a) ta có: x chia hết cho 25
\(\Rightarrow x\in B_{\left(25\right)}=\left\{25;50;...\right\}\)
b) ta có: 15 chia hết cho x + 13
\(\Rightarrow x+13\inƯ_{\left(15\right)}=\left\{\pm1;\pm3;\pm5;\pm15\right\}\)
nếu x + 13 = 1 => x = -12 (Loại)
..
bn tự xét tiếp nhé
\(a.x⋮25\)
\(\Rightarrow x\in B\left(25\right)\)
\(b.15⋮\left(x+13\right)\)
\(\Rightarrow\left(x+13\right)\in B_{\left(15\right)}=\left\{-15,-5,-3,-1,1,3,5,15\right\}\)
\(x+13=-15\Rightarrow x=-15-13=-28\)
\(x+13=-5\Rightarrow x=-5-13=-18\)
\(x+13=-3\Rightarrow x=-3-13=-16\)
\(x+13=-1\Rightarrow x=-1-13=-14\)
\(x+13=1\Rightarrow x=1-13=-12\)
\(x+13=3\Rightarrow x=3-13=-10\)
\(x+13=5\Rightarrow x=5-13=-8\)
\(x+13=15\Rightarrow x=15-13=2\)
\(x\in\left\{-28;-18;-16;-14;-12;-10;-8;2\right\}\)