The solution of mathematical binary linear equation In order to make two kinds of small rectangular boxes without cover, a paper factory uses leftover materials to cut out square and rectangular pieces of hard paper. The width of the rectangle is equal to the side length of the square. Now 150 pieces of square pieces of hard paper and 300 pieces of rectangular pieces of hard paper are all used to make these two kinds of small boxes. How many pieces of small boxes can be made There's no roof

Suppose that X type a cuboid boxes and Y type B cuboid boxes can be made,
According to the theme
x+2y=150------(1)
4x+3y=300---(2)
From (1), it is concluded that:
x=150-2y-----------(3)
Substituting (3) into (2) yields: 4 (150-2y) + 3Y = 300
-5y=-300
y=60
Substituting y = 60 into (1) yields x = 30
x=30
y=60
Let f (x) be a decreasing function defined in a positive infinite interval, f (XY) = f (x) + F (y). If f (- 3) = 2, the inequality f (x) + F (2-x) < 2 is solved
The answer is: - 1
The negative integer solutions of inequality systems 1-2 parts x ≥ 0 and X + 2 > - 2 parts 1 are
1-x/2≥0
x+2>-1/2
1≥x/2
x>-5/2
2≥x
x>-5/2
SO 2 ≥ x > - 5 / 2
So the negative integer solution is - 2, - 1
① So, X ≤ 2
② So, X ＞ - 5 / 2
The solution set of inequality system:
﹣5/2﹤x≤2
Negative integer solutions; - 2, - 1
From 1-x/2>=0 we can see that 2-x>=0, x-1/2 we can see that x>-2.5
The question requires x to be negative integer, we can work out that x=-2 or -1
It's embarrassing that
From 1-x/2>=0 we can see that 2-x>=0, x-1/2 we can see that x>-2.5
The question requires x to be negative integer, we can work out that x=-2 or -1
It's embarrassing that the computers in my school don't have the language of Chinese, so I can't answer in Chinese. Hope there's no limitation for you to get my point.
Any way, would you please accept my answer? For I worked hard on this
-2 -1
Mathematical quadratic equation of two variables
It is known that the square of the equation X-2 (m-1) x + M = 0
When m takes any value, the original equation has no roots
2. Take a suitable non-zero integer for m, and the normal school says that there are two real roots, and find the square sum of the two real roots
x^2-2(m-1)x+m^2=0
△=[2(m-1)]^2-4m^2=2-4m2-4m1/2
Therefore, when m > 1 / 2, the original equation has no real roots
(2) M = 0 has two roots
x1=0,x2=2
Sum of squares 4
How simple it is! The number obtained by subtracting 4ac from the square of B is greater than zero, two roots are equal to zero, one root is less than zero, and there is no root
The square sum of (x1 + x2) is equal to the square of X1 plus the square of x2 plus 2 x1x2
PS: where x is an unknown number and the multiplier sign is omitted
(1) When △ < 0, i.e. [2 (m-1)] and # 178; - 4m and # 178; < 0
The solution is m > 1 / 2
When m / 2 has no real root
(2)
X1 & # 178; + x2 & # 178; = (x1 + x2) &#... Expand
PS: where x is an unknown number and the multiplier sign is omitted
(1) When △ < 0, i.e. [2 (m-1)] and # 178; - 4m and # 178; < 0
The solution is m > 1 / 2
When m ＞ 1 / 2, the original equation has no real roots
(2)
X1²+X2² = （X1 + X2）² - 2 X1 X2 = 【2(M-1)】² - 2M ²=
2m & # 178; - 8m + 4
Given that function f (x) is an odd function defined on R, f (1) = 0, {XF'x-fx} x 2 > 0 (x > 0), then the solution of inequality x 2F x > 0 is obtained
Because {x f '(x) - f (x)} / x2 > 0 (x > 0), and X2 > 0, when x > 0, X f' (x) - f (x) > 0, so f '(x) > F (x) / X. moreover, f (x) is continuously differentiable when x > 0
Because f (1) = 0, f '(1) > F (1) / 1 = 0. That is, f (x) increases monotonically at the value of 1
Because x = 0 is not the solution of the last inequality, and when x is not equal to 0, X2 > 0, the actual inequality is f (x) > 0
Because f (x) is an odd function, it is positive and negative symmetric
The next two cases are discussed
Situation 1
If there exists P, 00. For any value Q in this interval, there is
So we get the contradiction between the value of F (q) and f (0.q) / F (0.1) / F in the above interval
In this case, if there is an n-value in the section (m, positive infinity), such that f (n) 0. For any value Q in this section, f '(q) < 0 < f (q) / Q is contradictory to the above f' (x) > F (x) / X. therefore, in the section (m, positive infinity), the function value f (x) > 0. Then in the section (1, positive infinity), f (x) > 0
To sum up:
On (0,1), f (x) 0;
Because it is an odd function, f (0) = 0;
From the symmetry property of odd function, we can see that:
On (negative infinity, 0), only the segment (- 1,0) f (x) > 0
So the results are (- 1,0) and (1, positive infinity)
... solve the inequality f (t-1) + F (T) < 0
Solving inequality f (t-1) + F (T) < 0
Let x = T-1, so t = x + 1
because
f(t)+f(t-1)
The analytic expression of F (t-1)? ~
Nothing else?
Xiaoming and Xiaoliang run along the 400m circular track. They start from somewhere at the same time. If they walk in the same direction, Xiaoming will catch up with Xiaoliang in 200s. If they walk in the opposite direction, they will meet in 40s, and the running speed of the two will be the same
(solved by quadratic equation of two variables)
Note: just give me a system of equations and I'll solve it myself~
Let Xiaoming speed be x m / s and Xiaoliang speed be y m / s
200x - 200y =400
40（x+y）=400
Let Xiao Ming's walking speed be x and Xiao Liang's walking speed be y,
Then: (x + y) 40 = 400
(x-y)200=400
（x+y）40=400
（x-y）200=400
Let the speed of two people be a and B respectively
If you go in the same direction, Xiaoming will catch up with Xiaoliang in 200s
So Xiao Ming ran one more lap than Xiao Liang at this time
200a-200b = 400
If you walk with your back, then after 40 seconds, they meet
(a+b)X40 = 400
These two equations form a system of equations
If the odd function f (x) defined on R is an increasing function on (0, + ∞), and f (- 3) = 0, then the solution set of the inequality XF (x) < 0 is ()
A. （-3，0）∪（0，3）B. （-∞，-3）∪（3，+∞）C. （-3，0）∪（3，+∞）D. （-∞，-3）∪（0，3）
It is shown that: F (- 3) = - f (3) = 0, f (3) = 0, and f (x) is an increasing function on (0, + ∞). When 0 ＜ x ＜ 3, f (x) ＜ 0, when x ＞ 3, f (x) ＞ 0, and f (x) is an odd function defined on R, f (- 3) = 0, when x ＜ - 3, f (x) ＜ 0, when - 3 ＜ x ＜ 0, f (x) ＞ 0. The picture is as follows: the solution set of inequality XF (x) ＜ 0 is So we choose a
Given f (x) = x ^ 2 - (a + 1 / a) x + 1, if the solution set of the inequality f (x) ≤ 0 about X is {X / 1 / 2 ≤ x ≤ 2}, find the value of A
(2) If a is greater than 0, the inequality f (x) about X is greater than or equal to 0
(3) If x is greater than 0, the minimum value of F (x) / X is not less than 4, try to find the value range of real number a
（1）
This is a quadratic function whose image opening is upward, from F (x) 0, then Ag (a) = ax ^ 2-xa ^ 2-x + a = - XA ^ 2 + (1 + x ^ 2) A-X = (- XA + 1) (A-X)
Two are x = A and x = 1 / A, and the opening of the image is downward, so the solution set is {x | 1 / A2
a+1/a
（1）f(x)=x^2-(a+1/a)x+1≤0
Then (x-a) (x-1 / a) ≤ 0
When a = 1 / 2, i.e
（x-1/2）（x-2）≤0
1/2 ≤x≤2
2）f(x)=x^2-(a+1/a)x+1≤0
Then (x-a) (x-1 / a) ≤ 0
When 0
linear equation in two unknowns
Xiao Li goes from a to B by bike, and Xiao Ming goes from B to a by bike. They both go at an average speed. It is known that they set out at 8 am at the same time. By 10 am, they are 36 km apart. By 1 noon, they are 36 km apart. The distance between a and B is calculated
Let Xiao Li's speed be V1 and Xiao Ming's speed be V2, then the following equation is established: (10-8) × (V1 + V2) + 36 = (13-8) × (V1 + V2) - 36, and the solution is (V1 + V2) = 24. Therefore, the distance is (10-8) × (V1 + V2) + 36 = 84 (km). Or, after walking for three hours, they walked a total of 72 km
According to the title, from 10:00 to 1:00, it took three hours to walk 36 + 36 = 72 km,
They set out at 8:00 at the same time. By 10:00 in the morning, they were 36 kilometers apart. They walked 2 * 72 / 3 = 48 in two hours,
The total length is 36 + 48
Let the sum of the velocities of Xiao Ming and Xiao Li be x and the distance between the two places be s
（10-8）x=s-36
（13-8）x=s+36
The solution is s = 84, and S is the distance between the two places.
Let the sum of the velocities of Xiao Ming and Xiao Li be x and the distance between the two places be s
（10-8）x=s-36
（13-8）x=s+36
The solution is s = 84, and S is the distance between the two places
What about the title??
Let the distance between a and B be x km
Then the speed of two people's cycling remains the same, and the equation is established
At 10, velocity = (x-36) / 2
At 1 noon, speed = (x + 36) / 5
That is, (x-36) / 2 = (x + 36) / 5
The solution is x = 84 (km)
Then shut up

RELATED INFORMATIONS