Factorization of mathematics in Grade Seven 1. ab-2a-2b+4 2. x³+6x²+11x+6 It's a process. It's a process

Factorization of mathematics in Grade Seven 1. ab-2a-2b+4 2. x³+6x²+11x+6 It's a process. It's a process

1.ab-2a-2b+4
=a（b-2）-2(b-2)
=(a-2)(b-2)
2.x³+6x²+11x+6
=x³+6x²+9x+2x+6
=x(x²+6x+9)+2(x+3)
=x(x+3)²+2(x+3)
=(x+3)(x²+3x+2)
=(x+1)(x+2)(x+3)
ab-2a-2b+4
=a（b-2）-2（b-2）
=（a-2）（b-2）
x³+6x²+11x+6
=x（x²+6x+9）+2x+6
=x（x+3）²+2（x+3）
=（x+3）（x²+3x+2）
=（x+3）（x+1）（x+2）
1. ab-2a-2b+4
=a(b-2)-2(b-2)
=(b-2)(a-2)
2. x³+6x²+11x+6
=x³+5x²+6x+x²+5x+6
=x(x+2)(x+3)+(x+2)(x+3)
=(x+2)(x+3)(x+1)
1.ab-2a-2b+4=a(b-2)-2(b-2)=(a-2)(b-2)
2. If I can't type out, I have to tell you the final result (x + 3) [x (x + 3) + 2]
Let's take a closer look. X3 + 6x2 + 9x + 2x + 6 = x (x + 3) 2 + 2 (x + 3) is equal to the formula above. The number after X is power
Classical arithmetic problems,
Cucumber 1 yuan 13 Jin, watermelon 3 yuan 1, balsam pear 1 yuan 3
My problem is that if the total quantity of three items is 100, the total price is 100
Cucumber 1 yuan 13, I was wrong.
Let cucumber, watermelon and balsam pear each buy x Jin, y pieces and Z pieces, x + y + Z = 100x / 13 + 3Y + Z / 3 = 100 to solve the equations, and eliminate y to get 38x / 13 + 8Z / 3 = 200. Let X / 13 = x, Z / 3 = Z. the original formula is 38x + 8Z = 200. Z = (200-38x) / 8 = 25-19x / 4. Here X and Z are positive integers, so only when x = 4, the equation holds, and the solution is Z =
x/13 + 3y + z/3 = 100
x + y + z = 100
x = 52
y = 30
z = 18
Unit: Units
Starting from the value of X, X can take 13, 26, 39, 52, 65, 78, 91
Although I'm not an expert, the problem is relatively simple
If x is the money for cucumber, 3Y is the money for watermelon, and Z is the money for balsam pear. (x, y, Z are all integers)
Equation 1: x + 3Y + Z = 100
Equation 2: 13X + y + 3Z = 100
To solve the system of equations, there are three positions and two constraints. The equation must have a solution, and then the integer solution is the solution of the problem.
You can also put two equations into rectangular coordinates, each equation represents a surface, and two equations represent a straight line, that is... Expansion
Although I'm not an expert, the problem is relatively simple
If x is the money for cucumber, 3Y is the money for watermelon, and Z is the money for balsam pear. (x, y, Z are all integers)
Equation 1: x + 3Y + Z = 100
Equation 2: 13X + y + 3Z = 100
To solve the system of equations, there are three positions and two constraints. The equation must have a solution, and then the integer solution is the solution of the problem.
In other words, the solution of the equations is on a straight line, and the integer point on the straight line is the solution of the problem.
For example, the following group of x = 4, y = 30, z = 6 ﹣ is folded
This still needs to go up mathematical intelligence expert???
Who can do it with primary school arithmetic without equations?
If "the sum of the three things is 100", which one of "Ge" and "Jin" refers to, or both?
If it is the latter, it can be referred to
Answer on the first floor. Note that the unknown on the first floor is money
I saw everyone's process, I think it is not perfect! Because the root number of cucumbers, the number of watermelons and balsam pear are integers, but the money for buying cucumbers, watermelons and balsam pear is not necessarily integers. The sum is 100. My views are as follows:
If x cucumbers, y watermelons and Z balsam pears were purchased, then:
X+Y+Z=100 ①
X/13+3Y+Z/3=100 ②
① 3 - 2
38x / 13 + 8Z / 3 = 200... Expansion
I saw everyone's process, I think it is not perfect! Because the root number of cucumbers, the number of watermelons and balsam pear are integers, but the money for buying cucumbers, watermelons and balsam pear is not necessarily integers. The sum is 100. My views are as follows:
If x cucumbers, y watermelons and Z balsam pears were purchased, then:
X+Y+Z=100 ①
X/13+3Y+Z/3=100 ②
① 3 - 2
38X/13+8Z/3=200 ③
It is known that: 8Z / 3
Given the function f (x) = 2-x2, G (x) = X. if f (x) * g (x) = min {f (x), G (x)}, then the maximum value of F (x) * g (x) is______ (Note: Min is the minimum value)
According to the meaning of the question, we can make a function image that meets the conditions, such as f (x) * g (x) = 2 − x2 & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; X ≤ − 2x & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; - 2 − x2 & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; X ≥ 1 x ＜ 1, which is known from the image, and the maximum value is 1
Given function f (x) = ax ^ 2 - │ x │ + 2a-1 (a is a real constant)
1. If a = 1, find the monotone interval of F (x);
2. If a > 0, let the minimum value of F (x) in the interval [1,2] be g (a), and find the expression of G (a)
3. Let H (x) = f (x) / x, if the function H (x) is an increasing function in the interval [1,2], find the value range of real number a
The discussion is divided into two situations, x0
2. If f (x) = ax ^ 2-x + 2a-1, the opening is upward and the minimum value is obtained, the relationship between the axis of symmetry and the interval should be studied
1）1/2a
For real numbers C and D, we can use min {C, D} to represent the smaller number of C and D, such as min {3, - 1} = - 1
If the image of the function y = min {2x square, a (x-t) square} of X is symmetric with respect to the line x = 3, then the values of a and t may be
The function f (x) = x ^ 2 (ax-3) defined on R is known, where a is a constant
If the function g (x) = f (x) + F '(x), X ∈ [0,2], get the maximum value at x = 0, find:
The value range of positive number a
F (x) = x ^ 2 (ax-3) f '(x) = 2x (ax-3) + ax ^ 2; G (x) = f (x) + F' (x) = x ^ 2 (ax-3) + 2aX ^ 2-6x + ax ^ 2 = ax ^ 3 + 3ax ^ 2-3x ^ 2-6x. G '(x) = 3ax ^ 2 + (6a-6) X-6 according to the title, X ∈ [0,2], the maximum value is obtained at x = 0, which indicates that the symmetry axis of G' (x) x = (1-A) / a > = 2, so the value range of: A is
f(x)=x^2(ax-3)
f'(x)=2x(ax-3)+ax^2;
g(x)=f(x)+f'(x)
=x^2(ax-3)+2ax^2-6x+ax^2
=ax^3+3ax^2-3x^2-6x.
g'(x)=3ax^2+(6a-6)x-6
x∈[0,2
X = 0 to get the maximum
The symmetry axis of G '(x) is x = (1-A) / a > = 2
The value range of a is as follows:
Zero
Let the function FX have f (x + y) = FX + FY for any real number x, y, and if x > 0, FX
f(0+0)=f(0)+f(0)
f(0)=0
0=f(0)=f(x+(-x))=f(x)+f(-x)
F (x) = - f (- x) is an odd function
f'(x)=f'(-x)
When x > 0, FX
Given that the function f (x) = x / (AX + b), (a, B are constants, and ab ≠ 0), and f (2) = 1, f (x) = x has a unique solution, then the analytic expression of y = f (x) is f (x) =?
If f (x) = x has a unique solution, then the discriminant (B-1) ^ 2 = 0 of the quadratic equation AX ^ 2 + (B-1) x = 0 obtained by x = x / (AX + b) transformation, and the solution is b = 1. Because f (2) = 1, that is, 2 / (2a + b) = 1, substituting B = 1, we can get a = 1 / 2, which is f (x) = 2x / (x + 2)
f(x)=2x/(x+2)
Given the function f (x), if x, y ∈ R, f (x + y) = f (x) + F (y), if x > 0, f (x) > 0, try to judge the monotonicity of F (x) on (0, + ∞)
Let 0 ＜ x1 ＜ X2, then f (x2) = f [X1 + (x2-x1)] = f (x1) + F (x2-x1) and ∵ x2-x1 ＞ 0, ∵ f (x2-x1) ＞ 0, ∵ f (x1) + F (x2-x1) ＞ f (x1), ∵ f (x2) ＞ f (x1), ∵ f (x) is an increasing function on (0, + ∞)
Given the function FX = SiNx + cosx [x ∈ R] (1), find the maximum value of function f (x) and the self variable x to obtain the maximum value
A collection of,
(2) Explain how the image of F (x) can be obtained from the image of y = SiNx
f(x)=√2[(√2/2)sinx＋(√2/2)cosx]=√2[sinxcos(π/4)＋cosxsin(π/4)]=√2sin(x＋π/4)
1. The maximum value is √ 2, then x + π / 4 = 2K π + π / 2, that is to say, the maximum value is obtained from the set of values: {x | x = 2K π + π / 4, K ∈ Z}
2. This function can shift π / 4 units from y = SiNx = = = = > > > > to the left [to get y = sin (x + π / 4)], and then the abscissa of all points on the function image is unchanged, and the ordinate is increased to 2 times of the original √ to get y = √ 2Sin (x + π / 4), that is, y = SiNx + cosx
f(x)=sinx+cosx=√2*[sin(x+π/4)]
F (x) can be obtained by sin, first translating π / 4 units to the left along the x-axis, then enlarging the scale of ordinate by √ 2 times and keeping the abscissa unchanged
The first question f (x) can be reduced to f (x) = radical two SiNx (x + π / 4)
So the maximum value is root 2
Second, the image of SiNx moves π / 4 units to the left, and the abscissa becomes one-half of the original root