Bài 1:
a) \(A=\sqrt{\dfrac{\left(a^2-3\right)^2+12a^2}{a}}+\sqrt{\left(a+2\right)^2-8a}\)
\(A=\sqrt{\dfrac{a^4-6a^2+9+12a^2}{a^2}}+\sqrt{a^2+4a+4-8a}\)
\(A=\sqrt{\dfrac{\left(a^2+3\right)^2}{a^2}}+\sqrt{\left(a-2\right)^2}\)
\(A=\dfrac{a^2+3}{\left|a\right|}+\left|a-2\right|\)
Nếu a < 0 thì \(A=\dfrac{-a^2-3-a^2+2a}{a}=\dfrac{-2a^2+2a-3}{a}\)
Nếu 0 < a ≤ 2 thì A = \(\dfrac{a^2+3-a^2+2a}{a}=\dfrac{2a+3}{a}\)
Nếu a > 2 thì A = \(\dfrac{a^2+3+a^2-2a}{a}=\dfrac{2a^2-2a+3}{a}\)
b) Với a ∈ Z thì | a - 2 | ∈ Z
Do đó: để A ∈ Z thì a2 + 3 ⋮ |a| hay 3 ⋮ |a|
Suy ra: a ∈ \(\left\{\pm1;\pm3\right\}\)
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