4a^2+4ab+b^2-64點因式分解?
請詳解!!
一元二次問題:(中學)
2006-10-18 8:00 am
回答 (3)
2006-10-18 8:03 am
✔ 最佳答案
4a^2+4ab+b^2-64=(2a+b)^2-8^2
=(2a+b+8)(2a+b-8)
2006-10-18 6:54 pm
4a^2+4ab+b^2-64
=(4a^2+4ab+b^2)-64
=[(2a)^2+2(2a)(b)+b^2]-64
=(2a+b)^2-64..............................................a^2+2ab+b^2=(a+b)^2
=(2a+b)^2-(8)^2
=[(2a+b)+8][(2a+b)-8]..................................a^2-b^2=(a+b)(a-b)
=(2a+b+8)(2a+b-8)
=(4a^2+4ab+b^2)-64
=[(2a)^2+2(2a)(b)+b^2]-64
=(2a+b)^2-64..............................................a^2+2ab+b^2=(a+b)^2
=(2a+b)^2-(8)^2
=[(2a+b)+8][(2a+b)-8]..................................a^2-b^2=(a+b)(a-b)
=(2a+b+8)(2a+b-8)
2006-10-18 9:03 am
4a^2 + 4ab + b^2 - 64
= (2a+b)^2 - 64
= (2a+b)^2 - 8^2
= (2a+b+8) (2a+b-8)
= (2a+b)^2 - 64
= (2a+b)^2 - 8^2
= (2a+b+8) (2a+b-8)
收錄日期: 2021-04-12 19:04:48
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