The seventh issue of tracking feedback challenges me:1.b2.d3.a4.d5.d6.a7.c8.a2, 1. 13a-5, 12x- 13y2. The answer is not unique. - 13 x2， 14 x3 4。 Four, four,-165. 1.5 (S-3)+66. (A+B) (A-B)。 (-2) Nxn+65438+。 (2)(4x- 1)； (3)(300+ 10n)； (4) X+Y2.2. No, the problem lies in ①. Correct solution: According to the original polynomial, if the highest term is -5xmy3, you can get m+3 = 6 and m = 3. So the original polynomial is-5x3x3+104x3-4xy2.3. (1. (2)56. Four. 1.( 1) 7; (2) 10; (3) 13; (4)3n+ 1.2。 The rope length of (1) method is 4a +4b +8c, that of (2) method is 4a +4b +4c, and that of (3) method is 6a +6b +4c, which shows that the rope length of (3) method is the longest. The rope of method (2) is the shortest and the lifting capacity is 1. This polynomial has at most five terms. For example, a polynomial that meets the requirements is a3+ a2b+ab2+ b3+ 1. If a and b satisfy a+b +(b- 1)2= 0, then a+b can be known. The solution is a =- 1, B =1.a3+a2b+ab2+B3+1= (-1) 3+(-1) 2×1+(-1)×/kloc. (2)399; (3)a = 5 or -7.
The eighth issue of tracking feedback challenges me:1.b2.b3.d4.c5.a6.b7.c8.d2,1.8a+10b2.5,43. - 1 4.m-N5.x3-6x26。 - 19 7.- 188.x2- 15x+9，-29x+ 15 3， 1。 (2)a2- 92 a+ 1.2。 Simplified result:-3x2+12; ; Evaluation result: 0.3. ( 1)0.5n+0.6； (2)4. 1 yuan 4, 1. Reason: The original formula = 2a3-3a2b-2ab2-a3+2ab2+B3+3a2b-a3-B3 = 0, which has nothing to do with the values of A and B, so their operation results are correct. (2)a+(a+d)+(a+2d)+…+(a+99d)= a+a+d+a+2d+…+a+99d =(a+a+…+a)+(0+d+2d+…+99d)= 100 a+99d×50 = 100 a+4 9 50d。 Self-improvement ability1.5 (2n+2)-10/kloc-0 =1010 = n.2. The area of the shaded part is equal to the area of the big rectangle minus three blank triangles. That is, s = 4x× 4y-12x4y× x-12x3x3y-12x3x3y = 5xy. The coefficient of this single item is 5 and the degree is 2.
Question 1 1: Follow-up feedback challenge ego 1, 1. C2 a3 . D4 . a5 . B6 . B7 . c8 . B2， 1.234，242.x = 2，X = 13.34。 -1 5.0.7x =1000 6.48 7.208.6x+6× 32x+120 = 720 III,1. When m = 43, we can get 2x+4x43 = 0 and x =-83 .2 by substituting into the original equation. Since 3 = 3 and -3 = 3, they are 1-2x = 3 or 1-2x =-3 respectively. So the solution of the equation 1-2x = 3 is x =- 1 or x = 2.3. ( 1)( 1+20% )x，2(x- 10)； (3) Substitute x = 25 into the left and right sides of the equation to get left =( 1+20% )× 25 = 30 and right = 2 ×(25- 10)= 30. Because left = right, x = 25 is the solution of the equation (1+20% )x = 2(x- 10). This means that the number of trees planted in Class B is indeed 25. As can be seen from the above inspection process, the number of trees planted in Class A should be 30. Not 35 trees. Fourth, 1. Guess: the solution of the equation x+mx = c+ mc about x is x 1= c, x2= mc. Verification: when x 1= c, x+mx = c+ mc, so x 1= c is the solution of the equation. Similarly, x2= mc is also the solution of the original equation. 2.(3x - 1)，x +(3x - 1)= 47。 Correct, because when Xiao Ming's age becomes twice that at this time, there will be x+x = 2x, and Xiao Ming's father's age will become. And 4x-1 = 2 ×2x-1, so the statement in the title is correct. The ability to improve is 1. (1) If the distance between two cars is 660km after x hours, then 72x+96x = 660-408; ; (2) If two cars meet x hours after the express train leaves, then 72+72x+96x = 408; ; (3) If the distance between the express train and the local train is 60km after x hours, then 408+72x-60 = 96x, or 408+72x = 96x-60.2. Subtract (3a+2b) from both sides of the equation to get 3b+2a- 1-(3a+2b) = 3a+2b- (. 0, so b > a.