\(\left\{{}\begin{matrix}x^2-y^2-xy=-11\\\left(x^2-y^2\right)\left(-xy\right)=-180\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x^2-y^2=a\\-xy=b\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a+b=-11\\a.b=-180\end{matrix}\right.\) \(\Rightarrow\) theo Viet đảo, a; b là nghiệm:
\(t^2+11t-180=0\Rightarrow\left[{}\begin{matrix}t=9\\t=-20\end{matrix}\right.\)
TH1: \(\left\{{}\begin{matrix}x^2-y^2=9\\xy=20\Rightarrow y=\frac{20}{x}\end{matrix}\right.\)
\(\Rightarrow x^2-\frac{400}{x^2}=9\Rightarrow x^4-9x^2-400=0\Rightarrow\left\{{}\begin{matrix}x=5;y=4\\x=-5;y=-4\end{matrix}\right.\)
TH2: \(\left\{{}\begin{matrix}x^2-y^2=-20\\xy=-9\Rightarrow y=\frac{-9}{x}\end{matrix}\right.\) \(\Rightarrow x^2-\frac{81}{x^2}=-20\Rightarrow x^4+20x^2-81=0\)
\(\Rightarrow\left[{}\begin{matrix}x=\sqrt{\sqrt{181}-10}\Rightarrow y=\frac{-9}{x}=...\\x=-\sqrt{\sqrt{181}-10}\Rightarrow y=\frac{-9}{x}=...\end{matrix}\right.\)
Lời giải:
Đặt \(x^2-y^2=a, xy=b\). Khi đó:
HPT \(\Leftrightarrow \left\{\begin{matrix} (x^2-y^2)-xy=-11\\ (x^2-y^2)xy=180\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} a-b=-11\\ ab=180\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} a-b=-11\\ a(-b)=-180\end{matrix}\right.\)
Theo định lý Vi-et đảo, $a,-b$ là nghiệm của PT \(X^2+11X-180=0\)
\(\Rightarrow (a,b)=(9,-20); (-20; 9)\)
TH1: Nếu \((a,b)=(9,-20)\Leftrightarrow \left\{\begin{matrix} x^2-y^2=9\\ xy=-20(1)\end{matrix}\right.\)
\(\Rightarrow \left\{\begin{matrix} x^2-y^2=9\\ x^2(-y^2)=-400\end{matrix}\right.\)
Theo định lý Vi-et đảo thì $x^2,-y^2$ là nghiệm của phương trình:
\(X^2-9X-400=0\)
\(\Rightarrow (x^2,-y^2)=(25,-16)\Rightarrow (x,y)=(\pm 5; \pm 4)\)
Kết hợp với (1) suy ra \((x,y)=(-5,4); (5,-4)\)
TH2: Nếu \((a,b)=(-20,9)\Leftrightarrow \left\{\begin{matrix} x^2-y^2=-20\\ xy=9(2)\end{matrix}\right.\)
\(\Rightarrow \left\{\begin{matrix} x^2-y^2=-20\\ x^2(-y^2)=-81\end{matrix}\right.\)
Theo định lý Vi-et đảo thì $x^2,-y^2$ là nghiệm của PT:
\(X^2+20X-81=0\)
\(\Rightarrow (x^2,-y^2)=(-10+\sqrt{181}; -10-\sqrt{181})\)
\(\Rightarrow (x,y)=(\pm \sqrt{-10+\sqrt{181}}; \pm \sqrt{10+\sqrt{181}})\)
Kết hợp với $(2)$ suy ra:
\((x,y)=(\sqrt{-10+\sqrt{181}}; \sqrt{10+\sqrt{181}}); (-\sqrt{-10+\sqrt{181}}; -\sqrt{10+\sqrt{181}})\)
Vậy............
Ta có :
\(\left\{{}\begin{matrix}x^2-xy-y^2=-11\\\left(x^2-y^2\right)xy=180\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2-y^2=xy-11\\\left(xy-11\right)xy=180\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2-y^2=xy-11\\x^2y^2-11xy-180=0\end{matrix}\right.\Rightarrow\left(xy-\frac{11}{2}\right)^2=\frac{841}{4}\)\
\(\Leftrightarrow\left[{}\begin{matrix}xy-\frac{11}{2}=\frac{29}{2}\\xy-\frac{11}{2}=-\frac{29}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}xy=20\\xy=-9\end{matrix}\right.\)
Với xy = 20 \(\Rightarrow\left\{{}\begin{matrix}x^2-y^2=20-11=9\\y=\frac{20}{x}\end{matrix}\right.\)
\(\Rightarrow x^2-\left(\frac{20}{x}\right)^2=9\) \(\Leftrightarrow\frac{x^4-400}{x^2}=9\Leftrightarrow x^4-9x^2-400=0\)
Giải phương trình : <=> x^3(x-5) + 5x^2(x-5) + 16x(x-5) + 80(x-5) = 0 \(\Leftrightarrow\left(x^3+5x^2+16x+80\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left(x^2+16\right)\left(x+5\right)\left(x-5\right)=0\Leftrightarrow x=\pm5\)
Do xy = 20 \(\Rightarrow y=\pm4\)
TH 2 : làm tương tự
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