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a) Vì x . y = 11
=> \(x;y\inƯ\left(11\right)=\left(\pm1;\pm11\right)\)
Ta có bang sau :
x | 1 | 11 | -1 | -11 |
y | 11 | 1 | -11 | -1 |
b) Vì x . y = 12
=> \(x;y\inƯ\left(12\right)=\left(\pm1;\pm12;\pm3;\pm4;\pm2;\pm6\right)\)
Ta có bảng sau:
x | 12 | 1 | -1 | -12 | 3 | 4 | -3 | -4 | 2 | 6 | -2 | -6 |
y | 1 | 12 | -12 | -1 | 4 | 3 | -4 | -3 | 6 | 2 | -6 | -2 |
a)x=1;y=11 hoacx=11;y=11
b)x=3;y=4
x=4;y=3
x=2;y=6
x=6;y=2
x=1;y=12
x=12;x=1
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Bài 1:
a, \(x^2\) +2\(x\) = 0
\(x.\left(x+2\right)\) = 0
\(\left[{}\begin{matrix}x=0\\x+2=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)
\(x\) \(\in\) {-2; 0}
b, (-2.\(x\)).(-4\(x\)) + 28 = 100
8\(x^2\) + 28 = 100
8\(x^2\) = 100 - 28
8\(x^2\) = 72
\(x^2\) = 72 : 8
\(x^2\) = 9
\(x^2\) = 32
|\(x\)| = 3
\(\left[{}\begin{matrix}x=-3\\x=3\end{matrix}\right.\)
Vậy \(\in\) {-3; 3}
c, 5.\(x\) (-\(x^2\)) + 1 = 6
- 5.\(x^3\) + 1 = 6
5\(x^3\) = 1 - 6
5\(x^3\) = - 5
\(x^3\) = -1
\(x\) = - 1