The inverse scale function y = K / X has a point P (m, n) on the image whose coordinates are the two roots of the quadratic equation T2 (square) - 3T + k = 0 with respect to t Why m + n = 3

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• 1. The image of the first-order function y = x + B intersects with the image of the inverse scale function y = K + 3 / X at point a (m, n), and m, n (m, n) is a quadratic equation of one variable with respect to X Two unequal real roots of KX ^ 2 + (2k-7) x + K + 3 = 0, where k is a non negative integer and m and N are constants (1) Finding the value of K (2) Find the coordinates of point a and the analytic expression of first-order function
• 2. There is a point P (m, n) on the image of the inverse scale function y = K / * (k not = 0), whose coordinates are the two roots of the quadratic equation t square-3t + k = 0 with respect to t The distance from the point P to the origin is the root 13, so the analytic expression of the inverse scale function is the process of writing out the solution
• 3. Which is a quadratic equation with one variable I remember the teacher said that unknowns can't be denominators, which is a quadratic equation? The square of x = - 4 The square of x-x / 1 = 4 Which one?
• 4. Given that x = 1 / 2 is the solution of the equation 2x-m / 4-1 / 2 = x-m / 3, find the value of 1 / 4 (- 4m & # 178; + 2m-8) - (1 / 2m-1)
• 5. Find the range of the real number m so that the equation x & # 178; + 2 (m-1) x + 2m + 6 = 0 (1) There are two real roots, one larger than 2 and the other smaller than 2 (2) There are two real roots, and both are larger than 1 (3) There are two real roots α and β, and 0 ＜ α ＜ 1 ＜ β is less than 4 (4) At least one positive root
• 6. The value range of real number m is the equation x ＾ + (m-1) x + 2m + 6 = 0 Find the range of the real number m so that the equation x ^ 2 + 2 (m-1) x + 2m + 6 = 0, (1) There are two real roots, one is greater than 2 and the other is less than 2; (2) Both solid roots are larger than 1; (3) Two real roots x1, X2 satisfy 0
• 7. Why? A. the real number - A ^ 2 is negative. B. the following sign (a ^ 2) = absolute value a The following statement is correct. Why? A. The real number - A ^ 2 is negative B. Follow sign (a ^ 2) = absolute value a C. The absolute value of - a must be positive D. The absolute value of real number - A is a
• 8. If a is known to be a real number, the absolute value of A. a B. - a C. A and the absolute value of D. negative a must be nonnegative
• 9. When the moving point a moves on the square of the circle x + the square of the circle y = 1, the trajectory equation of the midpoint of the line between the moving point a and the fixed point B (- 3,0) is
• 10. When the moving point a moves on the circle x2 + y2 = 1, the trajectory equation of the midpoint of its line with the fixed point B (3,0) is () A. （x+3）2+y2=4B. （x-3）2+y2=1C. （2x-3）2+4y2=1D. （x+3）2+y2=12
• 11. There is a point P (m, n) on the image of the inverse scale function y = K / x, whose coordinates are two roots of the quadratic equation T2 (square) - 3T + k = 0 with respect to t And the distance from P to the origin is the root sign 5 How to get m + n = 3 How to get m + n = 3 How to get m + n = 3 How to get m + n = 3 How to get m + n = 3
• 12. It is known that the equation MX2 - (2m-1) x + M = 0 has two unequal real roots. (1) find the value range of M; (2) find the two roots of the equation when m is the largest integer
• 13. If the equation (m-2) x ^ m ^ 2-2 + MX = 7 is a quadratic equation of one variable about X, find the value of M
• 14. Find the condition that the equation (m ^ 2-m-2) x ^ 2 + MX + M = 0 about X is a quadratic equation with one variable
• 15. If the equation (m ^ 2-4) x ^ 3 + (m-2) x ^ MX + m + 1 = 0 is a quadratic equation with one variable, what is the value of M? (m^2-4)x^3+(m-2)x^2-mx+m+1=0
• 16. If you add 5 meters to the length of a square, its area will be 15 square meters. If you add 3 meters to the width of a rectangle, its area will be 24 square meters What is the original area of this rectangle?
• 17. When a rectangular grassland is reduced by 5 m in length and increased by 3 m in width, a square grassland is obtained. The area of this square is equal to that of the original rectangle, Find the area of the original rectangle Let the length be x and the width be y The equations 3 (X-5) = 5Y (1) are given x-5=y+3② I want to ask: How did you get it?
• 18. Reduce one side of the square by 30% and increase the other side by 2m to get a rectangle, which is equal to the original square area. How many meters is the square area
• 19. Reduce one side of the square by 30% and increase the other side by 3 meters to get a rectangle, which is equal to the area of the original square______ Square meters
• 20. Reduce one side of a square by 20% and increase the other side by 2m to get a rectangle. This rectangle has the same area as the original square So what's the area of the original square? It's best to solve it by column arithmetic