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tìm x.y \(\in\)Z, thỏa mãn
a) 2.|3x+4|+|-5|=65 b) | x-5| - | -7|=|-12|
c)x.y+3y-2x=3 d) 2x.y +x -2y =5
tìm x.y \(\in\)Z, thỏa mãn
a) 2.|3x+4|+|-5|=65 b) | x-5| - | -7|=|-12|
c)x.y+3y-2x=3 d) 2x.y +x -2y =5
tìm x.y \(\in\)Z, thỏa mãn
a) 2.|3x+4|+|-5|=65 b) | x-5| - | -7|=|-12|
c)x.y+3y-2x=3 d) 2x.y +x -2y =5
a: \(\Leftrightarrow\left(x;y-3\right)\in\left\{\left(1;17\right);\left(17;1\right);\left(-1;-17\right);\left(-17;-1\right)\right\}\)
=>\(\left(x,y\right)\in\left\{\left(1;20\right);\left(17;4\right);\left(-1;-14\right);\left(-17;2\right)\right\}\)
b: \(\Leftrightarrow\left(x-1;y+2\right)\in\left\{\left(1;7\right);\left(7;1\right);\left(-1;-7\right);\left(-7;-1\right)\right\}\)
=>\(\left(x,y\right)\in\left\{\left(2;5\right);\left(8;-1\right);\left(0;-9\right);\left(-6;-3\right)\right\}\)
c: =>(y+1)(3x+1)=7
=>\(\left(3x+1;y+1\right)\in\left\{\left(1;7\right);\left(7;1\right)\right\}\)
=>\(\left(x,y\right)\in\left\{\left(0;6\right);\left(2;0\right)\right\}\)
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
b.y(x+3)-2x=3
y(x+3)-2x-6=-3
y(x+3)-(2x+6)=-3
y(x+3)-2(x+3)=-3
(y-2)(x+3)=-3