Question

[poppy Trang]

giải hpt:

\(\left\{{}\begin{matrix}x+\dfrac{1}{x}+y-\dfrac{1}{y}=3\\x^2+\dfrac{1}{x^2}+y^2+\dfrac{1}{y^2}=5\end{matrix}\right.\)

Answer 1

[Nguyễn Việt Lâm]

Đặt \(\left\{{}\begin{matrix}x+\dfrac{1}{x}=a\\y-\dfrac{1}{y}=b\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x^2+\dfrac{1}{x^2}=a^2-2\\y^2+\dfrac{1}{y^2}=b^2+2\end{matrix}\right.\)hệ đã cho tương đương:

\(\left\{{}\begin{matrix}a+b=3\\a^2+b^2=5\end{matrix}\right.\) \(\Rightarrow a^2+\left(3-a\right)^2-5=0\Rightarrow a^2-3a+2=0\)

\(\Rightarrow\left[{}\begin{matrix}a=1;b=2\\a=2;b=1\end{matrix}\right.\)

TH1: \(\left\{{}\begin{matrix}x+\dfrac{1}{x}=1\\y-\dfrac{1}{y}=2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x^2-x+1=0\left(vn\right)\\y^2-2y-1=0\end{matrix}\right.\) (loại)

TH2: \(\left\{{}\begin{matrix}x+\dfrac{1}{x}=2\\y-\dfrac{1}{y}=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x^2-2x+1=0\\y^2-y-1=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=1\\\left[{}\begin{matrix}y=\dfrac{1-\sqrt{5}}{2}\\y=\dfrac{1+\sqrt{5}}{2}\end{matrix}\right.\end{matrix}\right.\)

Vậy hệ đã cho có 2 cặp nghiệm:

\(\left(x;y\right)=\left(1;\dfrac{1-\sqrt{5}}{2}\right);\left(1;\dfrac{1+\sqrt{5}}{2}\right)\)

Answer 2

[tran nguyen bao quan]

Đặt \(a=x+\dfrac{1}{x}\Leftrightarrow a^2=x^2+\dfrac{1}{x^2}+2\Leftrightarrow x^2+\dfrac{1}{x^2}=a^2-2\)

\(b=y-\dfrac{1}{y}\Leftrightarrow b^2=y^2+\dfrac{1}{y^2}-2\Leftrightarrow y^2+\dfrac{1}{y^2}=b^2+2\)

Nên \(x^2+\dfrac{1}{x^2}+y^2+\dfrac{1}{y^2}=5\Leftrightarrow a^2-2+b^2+2=5\Leftrightarrow a^2+b^2=5\)Vậy ta có hệ phương trình \(\left\{{}\begin{matrix}a+b=3\\a^2+b^2=5\left(1\right)\end{matrix}\right.\)

Ta có a+b=3\(\Leftrightarrow b=3-a\)

Thay b=3-a vào (1)\(\Leftrightarrow a^2+\left(3-a\right)^2=5\Leftrightarrow a^2+9-6a+a^2=5\Leftrightarrow2a^2-6a+4=0\Leftrightarrow2\left(a^2-3a+2\right)=0\Leftrightarrow a^2-3a+2=0\Leftrightarrow a^2-a-2a+2=0\Leftrightarrow a\left(a-1\right)-2\left(a-1\right)=0\Leftrightarrow\left(a-1\right)\left(a-2\right)=0\Leftrightarrow\)\(\left[{}\begin{matrix}a-1=0\\a-2=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}a=1\\a=2\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}b=2\\b=1\end{matrix}\right.\)

TH1:\(\left\{{}\begin{matrix}a=1\\b=2\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x+\dfrac{1}{x}=1\\y-\dfrac{1}{y}=2\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x^2-x+1=0\\y^2-2y-1=0\end{matrix}\right.\)

Ta có \(x^2-x+1=x^2-2x.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)

Vậy phương trình (2) vô nghiệm

TH2: \(\left\{{}\begin{matrix}a=2\\b=1\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x+\dfrac{1}{x}=2\\y-\dfrac{1}{y}=1\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x^2-2x+1=0\\y^2-y-1=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}\left(x-1\right)^2=0\\\left(y-\dfrac{1}{2}\right)^2=\dfrac{5}{4}\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x=1\\y=\dfrac{1\pm\sqrt{5}}{2}\end{matrix}\right.\)

Vậy (x,y)={(\(1;\dfrac{1+\sqrt{5}}{2}\));(\(1;\dfrac{1-\sqrt{5}}{2}\))}