Given the equation 3x + 3 (x + 2) = 24, please write a mathematical problem (to be practical) and solve it

Given the equation 3x + 3 (x + 2) = 24, please write a mathematical problem (to be practical) and solve it

A and B set out at the same time from places a and B, which are 24 kilometers apart. They meet each other in three hours. It is known that B's speed is 2 kilometers faster than a's speed per hour. Let a's speed be x kilometers per hour. According to the meaning of the question: 3x + 3 (x + 2) = 24, the solution is: x = 3. A's speed is 3 kilometers per hour
Given the mapping f: a → B, where a = b = R, the corresponding relation F: X → y = - the square of X + 2x, for the real number k ∈ B, there are two different primitives in the set a
Then the value range of K is?
For K ∈ B, there are two different primitives in a
Then x & # 178; + 2x-k = 0 has two unequal real roots
Δ=4+4k>0
k>-1
A: the range of K is k > - 1
That is, for the real number k ∈ B, there are two different primitives in the set a,
∴y=-x2+2x=-（x2-2x+1）+1≤1，
When it is equal to 1, there are two identical x's,
∴k＜1，
From the problem, we can get that - X & # 178; + 2x = k has two unequal real roots
Draw an image of y = - X & # 178; + 2x, and use y = k to intercept the image, which has two intersections
Then K < 1
Factorization, (a2-a-72) + (AB + 8b)
Note: 2 after the first a is quadratic,
a^2-a-72+(ab+8b)=(a-9)(a+8)+b(a+8)=(a-9+b)(a+8)
a^2-a-72+(ab+8b)
=(a-9)(a+8)+b(a+8)
=(a+b)(a-9+b)
Try to draw a geometric figure whose area can represent the identity (2a-b) (a + 2b) = 2A ^ 2 + 3ab-2b ^ 2,
Identity (2a-b) (a + 2b) = 2A ^ 2 + 3ab-2b ^ 2
First draw a rectangle of length 2a and width a, then subtract B from length and add 2b to width, and get the area of the figure. The result is obvious. & nbsp; supplementary answer: & nbsp; as shown in the figure, ab = 2A, ah = a, BC = B, HF = 2B, then s rectangle AFHC = s rectangle abdh + s rectangle hfeg-s rectangle bced = 2A × a + (2a-b) × 2B AB = 2A & # 178; + 4ab-2b & # 178; - AB = 2A & # 178; + 3ab-2b & # 178;
On the multiplication, division and factorization of eighth grade integers
1. It is known that the n-th power of 3 + the m-th power of 11 can be divisible by 10. Please explain that the n-th power of 3 + the m-th power of 11 + 2 can also be divisible by 10
2. If the third power of 3 = - 8 times the sixth power of a times the ninth power of B, find the value of X
It's a mistake. If the third power of x = - 8 times the sixth power of a times the ninth power of B, find the value of X
First question:
N + 4 power of 3 + m + 2 power of 11 = 11 ^ 2 (3 ^ n + 11 ^ m) + (3 ^ 4-11 ^ 2) 3 ^ n
=121(3^n+11^m)-40*3^n
3 ^ n + 11 ^ m can be divided by 10, 40 can be divided by 10, so the N + 4 power of 3 + m + 2 power of 11 can also be divided by 10
Second, the landlord may have made a mistake. Where is x?
A. M = 2, n = 1, B.M = 2, n = 0, C.M = 4, n = 1, D.M = 4, n = 0___ I want the answer to the whole paper 1C 2C 3A 4B
(5a square - 3AB + b square) - (2a square + 3ab-2b Square), where a square + b square = 10, ab = 3
3a2-6ab+3b2
3（a2-2ab+b2）
3x（10-2x3）=12
Factorization exercise problem solving method (quick solution) (x ^ 2 + 3x) ^ 2-2x ^ 2-6x-8
(x^2+3x）^2-2x^2-6x-8
=(x^2+3x)^2-2(x^2+3x)-8
=(x^2+3x-4)(x^2+3x+2)
=(x+4)(x-1)(x+2)(x+1)
（x^2+3x）^2-2x^2-6x-8
=（x^2+3x）^2-2（x^2+3x）-8
=(x^2+3x-4)(x^2+3x+2)
=(x+4)(x-1)(x+1)(x+2)
(x^2+3x)^2-2(x^2+3x)-8
=(x^2+3x-4)(x^2+3x+2)
=(x+4)(x-1)(x+2)(x+1)
Find the value of 5A ^ 2-3ab ^ 2-2a ^ 2 3ab-2b ^ 2, where a ^ 2, B ^ 2 = 10, ab = 3
5a^2-3ab=5×10-3×3=50-9=41
b^2-2a^2=10-2×10=10-20=-10
3ab-2b^2=3×3-2×10=9-10=-1
Seeking three eight mathematical problems urgently
1. Given x + y = 0, x + 3Y = 1, find the value of 3x & sup2; + 12xy + 13y & sup2
2. Factorization: 4A & sup2; - 4AB + B & sup2; - 6A + 3b-4
3. Observe the following formulas: 1 × 2 × 3 × 4 + 1 = 5 & sup2;; 2 × 3 × 4 × 5 + 1 = 11 & sup2;; 3 × 4 × 5 × 6 + 1 = 19 & sup2;; judge whether the sum of the product of any four consecutive positive integers and 1 is the square of a certain positive integer, and explain the reason
The answer was added
1.3x & sup2; + 12xy + 13y & sup2; = (3x & sup2; + 12xy + 9y & sup2;) + 4Y & sup2; = 3 (x + 3Y) (x + y) + Y & sup2; = 0 + 4Y & sup2; = 4Y & sup2;. Is there a problem with the title? 2.4a & sup2; - 4AB + B & sup2; - 6A + 3b-4 = (2a-b) & sup2; + (- 6A + 3b) - 4 = (2a-b + 1) (2a-b-4)
Let 2A power = 5B power = m, and 1 / A + 1 / b = 2
Why not the plus and minus root sign ten
The a power of 2 is positive, not negative
The value of M can only be positive, that is, M = √ 10