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a) x + 7 = -12
x = (-12) - 7
x = 19
b) x - 15 = -21
x = (-21) + 15
x = -6
c) 13 - x = 20
x = 13 - 20
x = -7
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=>5/13>5/p>5/5q>5/16
=>13<p<5q<16
=>p=14; 5q=15
=>p=14; q=3
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a) 5 4 : 5 2 = 5 4 - 2 = 5 2
b) 11 4 : 11 2 = 11 2
c) 10 7 : 10 2 : 10 3 = 10 2
d) a 11 : a 7 : a = a 3
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\(A=\frac{3}{n+5}\left(n\inℤ\right)\)
A là số nguyên \(\Leftrightarrow\frac{3}{n+5}\)là số nguyên
\(\Leftrightarrow3⋮\left(n+5\right)\)
\(\Leftrightarrow n+5\inƯ(3)\in\left\{\pm1;\pm3\right\}\)
Ta có bảng sau:
n+5 | -3 | -1 | 1 | 3 |
n | -8 | -6 | -4 | -2 |
Vì \(n\inℤ\Leftrightarrow n\in\left\{-8;-6;-4;-2\right\}\)
Vậy \(n\in\left\{-8;-6;-4;-2\right\}\)thì A là số nguyên.
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\(a^{11}:a^7:a=a^{11-7-1}=a^3\)