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First, we have
(1 + 2/x²) (2x - 3/x)⁵ = (2x - 3/x)⁵ + 2/x² (2x - 3/x)⁵
In the expansion of (2x - 3/x)⁵, there is no x² term. Each term takes the form
[tex]c (2x)^{5 - i} \left(-\dfrac3x\right)^i[/tex]
where c is a binomial coefficient, and i is taken from the range {0, 1, 2, 3, 4, 5}. Looking at just the power of x in the product, we have
[tex]x^{5 - i} \left(\dfrac1x\right)^i = x^{5 - 2i}[/tex]
and 5 - 2i = 1 only if i = 2. By the binomial theorem, this term is given by
[tex]\dbinom52 (2x)^{5-2} \left(-\dfrac3x\right)^2 = \dfrac{5!}{2!(5-2)!} \cdot 2^3 \cdot (-3)^2 x = 720x[/tex]
In the other expansion, we have an additional factor of 1/x², so that any given them in the expansion contains a power of x of the form
[tex]\dfrac1{x^2} x^{5 - i} \left(\dfrac1x\right)^i = x^{3 - 2i}[/tex]
and 3 - 2i = 1 only if x = 1. This term in the expansion is
[tex]\dfrac2{x^2} \dbinom51 (2x)^{5-1} \left(-\dfrac3x\right)^1 = \dfrac{5!}{1!(5-1)!} \cdot 2^5 \cdot (-3)^1 x = -480x[/tex]
Then the coefficient of the x term in the whole expansion is 720 - 480 = 240.