因?yàn)?方程 (a + 1)x^|a + 2| - 2 = 0是一元一次方程
所以,｜ a + 2 | = 1
解得：a = - 1 或 a = - 3
當(dāng) a = - 1時(shí),方程系數(shù) a + 1 = 0 ,方程無(wú)解,故應(yīng)舍去
因此,3/(2a^2 - a) - 2/( a^2 - a) - 6/(a^2 + 5a)
=3/a(2a - 1) - 2/a(a - 1) - 6/a(a + 5)
= 3(a + 5)/a(2a -1)(a + 5) - 2(a + 5)/a(2a - 1)(a + 5) + 6(2a - 1)/a(2a- 1)(a + 5)
= [ 3(a + 5) - 2(a + 5) + 6(2a - 1)]/a(2a - 1)(a + 5)
= (a + 5 + 12a - 6)/a(2a - 1)(a + 5)
= ( 13a - 1)/a(2a - 1)(a + 5)
把a(bǔ) = - 3代入上式得：
原式 = （13 × (-3) - 1 )/(-3)(2×(-3) - 1)(-3 + 5)
= ( - 40)/42
= - 20/21