Question

已知（a-c)²=4（a-b)(b-c)，求（a-2b+c)²的值

Answer 1

（a-c)²=4（a-b)(b-c)
【(a-b)+(b-c)】²=4（a-b)(b-c)
(a-b)²+2(a-b)(b-c)+(b-c)²=4(a-b)(b-c)
(a-b)²-2(a-b)(b-c)+(b-c)²=0
[(a-b)-(b-c)]²=0
即 （a-2b+c)²=0

Answer 2

（a-c)²=[（a-b）+(b-c)]²=(a-b)²+(b-c)²+2(a-b)(b-c)=4（a-b)(b-c)
则(a-b)²+(b-c)²=2(a-b)(b-c)
（a-2b+c)²=[(a-b)-(b-c)]²=(a-b)²+(b-c)²-2(a-b)(b-c)=0

Answer 3

(a－c)²＝4(a－b)(b－c)
[(a－b)＋(b－c)]²－4(a－b)(b－c)＝0
(a－b)²＋(b－c)²＋2(a－b)(b－c)－4(a－b)(b－c)＝0
(a－b)²＋(b－c)²－2(a－b)(b－c)＝0
[(a－b)－(b－c)]²＝0
(a－2b＋c)²＝0

Answer 4

∵(a－c)²＝4(a－b)(b－c)
∴[(a－b)＋(b－c)]²－4(a－b)(b－c)＝0
∴(a－b)²＋(b－c)²＋2(a－b)(b－c)－4(a－b)(b－c)＝0
∴(a－b)²＋(b－c)²－2(a－b)(b－c)＝0
∴[(a－b)－(b－c)]²＝0
∴(a－2b＋c)²＝0

Answer 5

（a-c)²=4（a-b)(b-c)
（a-c)²-4（a-b)(b-c)=0
a²-4ab+4b²+2ac-4bc+c²=（a-2b+c)²=0