Question

(2n²–3n)–[2(n–1)²-3(n–1)]怎么化简

Answer 1

(2n^2–3n)–[2(n–1)^2-3(n-1)]
=(2n^2-3n)-(2n^2-4n+2-3n-3)
=(2n^2-3n)-(2n^2-7n-1)
=4n+1

追问

最后是4n–5 不是4n＋1

追答

(2n^2–3n)–[2(n–1)^2-3(n-1)]
=(2n^2-3n)-(2n^2-4n+2-3n+3)
=(2n^2-3n)-(2n^2-7n+5)
=4n-1

追问

4n-5啊 好不好

追答

(2n^2–3n)–[2(n–1)^2-3(n-1)]
=(2n^2-3n)-(2n^2-4n+2-3n+3)
=(2n^2-3n)-(2n^2-7n+5)
=4n-5

追问

哦可

Answer 2

(2n²–3n)–[2(n–1)²-3(n–1)]
=[2n²–2(n–1)²]+[3(n–1)–3n]
=4n-2-3
=4n-5

追问

4n咋来的

在吗

追答

a²–b²=(a-b)(a+b)
2n²–2(n–1)²=2[n-(n-1)][n+(n-1)]=2(2n-1)=4n-2

追问

到底是4n-5 还是4n–2