If the quadratic power of polynomial x + ax + 16 is the square of a polynomial, find the value of A

If the quadratic power of polynomial x + ax + 16 is the square of a polynomial, find the value of A

8 or - 8

How much is 0.36 to 0.12

36 vs. 0. 12
=36:12
=3:1

Kneel down to seek the courseware (or explanation) and exercises of sequence splitting cancellation, dislocation subtraction and reverse order addition!
Hope to get the courseware (or explanation) and exercises about the sum method of the fifth sequence unit of the compulsory mathematics in senior one, such as the split term elimination method, the dislocation subtraction method and the reverse order addition method

Adding in reverse order is like Gaussian algorithm
Generally used for summation of arithmetic sequence
For example, 1 + 2 + 3 + 4 +. + 99 + 100 is written as 100 + 99 + 98 + 97 +... + 2 + 1
It's 100 101's, and then it's divided by two
This method is used in a narrow range, unless there is a special sequence
For example, an + A1 = constant
Split term cancellation
This type of question is generally used in the case of arithmetic sequence multiplication
as
An = 1 / N * (n + 1) so an = ((n + 1) - n) / N * (n + 1) = 1 / N - 1 / (n + 1)
An = 1 / N * (n + k) k is a constant
Multiply the numerator and denominator by K, that is, an = K / k * n * (n + k) = (1 / k) * (n + k - n) / (n * (n + k))
=(1/k)*(1/n - 1/(n+k) )
An=1/n*(n+k)(n+2k)
K is a constant
Multiply the numerator and denominator by 2K
That is, an = 2K / 2K * n * (n + k) (n + 2K)
=(1/2k)*(n+2k - n)/n*(n+k)(n+2k)
=(1/2k)*(1/n*(n+k) - 1/(n+k)(n+2k)
In the next four and five, we'll see less
For other split terms
as
When (an + 1 - an) / an + 1 appears, it can also be considered to change it into 1 / an + 1 - 1 / an, and then 1 / an can be regarded as a new sequence
There is also a forced split term
An=n*(2^n)
Let an = BN + 1 - BN, then Sn = a1 + A2 +... + an = (b2-b1) + (b3-b2) + (BN + 1 - BN)
=Bn+1 - Bn
If we observe that there is a 2 ^ n after an, we can be sure that there is also a 2 ^ n after BN
Let BN = (KN + T) 2 ^ n, then BN + 1 = (K (n + 1) + T) 2 ^ (n + 1)
Write 2 ^ (n + 1) as 2 * 2 ^ n and multiply 2
Bn+1 = (2K(n+1)+2T)2^n=（2Kn+2K+2T)2^n
An=Bn+1 - Bn =(2Kn+2K+2T -Kn - T)2^n=(Kn+2K+T)2^n
Compared with an
K = 1, 2K + T = 0, so t = - 2
Bn=(n-2)*2^n
Sn=Bn+1 - B1 =(n-1)2^（n+1）+2
An = n * (2 ^ n) can also be calculated by the following dislocation subtraction
But for example, an = (n ^ 2 + 1) 2 ^ n dislocation subtraction takes two times, which is very complicated. It is easy to use the split term
Let BN = (KN ^ 2 + TN + C) 2 ^ n, and then follow the above steps
Dislocation subtraction
It is mainly used in the case that the equal ratio sequence and the equal difference sequence want to multiply. The method is to multiply the common ratio and then dislocation
For example, an = 1 / 2 ^ n
Let s = 1 / 2 + 1 / 4 + 1 / 8 +. + 1 / 2 ^ n
2S=1+1/2 + 1/4 +1/8 + .+1/2^（n-1）
S = 1-1 / 2 ^ n
An=n/2^n
Let s = 1 / 2 + 2 / 4 + 3 / 8 +. N / 2 ^ n
2S= 1 + 2/2 + 3/4+.n/2^(n-1)
After dislocation cancellation
S=（1+1/2+1/4.+1/2^(n-1) )-n/2^n
=2- 1/2^(n-1)-n/2^n

Given that the image of the function y = x2 + 2 (a + 2) x + A2 has two intersections with the x-axis, and both are on the negative half axis of the x-axis, then the value range of a is______ ．

Let the abscissa of the intersection of the parabola and the x-axis be x1, x2. According to the meaning of the question, we can get X1 + x2 = - 2 (a + 2) < 0. ① x1 · x2 = A2 > 0. ② and △ = b2-4ac > 0, ∧ a + 1 > 0 & nbsp; ③ the solution is: a > - 1 and a ≠ 0. So fill in the blanks: a > - 1 and a ≠ 0

Finding the greatest common factor by short division
5,9
39,42
18,36
24,16

The greatest common factor of 5.9 is 1
39,42 greatest common factor 3
18,36 greatest common factor 18
24,16 greatest common factor 8

Given a > 0b > 0, prove a ^ 3 + B ^ 3 > = a ^ 2B + AB ^ 2

∵a>0,b>0
∴(a-b)^2≥0
That is, a ^ 2-2ab + B ^ 2 ≥ 0
That is, a ^ 2-AB + B ^ 2 ≥ ab
And ∵ a > 0, b > 0
∴a+b>0
∴(a+b)(a^2-ab+b^2)≥(a+b)ab
That is, a ^ 3 + B ^ 3 ≥ a ^ 2B + AB ^ 2

Sum: SN = x + 2x ^ 2 + 3x ^ 3 + +nx^x

Analysis: multiply both sides by xxsn = x ^ 2 + 2x ^ 3 + nx^(n+1).①Sn=x+2x^2+3x^3+…… +Then (1-x) * Sn = x + x ^ 2 + x ^ 3 + +x^n-nx^(n+1)Sn=(x+x²+x³+...+x^n)/(1-x)=[ x(1-x^n)/(1-x)]/(1-x)=x(1-x^n)/(1-x)²...

A simple method to calculate 4.28 * 12.5 * 0.8 =?

4.28*12.5*0.8
=4.28*（12.5*0.8）
=4.28*10
=42.8

Subtract 4 from 9 times a number and divide it by 5. The quotient is 10. Solve the equation

9x-4/5=10
9x-4=50
x=6

2(0.3x+4）-5（0.2x-7）=9

2(0.3x+4）-5（0.2x-7）=9
0.6x+8-x+35=9
-0.4x=-34
x=85