Lấy 3 điểm bất kì thuộc đường tròn đó, nối 3 điểm lại với nhau ta được 1 tam giác. Giao điểm 3 đường trung trực của tam giác chính là tâm đường tròn đó.
a) (x - 2)(x + 1) = 0\(\Rightarrow\orbr{\begin{cases}x-2=0\\x+1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2\\x=-1\end{cases}}}\)
b) (x2 + 7)(x2 - 49) < 0
\(x^2\ge0\Rightarrow x^2+7>0\Rightarrow\hept{\begin{cases}x^2+7>0\\x^2-49< 0\end{cases}\Rightarrow-7< x^2< 49\Rightarrow x^2\in\left\{1;4;9;16;25;36\right\}}\)
\(\Rightarrow x\in\left\{-6;-5;-4;...;4;5;6\right\}\)
c) (x2 - 7)(x2 - 49) < 0 => x2 - 7 và x2 - 49 khác dấu mà x2 - 7 > x2 - 49
\(\Rightarrow x^2-7>0>x^2-49\Rightarrow7< x^2< 49\Rightarrow x^2\in\left\{9;16;25;36\right\}\Rightarrow x\in\left\{-6;-5;-4;-3;3;4;5;6\right\}\)
x(y + 2) + y = 1
=> x(y + 2) + (y + 2) = 3
=> (x + 1)(y + 2) = 3.Ta có bảng sau :
| x + 1 | -3 | -1 | 1 | 3 |
| y + 2 | -1 | -3 | 3 | 1 |
| x | -4 | -2 | 0 | 2 |
| y | -3 | -5 | 1 | -1 |
Vậy (x ; y) = (-4 ; -3) ; (-2 ; -5) ; (0 ; 1) ; (2 ; -1)
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mk ko hỉu
Thịnh giải cái j mà có câu a, câu b vậy
