Decomposition factor: 1-x2 + 2xy-y2=______ .
The original formula is 1 - (x2-2xy + Y2) = 1 - (X-Y) 2 = (1 + X-Y) (1-x + y), so the answer is (1 + X-Y) (1-x + y)
RELATED INFORMATIONS
- 1. Factorization factor 1-x2 + 2xy-y2
- 2. Given x + y = 1, find the value of x2 + 2XY + Y2
- 3. If x2-2xy-y2-x + Y-1 = 0, find the value of X-Y
- 4. Given that X-Y + 1 and x 2 + 8 x + 16 are opposite to each other, find the value of x 2 + 2XY + y 2
- 5. -Is x2 + 2xy-y2 a complete square If it can be used, does it mean it can be used? that. 4Y & # 178 / / X & # 178; - XY + Y & # 178; this should not work, right?
- 6. If x and y satisfy x2 + y2 = 1, then the minimum value of (Y-2) / (x-1) is; the maximum value of X / 3 + Y / 4 is I don't understand why (Y-2) / (x-1) can be regarded as the slope of the line between any point (x, y) and point (1,2) on a circle
- 7. If x2 + y2 = 1, find the maximum and minimum of y = 3x2 + 4y2-2x-5 Specific steps should be taken
- 8. Given 3x2 + 2Y2 = 9x, find the maximum value of x2 + Y2 3x2 means 3x square, and so on Urgent, Can we use the monotonicity and the maximum (minimum) value of the function of higher one to answer?
- 9. Given 3x2 + 2Y2 = 2x, find the maximum value of x2 + Y2
- 10. If | x + Y-2 | + [2xy-1] 2 [this 2 is square] = 0, find the value of x2 + 6xy + Y2
- 11. Can x2-2xy + y2 = 1 be factorized
- 12. -2xy-x2-y2 factorization
- 13. What is the cubic power of AX2 + 2a2x + A and - 2xy-x2-y2,
- 14. If y = √ 1-2x + √ 2x_ 1 + 1 / 2, find √ x2 + Y2 - √ x2 + Y2 + 2XY =?
- 15. Given x2 + y2 = 2, find the value range of x2-2xy-y2
- 16. What is the sum of [1 / 2XY + Y2 + 1] + [x2-1 / 2xy-2y2-1]
- 17. 2xy-x2-y2+9 Decomposition cause death form
- 18. (x2-2xy + Y2) - (x2 + 2XY + Y2) = what
- 19. If there is a point on the right branch of hyperbola x2a2 − y2b2 = 1 (a > 0, b > 0), and the distance from it to the right focus and the left quasilinear is equal, then the value range of eccentricity of hyperbola is () A. (1,2]B. [2,+∞)C. (1,2+1]D. [2+1,+∞)
- 20. If there is a point on the right branch of hyperbola x2a2 − y2b2 = 1 (a > 0, b > 0), and the distance from it to the right focus and the left quasilinear is equal, then the value range of eccentricity of hyperbola is () A. (1,2]B. [2,+∞)C. (1,2+1]D. [2+1,+∞)