First, multiple-choice questions (***5 questions, 7 points for each question, out of 35 points) give four options for each question below, code A, B, C and D, of which one and only one option is correct. Please fill in the code of the correct option in the brackets after the question, if it is left blank, it will be 0 if it is overfilled or misspelled) 1. If the non-zero real numbers A and B are known, then a+b is equal to () A,-1 B, 0 C, 1 D, 2. If the problem is solved, it is assumed that a≥3.

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2. As shown in the figure, if the side length of diamond-shaped ABCD is A, point O is a little on diagonal AC, OA=a, OB=OC=OD= 1, then A is equal to () A, 5+ 12 B, 5- 12 C,12c.

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3. Throw a square dice with six sides numbered 1, 2, 3, 4, 5, and 6 twice. Remember that the number of points thrown for the first time is A, and the number of points thrown for the second time is B. Then the probability that the equations of X and Y have only positive solutions is () A, 1 12 B, 29. When 2a-b=0, the equation has no solution. When 2a-b≠0, the solution of the equation consists of a and b 0.

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4. As shown in figure 1, in the right-angled trapezoid ABCD, AB‖CD, ∠ B = 90, the moving point P starts from point B and moves from B→C→D→A along the side of the trapezoid. The moving distance of point P is X, the area of △ABP is Y, and Y is regarded as a function of X. The function image is shown in Figure 2. Then the area of △ABC is () a, 10 B, 16 C, 18 D, 32. According to the image, BC=4, CD=5, DA=5, and then AB=8, so S △ ABC = 12× 8 ×.

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5. The number of integer solutions (x, y) of the equation x2+xy+y2=29 about x and y is () a, 2 groups of b, 3 groups of c, 4 groups of d and infinitely many groups of solutions. The original equation can be regarded as a quadratic equation about X, which is transformed into x2+yx+(2y2-29)=0. Because this equation has integer roots, according to. And the complete square number is △=y2-4(2y2-29). = -7y2+ 1 16≥0，Y2≤ 167≈ 16.57 y20 149 16△ 165438+。 Meet the requirements. When y=4, the original equation is x2+4x+3=0, at this time, x1=-kloc-0/,x2=-3. When y=-4, the original equation is x2-4x+3=0, then x3= 1 and x4=3. Therefore, the original equation

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6, a bicycle tire, if installed on the front wheel, the bicycle will be scrapped after driving 5000km; If it is installed on the rear wheel, the bicycle will be scrapped after driving for 3000km, and the front and rear tires can be replaced after driving for a certain distance. If the front and rear tires are interchanged and a pair of new tires of a bicycle are scrapped at the same time, then the bicycle can run; Assuming that the total wear amount of each tire is K, the wear amount of the tire mounted on the front wheel is k5000 per 1km, and the wear amount of the tire mounted on the rear wheel is k3000 per 1km. When a pair of new tires are set to exchange positions, they will go xkm ahead and ykm behind, respectively, taking the total wear amount of one tire as an equation, sometimes X+Y = 3000.

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7. It is known that the midpoint of line segment AB is C, with point C as the center and the length of AB as the radius. Take point D on the extension line of line AB, make BD = AC, then take point D as the center, the long radius of DA as a circle, intersect with ⊙A at point F and point G respectively, and connect FG with AB at point H, then the value of AHAB is; The solution is shown in the figure. The extended AD and ⊙ D intersect at point E, connecting AF and EF. According to the topic, AC= 13 AD, AB= 13 AE, where △FHA and △EFA, ∠ EFA = ∠ FHA = 90, ∠fah =∠EAF∴rt△FHA∞

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8. It is known that a 1, a2, a3, a4 and a5 are five different integers satisfying the condition a 1+a2+a3+a4+a5=9. If b is the equation about x (X-A 1) (X-A2) (X-A3) (X-A4). The solution ∵ (b-a 1) (b-a2) (b-a3) (b-a4) (b-a5) = 2009, and a1,a2, a3, a4 and a5 are five different integers.

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9. As shown in the figure, in △ABC, CD is the bisector of height and CE is ∠ACB. If AC= 14, BC=20 and CD= 12, the length of CE is equal to the solution as shown in the figure, and Pythagorean theorem shows that AD=9 and BD= 16. Let EF=x, from ∠ ECF = 12 ∠ ACB = 45, we get CF=x, so BF=20-x, and because EF ∠ AC, EF‖AC = BFBC, that is, x15.

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10 and 10 people form a circle to play games. The rules of the game are as follows: everyone thinks of a number in his mind, tells his thoughts to two people on both sides truthfully, and then everyone reports the average of the numbers told to him by two people on both sides. If the number is as shown in the figure, the number in the mind of the person who reported 3 is; If the number in the mind of the person signing for 3 is X, and the number in the mind of the person signing for 5 should be 8-x, then the number in the mind of the person signing for 7 is 12-(8-x)=4+x, and the number in the mind of the person signing for 9 is16-(4+x) =12-.

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13, as shown in the figure, given acute angles △ABC, BC < Ca, AD and BE are its two heights, the intersection point C is the tangent L of the circumscribed circle of △ABC, the overcurrent D and E are the vertical lines of L, and the vertical feet are F and G respectively. Try to compare the sizes of line segments DF and EG to prove your conclusion? The conclusion of the solution 1 is DF=EG, and the following proof is given. ∫∠fcd =∠eab, ∴Rt△FCD∽Rt△EAB, so we can get ... 5 points DF=BE? CDAB can also get EG=AD? ceab…… 10 and∵tan∠ACB = adcd = bece ,∴be？ CD=AD？ CE, so we can get that DF=EG ... The conclusion of decomposition method 2 is DF = eg, and give the following proof ... 5 points connect DE, ∫∠ADB =∠aeb = 90°, ∴A, B, D, E * * circle, so ∝.