Câu hỏi của Phùng Gia Bảo

Question

Giải hệ phương trình $\hept{\begin{cases}x^2+y^2=11\x+xy+y=3+4\sqrt{2}\end{cases}}$

ST · Accepted Answer

$HPT\Leftrightarrow\hept{\begin{cases}\left(x+y\right)^2-2xy=11\\left(x+y\right)+xy=3+4\sqrt{2}\end{cases}}$Đặt x+y=a;xy=b thì hệ trở thành: $\hept{\begin{cases}a^2-2b=11\a+b=3+4\sqrt{2}\end{cases}\Leftrightarrow\hept{\begin{cases}a^2-2b=11\b=3+4\sqrt{2}-a\end{cases}}}$=> $a^2-2\left(3+4\sqrt{2}-a\right)=11$<=>$a^2-6-8\sqrt{2}+2a-11=0$$\Leftrightarrow a^2+2a-17-8\sqrt{2}=0$$\Leftrightarrow a^2+2a-\left(16+2.4.\sqrt{2}+2-1\right)=0$$\Leftrightarrow\left(a+1\right)^2-\left(4+\sqrt{2}\right)^2=0$$\Leftrightarrow\left(a+1-4-\sqrt{2}\right)\left(a+1+4+\sqrt{2}\right)=0$$\Leftrightarrow\left(a-3-\sqrt{2}\right)\left(a+5+\sqrt{2}\right)=0$$\Leftrightarrow\orbr{\begin{cases}a-3-\sqrt{2}=0\a+5+\sqrt{2}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}a=3+\sqrt{2}\a=-5-\sqrt{2}\end{cases}}}$=>$\orbr{\begin{cases}b=3+4\sqrt{2}-3-\sqrt{2}=3\sqrt{2}\b=3+4\sqrt{2}+5+\sqrt{2}=8+5\sqrt{2}\end{cases}}$- Với $a=3+\sqrt{2},b=3\sqrt{2}$,ta có: $x+y=3+\sqrt{2}\Rightarrow y=3+\sqrt{2}-x$ (1)Thay (1) vào $xy=3\sqrt{2}=b\Rightarrow x\left(3+\sqrt{2}-x\right)=3\sqrt{2}$$\Leftrightarrow3x+x\sqrt{2}-x^2=3\sqrt{2}\Leftrightarrow x\left(3-x\right)+\sqrt{2}\left(x-3\right)=0$$\Leftrightarrow\left(x-3\right)\left(\sqrt{2}-x\right)=0\Leftrightarrow\orbr{\begin{cases}x=3\x=\sqrt{2}\end{cases}}\Rightarrow\orbr{\begin{cases}y=\sqrt{2}\y=3\end{cases}}$- Với $a=-5-\sqrt{2},b=8+5\sqrt{2}$, ta có: $x+y=-5-\sqrt{2}\Rightarrow y=-5-\sqrt{2}-x$(2)Thay (2) vào $xy=8+5\sqrt{2}=b\Rightarrow x\left(-5-\sqrt{2}-x\right)=8+5\sqrt{2}$$\Leftrightarrow-x^2-5x-x\sqrt{2}-8-5\sqrt{2}=0$$\Leftrightarrow-x^2+\left(-5-\sqrt{2}\right)x+\left(-8-5\sqrt{2}\right)=0$(3)$\Delta=\left(-5-\sqrt{2}\right)^2-4.\left(-1\right).\left(-8-5\sqrt{2}\right)$$=27+10\sqrt{2}-32-20\sqrt{2}=-5-10\sqrt{2}< 0$=>pt (3) vô nghiệmVậy $\left(x;y\right)=\left(3;\sqrt{2}\right)$ hoaojwc $\left(\sqrt{2};3\right)$Câu trả lời: $HPT\Leftrightarrow\hept{\begin{cases}\left(x+y\right)^2-2xy=11\\left(x+y\right)+xy=3+4\sqrt{2}\end{cases}}$Đặt x+y=a;xy=b thì hệ trở thành: $\hept{\begin{cases}a^2-2b=11\a+b=3+4\sqrt{2}\end{cases}\Leftrightarrow\hept{\begin{cases}a^2-2b=11\b=3+4\sqrt{2}-a\end{cases}}}$=> $a^2-2\left(3+4\sqrt{2}-a\right)=11$<=>$a^2-6-8\sqrt{2}+2a-11=0$$\Leftrightarrow a^2+2a-17-8\sqrt{2}=0$$\Leftrightarrow a^2+2a-\left(16+2.4.\sqrt{2}+2-1\right)=0$$\Leftrightarrow\left(a+1\right)^2-\left(4+\sqrt{2}\right)^2=0$$\Leftrightarrow\left(a+1-4-\sqrt{2}\right)\left(a+1+4+\sqrt{2}\right)=0$$\Leftrightarrow\left(a-3-\sqrt{2}\right)\left(a+5+\sqrt{2}\right)=0$$\Leftrightarrow\orbr{\begin{cases}a-3-\sqrt{2}=0\a+5+\sqrt{2}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}a=3+\sqrt{2}\a=-5-\sqrt{2}\end{cases}}}$=>$\orbr{\begin{cases}b=3+4\sqrt{2}-3-\sqrt{2}=3\sqrt{2}\b=3+4\sqrt{2}+5+\sqrt{2}=8+5\sqrt{2}\end{cases}}$- Với $a=3+\sqrt{2},b=3\sqrt{2}$,ta có: $x+y=3+\sqrt{2}\Rightarrow y=3+\sqrt{2}-x$ (1)Thay (1) vào $xy=3\sqrt{2}=b\Rightarrow x\left(3+\sqrt{2}-x\right)=3\sqrt{2}$$\Leftrightarrow3x+x\sqrt{2}-x^2=3\sqrt{2}\Leftrightarrow x\left(3-x\right)+\sqrt{2}\left(x-3\right)=0$$\Leftrightarrow\left(x-3\right)\left(\sqrt{2}-x\right)=0\Leftrightarrow\orbr{\begin{cases}x=3\x=\sqrt{2}\end{cases}}\Rightarrow\orbr{\begin{cases}y=\sqrt{2}\y=3\end{cases}}$- Với $a=-5-\sqrt{2},b=8+5\sqrt{2}$, ta có: $x+y=-5-\sqrt{2}\Rightarrow y=-5-\sqrt{2}-x$(2)Thay (2) vào $xy=8+5\sqrt{2}=b\Rightarrow x\left(-5-\sqrt{2}-x\right)=8+5\sqrt{2}$$\Leftrightarrow-x^2-5x-x\sqrt{2}-8-5\sqrt{2}=0$$\Leftrightarrow-x^2+\left(-5-\sqrt{2}\right)x+\left(-8-5\sqrt{2}\right)=0$(3)$\Delta=\left(-5-\sqrt{2}\right)^2-4.\left(-1\right).\left(-8-5\sqrt{2}\right)$$=27+10\sqrt{2}-32-20\sqrt{2}=-5-10\sqrt{2}< 0$=>pt (3) vô nghiệmVậy $\left(x;y\right)=\left(3;\sqrt{2}\right)$ hoaojwc $\left(\sqrt{2};3\right)$

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