Decomposition factor: 16A ^ 2B + 16A ^ 3-4ab ^ 2 Urgent

Decomposition factor: 16A ^ 2B + 16A ^ 3-4ab ^ 2 Urgent

Original form=
4a(4ab-4a^2-b^2)
=-4a(b^2-4ab+4a^2)
=-4a(b-2a)^2

If AB + 1 = 0, find - AB by factorization (a, b5-ab3-b)

AB + 1 = 0
Required AB ^ 2 = - 1
-AB (a power B5 power-ab3 power-b)
=-ab[ab^2(ab^3-b)-b]
=-ab(-ab^3+b-b)
=-ab*(-ab^3)
=a^2b^4
=(-1)^2
=1

It is known that: a + B = 6, the negative power of (AB) = - 1 / 7, find A-B / 3AB × (A & # 178; - B & # 178; / A & # 178; + 2Ab+
²)

Then: ab = - 7 (a-b) (a-b) = (a + b) (a + b) - 4AB = 36 + 28 = 64, the original formula = 64 / 3 * (- 7) = - 64 /

1. Give a group of numbers: 1, √ 2 / 2, √ 3 / 3 ····· √ 99 / 99, √ 100 / 100
If we choose several numbers from them in order to make their sum greater than 3, then at least how many numbers should we choose?

Because √ n / N = √ (1 / N), with the increase of N, the smaller the value of √ n / N is, so the preceding number should be taken
If we take 1, √ 2 / 2, √ 3 / 3 ······ the sum of the first five numbers is greater than 3, if we take 1, √ 2 / 2, √ 3 / 3 ······ the sum of the first four numbers is less than 3
So we should take at least five numbers
I asked our teacher,