It is known that (2x-21) (3x-7) (3x-7) (X-13) can be decomposed into (3x 10a) (X-B), where a and B are integers, then a + 3b is equal to? AB?

It is known that (2x-21) (3x-7) (3x-7) (X-13) can be decomposed into (3x 10a) (X-B), where a and B are integers, then a + 3b is equal to? AB?

Simple calculation method of 22 times 98 plus 44

22 times 100 minus 22 times 2 plus 44 equals 22 times 100,

If n is a positive integer, the value of (n + 11) ² - N & #178; can always be divisible by K, then K is equal to the odd multiple of a 11 b 22 C 11 or 22 D11

(n+11)²-n² =11*(2n+11)
11, 2n + 11 are not divisible by 2, so only k = 11 can satisfy the condition of the problem

16 (3x + 1) = 9 (x + 2) to solve the equation

16(3x+1)=9(x+2)
48x+16=9x+18
48x-9x=18-16
39x=2
x=2/39

Can you calculate 4.8 * 0.25 + 0.48 * 5.2 + 0.048 * 23 in a simple way?

4.8x(0.25+0.52+0.23)=4.8

Let the sum of the first n terms of the sequence an, Sn = 1 + (1 / 16), be multiplied by an to the power of R, and the value range of r that can make limsn = 1 be obtained
The answer is R ＞ 1 / 4. It's better to photograph it for me,

Sn=1+(1/16)^r*an
When n > 1, s (n-1) = 1 + (1 / 16) ^ R * a (n-1)
By subtracting the two formulas, an = (1 / 16) ^ R * (an-a (n-1))
The result is: an = a (n-1) / (1-16 ^ R)
In other words, an is an equal ratio sequence with 1 / (1-16 ^ R) as the common ratio
Then let n = 1 substitute Sn = 1 + (1 / 16) ^ R * an to get A1 = 16 ^ R / (1-16 ^ R)
If the limit of Sn = 1, the absolute value of the common ratio must be less than 1
I.e. - 11 / 4

Find the remainder of 1992 × 59 divided by 7

1992 △ 7 = 284 more than 4
1992×59
=(284×7+4）×59
=284×7×59+4×59
The remainder of 7 is the same as that of 4 × 59 divided by 7
4×59=236
236 ÷ 7 = 33 + 5
The remainder of 1992 × 59 divided by 7 is 5

How to calculate (80 + 8 + 0,8) times 1,25

80X1.25+8X1.25+0.8X1.25
=100+10+1
=111

Find the general solution of the first order differential equation y '= e ^ 2x-4y,

y'=e^(2x-4y)
dy/dx=e^2x/e^4y
e^4ydy=e^2xdx
General solution e ^ 4Y = 2E ^ 2x + C

Given f (x) = x ^ 2 - (a + 1 / a) x + 1, if a is greater than 0, the inequality f (x) about X is less than or equal to 0

f(x)=x^2-(a+1/a)x+1
=(x-a)(x-1/a)
f(x)≤0
(x-a)(x-1/a)≤0
The discussion is divided into three situations
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