Factorization 4 (a square + 1) square - 16A square

Factorization 4 (a square + 1) square - 16A square

4(a^2+1)^2-16a^2
=4[(a^2+1)-4a^2]
=4[(a^2+1-2a)(a^2+1-2a)]
=4(a^2+1-2a)(a^2+1-2a)

Factorization: square of a - square of 9b + a-3b =?

Factorization: (- 2Ab) (3a ^ 2-2ab-4b ^ 2) =?

(-2ab)(3a²-2ab-4b²)=-6a³b+4a²b²+8ab³
I understand. Please accept,
If you have any new questions, please ask for help,

If Sn = 48, s2n = 60, then s3n = ()
A. 63B. 64C. 66D. 75

∵ sequence {an} is an equal ratio sequence; Sn, s2n Sn, s3n-s2n is an equal ratio sequence; s3n − s2n & nbsp; s2n − Sn = s2n − snsn, that is, s3n − 6060 − 54 = 60 − 5454 − s3n = 63, so select a

1. The radius of a circle is increased by 1 / 5, and the area is increased by what percentage?

44 percent

The section of the reservoir dam is trapezoidal ABCD, the crest ad is 6 meters, the slope length CD is 8 meters, the slope bottom BC is 30 meters, and the angle ADC is 135 degrees
Find the size of angle ABC?
If the dam is 100 meters long, how many earth and stone materials are needed to build the dam?

For de ⊥ BC, AF ⊥ BC, e and f respectively, 〈 ADC = 135 °, then 〈 C = 45 °, de = dcsin, 45 ° = 8 * {2 / 2 = 4} 2, CE = de = 4} 2, Fe = ad = 6, BF = bc-ce-ef = 30-6-4} 2 = 24-4} 2, AF = de = 4} 2, tanb = AF / BF = 4} 2 / (24-4} 2) = (3} 2 + 1) / 17 ≈ 0.30835, 〈 B ≈ 17.1

Please help me to solve the math problems of grade four in primary school
Grandfather's age this year is six times that of Xiao Ming. In a few years, grandfather's age will be five times that of Xiao Ming. In a few years, grandfather's age will be four times that of Xiao Ming. Do you know how old Xiao Ming's grandfather is this year?

My grandfather is 72 years old and Xiao Ming is 12 years old
Three years later, my grandfather is 75 years old, and Xiao Ming is 15 years old. My grandfather is five times as old as Xiao Ming;
Five years later, my grandfather is 80 years old, and Xiao Ming is 20 years old. My grandfather is four times as old as Xiao Ming;
According to the first sentence "grandfather's age this year is six times the age of Xiaoming", we can infer that grandfather's age this year can be divided by six. Combined with the actual life, we can infer that grandfather's age is 48, 54, 60, 66, 72 and 78 years old
Take the 54 year old grandfather as an example: then Xiaoming's age is 54 / 6 = 9 years old,
After X years, my grandfather will be five times as old as Xiao Ming,
So we have (54 + x) / (9 + x) = 5,
If the solution x is not an integer, it means that grandfather's age assumption is wrong, change the age to continue the calculation
If x is an integer, it means that grandfather's age hypothesis may be correct. Continue to verify as follows: take grandfather 48 as an example, then Xiao Ming is 8 years old
After another year y, my grandfather will be four times as old as Xiao Ming,
Then, (48 + X + y) / (8 + X + y) = 4

If the determinants of two matrices of order 2 are reciprocal to each other, are these two matrices invertible?
Can you give me a matrix whose determinants are reciprocal but they are not? thank you

It is invertible, indicating that determinant ≠ 0
Remember, it's not reciprocal
as
（1 0
0 1）
Its inverse matrix is itself
（2,0
0,1/2)
Its inverse matrix is:
(1/2 0
0 2)
Their determinants are equal to 1, reciprocal
This is not called reciprocal, they are invertible matrices
Reciprocal means: ab = E
A and B are inverse to each other

1 2 3 4 5 6 7 8 9 10 add, subtract, multiply and divide to be 203

1 X 2 X 3 X 4 X 5 - 6 + 7 - 8 + 9 X 10=203
Ideas
Construct the number closest to 200. According to the order, you can only multiply it. When you multiply it to 5, it will be 120, and then it will be 720. At this time, you stop, leaving 678910. Observe, the product of the last two digits is 90120 + 90, which is 210. It's very close. Try to combine 678 into - 7, and it's successful
If we multiply every two numbers into a group and add and subtract them when we have 678910 left, we can't get the number closest to 203 in any combination

A is a square matrix of order r, B is an R * n matrix, rank (b) = R, ab = 0, prove a = 0
Come on

From ab = 0, and the number of columns of a, the number of rows of B is r, we get: rank (a) + rank (b) ≤ R, and rank (b) = R, so rank (a) = 0, so a = 0