GTNN (X^2-X+1)/(X^2+4X+4)
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AH
Akai Haruma
Giáo viên
11 tháng 3 2024
Lời giải:
Vì $y^2\geq 0$ với mọi $y$ nên $14(x-2023)^2=26-3y^2\leq 26$
$\Rightarrow (x-2023)^2\leq \frac{26}{14}< 2$
Mà $(x-2023)^2$ là scp nên $(x-2023)^2=0$ hoặc $(x-2023)^2=1$
Nếu $(x-2023)^2=0$ thì: $26-3y^2=0\Rightarrow y^2=\frac{26}{3}$ (vô lý - loại)
Nếu $(x-2023)^2=1$ thì:
$x-2023=\pm 1\Rightarrow x=2022$ hoặc $x=2024$
$26-3y^2=14\Rightarrow 3y^2=12\Rightarrow y^2=4\Rightarrow y=\pm 2$
Vậy $(x,y)=(2022, 2), (2022, -2), (2024,2), (2024,-2)$
SV
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SV
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