thay vào pt thứ nhất:

\(4x=\frac{1}{x^2}\Rightarrow 4x^3=1\Rightarrow x=\sqrt[3]{\frac{1}{4}}=y\)

pt1:

(x-1)^3-12x+12=(y+1)^3-12y-12

<=> (x-1)^3 -12(x-1)=(y+1)^3-12(y+1). đặt x-1=a ; y+1=b

a^3-12a=b^3-12b

=>(a-b) (a^2+ab+b^2-12)=0

với a^2+b^2+ab=12

(x-1)^2+(y+1)^2+(x-1)(y+1)=12

x^2+y^2+xy-x+y=11(1)

kết hợp pt2 x^2+y^2+y-x=1/2 thay vào (1)

xy=21/2 từ đây thế x theo y dễ dàng giải nghiệm

với a=b thì x-1=y+1

=>từ đây thế x theo y vào pt2 ta dễ dàng giải

1.

\(\left\{{}\begin{matrix}x^3+y^3+x^3y^3=17\\x+y+xy=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)^3-3xy\left(x+y\right)+x^3y^3=17\\x+y+xy=5\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}x+y=a\\xy=b\end{matrix}\right.\left(a^2\ge4b\right)\)

Hệ phương trình trở thành \(\left\{{}\begin{matrix}a^3-3ab+b^3=17\\a+b=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(a+b\right)^3-3ab\left(a+b+1\right)=17\\a+b=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}ab=6\\a+b=5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}a=2;b=3\left(l\right)\\a=3;b=2\end{matrix}\right.\)

\(\left\{{}\begin{matrix}a=3\\b=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+y=3\\xy=2\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\\\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\end{matrix}\right.\)

2.

\(\left\{{}\begin{matrix}x^3+y^3=2\\xy\left(x+y\right)=2\end{matrix}\right.\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)^3-6=2\\xy\left(x+y\right)=2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)^3=8\\xy\left(x+y\right)=2\end{matrix}\right.\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}x+y=2\\xy=1\end{matrix}\right.\)

\(\Leftrightarrow x=y=1\)

3.

\(\left\{{}\begin{matrix}x^4+y^4+x^2y^2=481\\x^2+y^2+xy=37\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x^2+y^2\right)^2-x^2y^2=481\\x^2+y^2+xy=37\end{matrix}\right.\)

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

Hệ phương trình trở thành \(\left\{{}\begin{matrix}a^2-b^2=481\\a+b=37\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(37-b\right)^2-b^2=481\\a=37-b\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}74b=888\\a=37-b\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=12\\a=25\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}xy=12\\x^2+y^2=25\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2xy=24\\x^2+y^2=25\end{matrix}\right.\)

\(\Rightarrow\left(x+y\right)^2=49\)

\(\Leftrightarrow x+y=\pm7\)

TH1: \(\left\{{}\begin{matrix}xy=12\\x+y=7\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=3\\y=4\end{matrix}\right.\\\left\{{}\begin{matrix}x=4\\y=3\end{matrix}\right.\end{matrix}\right.\)

TH2: \(\left\{{}\begin{matrix}xy=12\\x+y=-7\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=-3\\y=-4\end{matrix}\right.\\\left\{{}\begin{matrix}x=-4\\y=-3\end{matrix}\right.\end{matrix}\right.\)