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\(\left\{{}\begin{matrix}xy-x-y+1=6\\\dfrac{1}{\left(x-1\right)^2-1}+\dfrac{1}{\left(y-1\right)^2-1}=\dfrac{2}{3}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\left(y-1\right)-\left(y-1\right)=6\\\dfrac{1}{\left(x-1\right)^2-1}+\dfrac{1}{\left(y-1\right)^2-1}=\dfrac{2}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)\left(y-1\right)=6\\\dfrac{1}{\left(x-1\right)^2-1}+\dfrac{1}{\left(y-1\right)^2-1}=\dfrac{2}{3}\end{matrix}\right.\) \(\Rightarrow\) đặt \(\left\{{}\begin{matrix}x-1=a\\y-1=b\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a.b=6\Rightarrow b=\dfrac{6}{a}\\\dfrac{1}{a^2-1}+\dfrac{1}{b^2-1}=\dfrac{2}{3}\end{matrix}\right.\) \(\Rightarrow\dfrac{1}{a^2-1}+\dfrac{1}{\dfrac{36}{a^2}-1}=\dfrac{2}{3}\)
\(\Rightarrow\dfrac{1}{a^2-1}+\dfrac{a^2}{36-a^2}=\dfrac{2}{3}\Rightarrow a^4-16a^2+36=0\)
\(\Rightarrow\left[{}\begin{matrix}a^2=8+2\sqrt{7}\\a^2=8-2\sqrt{7}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}a=\pm\sqrt{8+2\sqrt{7}}=\pm\left(\sqrt{7}+1\right)\\a=\pm\sqrt{8-2\sqrt{7}}=\pm\left(\sqrt{7}-1\right)\end{matrix}\right.\)
\(x=a+1\Rightarrow\left[{}\begin{matrix}x=2+\sqrt{7}\\x=-\sqrt{7}\\x=\sqrt{7}\\x=2-\sqrt{7}\end{matrix}\right.\) \(\Rightarrow y=\dfrac{6}{a}+1=\left[{}\begin{matrix}\sqrt{7}\\2-\sqrt{7}\\2+\sqrt{7}\\-\sqrt{7}\end{matrix}\right.\)
Vậy hệ đã cho có 4 cặp nghiệm thỏa mãn:
\(\left(x;y\right)=\left(2+\sqrt{7};\sqrt{7}\right)\),\(\left(-\sqrt{7};2-\sqrt{7}\right)\),\(\left(\sqrt{7};2+\sqrt{7}\right)\) ,\(\left(2-\sqrt{7};-\sqrt{7}\right)\)
a: \(\Leftrightarrow\left\{{}\begin{matrix}\left(x+2\right)\left(y+3\right)-xy=100\\xy-\left(x-2\right)\left(y-2\right)=64\end{matrix}\right.\)
=>xy+3x+2y+6-xy=100 và xy-xy+2x+2y-4=64
=>3x+2y=94 và 2x+2y=68
=>x=26 và x+y=34
=>x=26 và y=8
b: \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3x+3+2}{x+1}-\dfrac{2}{y+4}=4\\\dfrac{2x+2-2}{x+1}-\dfrac{5y+20-11}{y+4}=9\end{matrix}\right.\)
=>\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{x+1}-\dfrac{2}{y+4}=4-3=1\\\dfrac{-2}{x+1}+\dfrac{11}{y+4}=9+5-2=12\end{matrix}\right.\)
=>x+1=18/35; y+4=9/13
=>x=-17/35; y=-43/18
a: \(\Leftrightarrow\left\{{}\begin{matrix}\left(x+2\right)\left(y+3\right)=xy+100\\\left(x-2\right)\left(y-2\right)=xy-64\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x+2y=94\\-2x-2y=-68\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=26\\y=8\end{matrix}\right.\)
b: \(\Leftrightarrow\left\{{}\begin{matrix}-3x+2y=0\\-x+y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=3\end{matrix}\right.\)
c: \(\Leftrightarrow\left\{{}\begin{matrix}xy-2x=xy-4x+2y-8\\2xy+7x-6y-21=2xy+6x-7y-21\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-2y=-8\\x+y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=2\end{matrix}\right.\)
c: =>3x^2+3y^2=39 và 3x^2-2y^2=-6
=>5y^2=45 và x^2=13-y^2
=>y^2=9 và x^2=4
=>\(\left\{{}\begin{matrix}x\in\left\{2;-2\right\}\\y\in\left\{3;-3\right\}\end{matrix}\right.\)
d: \(\Leftrightarrow\left\{{}\begin{matrix}5\sqrt{x}=5\\\sqrt{x}-\sqrt{y}=-\dfrac{11}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\\sqrt{y}=1+\dfrac{11}{2}=\dfrac{13}{2}\end{matrix}\right.\)
=>x=1 và y=169/4
b: \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3x+3-3}{x+1}-\dfrac{2}{y+4}=4\\\dfrac{2x+2-2}{x+1}-\dfrac{5}{y+4}=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{3}{x+1}-\dfrac{2}{y+4}=4-3=1\\-\dfrac{2}{x+1}-\dfrac{5}{y+4}=9-2=7\end{matrix}\right.\)
=>x+1=11/9 và y+4=-11/19
=>x=2/9 và y=-87/19
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x+y-2}+1+\dfrac{4}{x+2y}=3\\\dfrac{x+y-2+2}{x+y-2}-\dfrac{8}{x+2y}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x+y-2}+\dfrac{4}{x+2y}=2\\\dfrac{2}{x+y-2}-\dfrac{8}{x+2y}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x+y-2}=1\\\dfrac{1}{x+2y}=\dfrac{1}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=3\\x+2y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)
phương trình 2 ⇔\(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{2}{xy}=7-3xy\)⇔\(\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^2=7-3xy\)
đoạn sau bạn tự giải nha