Given that | a | = 2, B = - 7, the opposite number of C is - 5, find the value of the algebraic formula a + (- b) + (- C)

Given that | a | = 2, B = - 7, the opposite number of C is - 5, find the value of the algebraic formula a + (- b) + (- C)

When a = - 2, B = - 7, C = 5. Original = - 2 + 7-5 = 0
When a = 2, primitive = 2 + 7-5 = 4

2. When x = 2, the square of the algebraic formula ax + BX is 12. When x = - 2, the value of the algebraic formula is 4. Find the value of a and B

4a＋2b=12
4a-2b=4
Additive, 2
4b=16
b=4
therefore
4a＋8=12
4a=4
a=1
Namely
a=1,b=4

It is known that for the algebraic formula AX2 + BX + C, when x = 1, y = - 4; when x = - 1, y = - 12; when x = 3, y = - 20

When x = 1, y = - 4; when x = - 1, y = - 12; when x = 3, y = - 20 is substituted into the algebraic expression AX2 + BX + C to get: a + B + C = - 4 & nbsp; ① a − B + C = - 12 & nbsp; ② 9A + 3B + C = - 20 ③, ① - ② to get: 2B = 8, B = 4, ③ - ② to get: 8A + 4B = - 8 ④, B = 4 to get: a = - 3, a = - 3, B = 4 to get: C = - 5; then the values of a, B, C are - 3, 4, - 5 respectively

For the algebraic expression ax 5 square + BX 3 square + CX + 8, we know that when x = 3, its value is 69, then when x = - 3, what is its value equal to

When x = 3,
5 square of AX + 3 square of BX + CX + 8
=a*3^5+b*3^3+3c+8
=69
a*3^5+b*3^3+3c=69-8=61
When x = - 3,
5 square of AX + 3 square of BX + CX + 8
=a*(-3)^5+b*(-3)^3+(-3)c+8
=-(a*3^5+b*3^3+3c)+8
=-61+8
=-53

When x is equal to 5, the value of the algebraic formula ax ^ 5 + BX + 2 is 18. When x = - 5, the value of the algebraic formula ax ^ 5 + BX ^ 3-cx + 2 is obtained

Let f (x) = ax ^ 5 + BX, then f (- x) = - f (x)
When x = 5, f (5) + 2 = 18, f (5) = 16
When x = - 5, there is:
ax^5+bx+2=f(-5)+2
=-f(5)+2
=-16+2
=-14

If the value of algebraic formula 2 times the square of X + 3x + 7 is 12, find the value of algebraic formula 4 times the square of X + 6x-10

2 times the square of X + 3x + 7 = 12
2x^2+3x=5
4 times the square of X + 6x-10
2*5-10=0

Let n be a perfect square number, n be at least three digits, its last two digits are not 00, and after removing these two digits, the remaining number is still a perfect square number, then the maximum value of n is zero______ ．

Let n = X2 (x is a natural number), and the last two digits of N form an integer y. remove these two digits and get the integer m, M = K2 (k is a natural number), then 1 ≤ y ≤ 99, X2 = 100k2 + y, y = x2-100k2 = (x + 10K) (x-10k). Let x + 10K = a, x-10k = B, then B ≥ 1, K ≥ 1, x = 10K + B ≥ 11, a = x + 10K

A mathematical proof of the square number
Given that the integer P is equal to the sum of the squares of two adjacent natural numbers, find the least three digit P
incidentally..
Prove: divide all square numbers into two groups, there must be a group in which the sum of two numbers is a square number

① It is known that the integer P is equal to the sum of the squares of two adjacent natural numbers, the least three digit P
Let two adjacent natural numbers be n, N + 1
Then p = n ^ 2 + (n + 1) ^ 2 ≥ 100
2n^2+2n+1≥100
n^2+n≥49.5
(n+0.5)^2≥49.75
n≥√49.75-0.5
n>6.553
The least three digits P = 7 ^ 2 + 8 ^ 2 = 49 + 64 = 113
② Prove: divide all square numbers into two groups, there must be a group in which the sum of two numbers is a square number
As long as we can find out that the sum of any two squares is three squares of the square,
Then divide all square numbers into two groups. There must be a group in which the sum of two numbers is square
∵44^2=1936,117^2=13689,240^2=57600
1936+13689=15625=125^2
13689+57600=71289=267^2
57600+1936=59536=244^2
If we divide all the squares into two groups, 19361368957600, two of the three squares must be in the same group,
The sum of the two numbers in the same group is the square
If we divide all the square numbers into two groups, there must be a group in which the sum of two numbers is the square number

6 ^ m + 2 ^ n + 2 (m, n is a natural number) is a complete square number. Find all possible values of M, n

6^m+2^n+2
=2^m*3^m+2^n+2
=2[2^(m-1)3^m+2^(n-1)+1]
It can be seen that when the index > 0
2 ^ (m-1) 3 ^ m + 2 ^ n is an even number
So [2 ^ (m-1) 3 ^ m + 2 ^ n + 1] is an odd number
It can't be a perfect square after multiplying 2
So only when m = n = 0
The original formula is 4

Given that X and y are opposite to each other, and (x + y + 3) (x + 4-y) = 6, then (# 178; y) [← square] of - x=
[selection] if the square of equation (m-2) x (M & # 178; - 3) = 5 is a linear equation of one variable with respect to x, then the value of M is ()
a.+-2 b.-2 c.2 d.4
The following statement is correct ()
a. If a = B, then a / (1 + m) = B / (1 + m) B. If x-3 = Y-3, then X-Y = 0
c. If a & # 178; = B & # 178;, then a = B, if MX = my, then x = y
There are m buses and n people. If each bus takes 40 people, there are Baidu search this topic
N-10 / 40 = n + 1 / 43

The first question x = 1, y = - 1, the following formula is not understood