數學恆等式求解 展開-4(a-5b)(-5b-a)?
回答 (6)
2016-10-15 8:57 am
✔ 最佳答案
-4(a-5b)(-5b-a)= 4(a-5b)(a+5b)
= 4(a²-25b²)
= 4a²-100b²
我先把第二個括號內的兩個負號一起拿出來，
-4(a-5b)(-5b-a) = -(-4)(a-5b)(5b+a)=4(a-5b)(a+5b)
再利用公式x²-y²=(x+y)(x-y)，得到：
4(a-5b)(a+5b)=4[ (a)² - (5b)² ] = 4(a²-25b²)
最後再把4倍乘進去，得
4(a²-25b²)=4a²-100b²
有問題可以再問我！
2016-10-24 3:30 pm
-4(a-5b)(-5b-a)
=(-4a+20b)(-5b-a)
=(20ab)+(4a^2)-(100b^2)-(20ab)
=4a^2-100^2
=(-4a+20b)(-5b-a)
=(20ab)+(4a^2)-(100b^2)-(20ab)
=4a^2-100^2
2016-10-23 12:14 pm
-4(a-5b)(-5b-a)
=-(-4) (a-5b) (a+5b)
=4(a*a-25(b)(b))
=4a*a-100b*b
=4a²-100b²
=-(-4) (a-5b) (a+5b)
=4(a*a-25(b)(b))
=4a*a-100b*b
=4a²-100b²
2016-10-20 12:51 am
4(a-5b)^2
=4(a^2-10ab+25b^2)
=4a^2-40ab+100b^2
=4(a^2-10ab+25b^2)
=4a^2-40ab+100b^2
2016-10-16 1:05 am
-4(a-5b)(-5b-a)
=4(a-5b)(a+5b)
=4(a^2 -25b^2)
=4a^2-100b^2
=4(a-5b)(a+5b)
=4(a^2 -25b^2)
=4a^2-100b^2
2016-10-15 8:54 am
Sol
－4(a－5b)(－5b－a)
=4(a－5b)(a+5b)
=4(a^2－25b^2)
=4a^2－100b^2
－4(a－5b)(－5b－a)
=4(a－5b)(a+5b)
=4(a^2－25b^2)
=4a^2－100b^2
收錄日期: 2021-04-18 15:50:16
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20161015003334AAYc7WY