How to solve the sum of X + 1 of quadratic inequality x with one variable less than or equal to 3?

X (x + 1) ≤ 3, the two solutions are x = (1 + √ 13) / 2, x = (1 - √ 13) / 2, so (1 - √ 13) / 2 ≤ x ≤ (1 + √ 13) / 2

It is known that the solution set of quadratic inequality f (x) > 0 is (1,3), and the solution set of quadratic inequality g (x) less than or equal to 0 is empty,
Then the solution set of inequality f (x) / g (x) less than or equal to 0 is?

Solution
f(x)/g(x)

Solve the inequality x ^ 2-2x - 3 > = 0 (quadratic inequality of one variable,)

-The square of 2 minus 4 × 1 × - 3 equals 16 X1 = 1 x2 = 3, X is less than 1 or X is greater than 3
Remember to adopt it

How to solve the quadratic inequality x ^ 2-2x-3 with one variable?

x^2-2x-3=(x-3)*(x+1)
In addition, it doesn't seem that it's not equal to trying. Why isn't it unequal

Solving quadratic inequality (x-1) (2x-3) ≥ 1
After I calculate it, it's (X-2) (2x-1) ≥ 0. It's not the final solution to be x ≥ 2 and X ≥ 1 / 2, X ≥ 2 or X ≤ 1 / 2. What's the matter with me~

Multiplication greater than or equal to 0
So both brackets are greater than or equal to 0 or less than or equal to 0
If all are greater than or equal to 0
Then x ≥ 2 and X ≥ 1 / 2
So x ≥ 2
If all are less than or equal to 0
Then x ≤ 2 and X ≤ 1 / 2
So x ≤ 1 / 2
So x ≥ 2 or X ≤ 1 / 2

And inequality | 2x-3 | + | x + 1|

|2x-3|+|x+1| - 1
∴ -1 < x ≤ 3/2
When x > 3 / 2
2x - 3 +x+1 < 5
x < 7/3
∴ 3/2 < x < 7/3
To sum up, - 1 < x < 7 / 3
The quadratic inequality of one variable can be (x + 1) (3x - 7) < 0

First degree inequality of one variable and first degree function
1. Supermarket a and supermarket B sell the same goods at the same price. In order to attract customers, they offer different preferential schemes: after supermarket a has purchased more than 300 yuan, the excess part will be given a 20% discount at the original price; after supermarket B has purchased more than 200 yuan, the excess part will be given a discount at the original price of 8.5 yuan
(1) Please use the algebraic formula containing x to represent the expenses paid by customers in two supermarkets;
(2) Which supermarket is more favorable for customers? Explain your reasons?
2. A residential area sells houses in the form of installment payment, and the government gives a certain discount. Xiaoming family buys a house with a current price of 120000 yuan, and the down payment (the first year) is 30000 yuan. From the second year, the sum of 5000 yuan and the interest of the last year's surplus is set as the annual interest rate of the surplus is 0.4%
(1) If x (x ≥ 2) year is y yuan, the function relation between Y yuan and X (year) is obtained;
(2) It is estimated that the amount of house payment payable in the third and the tenth year;
(3) From what year, Xiaoming's annual housing payment is less than 5100 yuan?

How much does a customer spend on shopping in supermarket a
300+（X-300）*0.8＝60+0.8X
The cost of shopping in supermarket B:
200+（X-200）*0.85＝30+0.85X
After analysis, when the customer's shopping is less than 600 yuan, it is more favorable in supermarket a; when the customer's shopping is more than 600 yuan, it is more favorable in supermarket B

First degree inequality of one variable and first degree function
1. If the line y = - 2x-1 intersects the line y = 3x + m in the third quadrant, please determine the value range of real number M
2 given the function y = ax (a < 0), if a (x1, Y1) and B (X2, Y2) are two points on the line y = ax, and X2 > x1, then the relationship between Y1 and Y2 is ()
3 if the intersection of the image of the linear function y = (m-1) x-m + 4 and the Y axis is above the X axis, then the value range of M is ()

1.
y＝-2x-1
y＝3x＋m
Simultaneous solution
x=-（m+1）/5,y=（2m-3）/5
I.e. intersection [- (M + 1) / 5, (2m-3) / 5]
Because the intersection is in the third quadrant
So - (M + 1) / 5 < 0, (2m-3) / 5 < 0
The solution is m ＞ - 1 and m ＜ 3 / 2
So the value range of real number m is - 1 < m < 3 / 2
two
Because a < 0
In the function y = ax, y decreases with the increase of X
And a (x1, Y1) and B (X2, Y2) are two points on the straight line y = ax, and X1 < x2
So Y1 > Y2
three
Because the intersection of the image of the linear function y = (m-1) x-m + 4 and the Y axis is above the X axis
So - M + 4 > 0, and M-1 ≠ 0
The solution is m < 4 and m ≠ 1
That is, the value range of M is m < 4, and m ≠ 1

Mathematics of grade two in junior high school
1. If the relation between two variables X and y can be expressed as_____ Y is said to be a linear function of X
2. The coordinates of the intersection of the image of the first-order function y = - 2x + 3 and the x-axis are When the function value is greater than 0, the value range of X is_ When the function value is less than 0, the value range of X is_ .
3. If a jacket with a price of M yuan is sold at a 20% discount, it is more profitable for the seller than the one with a price reduction of 32 yuan
4. If the average speed of the vehicle is 50km / h, the relationship between the distance s (km) and the driving time t (H) is The value range of the independent variable t is_ .
5. Given that Y1 = 3x + 4, y2 = 2x-8, when_ Y1 > Y2
6. Given the function y = ax + 2 (a < 0), if a (x1, Y1), B (X2, Y2) are two points on the straight line y = ax + 2, and X2 > x1, then the relationship between Y1 and Y2 is .
7. If the intersection of the image of the first-order function y = (m-1) x-m + 4 and the Y axis is above the X axis, then the value range of M is____ .

1.y=kx+b（k≠0）
2.(3/2,0) x3/2
3.m-12
6.y1>y2
7.m

Given that the image a of a function of a degree passes through the point m (- 1, - 4.5), n (1, - 1,5) (1), find the analytic expression of the function, and draw the image (2), find out the focus a of x-axis and y-axis
(3) if P (4, m) and m, n are collinear, find M
(4) If the lines a and B intersect at the above point P, the area of PAC of the triangle formed by A. B and X axis is 9

The image a of a given function of degree passes through points m (- 1, - 4.5), n (1, - 1,5)
(1) Find the analytic expression of this function and draw the image
Let the first order function be y = KX + B
Substituting the values of points m (- 1, - 4.5) and n (1, - 1,5) into
-4.5=-k+b,-1.5=k+b
The solution is k = 1.5, B = - 3
The analytic expression of the function is y = 1.5x-3
In the rectangular coordinate system, make points m (- 1, - 4.5), n (1, - 1,5), the line connecting two points is the image of the function
(2) Find out the coordinates of the focus ab of x-axis and y-axis
The coordinates of the intersection ab of y = 1.5x-3 and x-axis and y-axis are (0, - 3) and (2,0)
(3) If P (4, m) and m, n are collinear, then M
Substituting the value of P (4, m) into the analytic formula y = 1.5x-3, M = 3
(4) If the lines a and B intersect at the above point P, the area of PAC of the triangle formed by A. B and X axis is 9
If the coordinate of point C is (P, 0), then there is
The area of triangle PAC = | P-2 | * 3 / 2 = 9
We get P = - 4 or P = 8
The coordinates of point C are (- 4,0) or (8,0)

How to solve the sum of X + 1 of quadratic inequality x with one variable less than or equal to 3?