1.A 2。 D 3。 D 4。 D 5。 C 6。 B 7。 An eight. B 9。 B 10。 D
Second, fill in the blanks
1 1.3 12. 13.- 1 14.=
Third, ellipsis
Fourth,
17. The other root of the equation is that its value is 4.
18. Because a+b=2++2-=4, a-b=2+-(2-)=2,
ab=(2+)(2-)= 1
Five,
19. Solution: Let the total output of crop straw in our province be A and the growth rate of rational utilization be X, which is deduced from the meaning of the question:
30%a( 1+x)2=60%a, that is, (1+x)2=2.
∴x 1≈0.4 1, x2≈-2.4 1 (it doesn't matter).
∴x≈0.4 1。
In other words, the annual growth rate of rational utilization of straw in our province is about 4 1%.
20. Solution: (1)∵ The equation has a real root ∴ δ = 22-4 (k+ 1) ≥ 0.
The solution shows that k≤0, and the range of K is k≤0 (5 points).
(2) According to the relationship between the root and the coefficient of a quadratic equation, X 1+X2 =-2, X 1x2 = K+ 1.
x 1+x2-x 1x 2 =-2+k+ 1
Known -2+k+ 1 <-1 k= "">-2.
And (1)k≤0 ∴ -2.
∫k is an integer ∴, and the values of k are-1 and 0. (5 points)
Six,
2 1.( 1) The solution can be drawn from the meaning of the question.
∴(3 integral)
And point A is on the function, so the solution is like this.
solve an equation
So the coordinate of point B is (1, 2) (8 o'clock).
(2) When 02, y 1
When1y2;
When x= 1 or x=2, y 1=y2. (12)
Seven,
22. Solution: (1) Let the width be x meters, then: x(33-2x+2)= 150,
Solution: x 1= 10, x2= 7.5.
When x= 10, 33-2x+2 = 15.
When x=7.5, 33-2x+2 = 20 >; 18, irrelevant, give up.
∴ Chicken farm length15m, width10m. (5 points) (2) If the width is x meters, then: x(33-2x+2)=200,
That is x2-35x+200=0.
δ=(-35)2-4×2×200 = 1225- 1600 & lt； 0
The equation has no real number solution, so the chicken farm area can't reach 200 square meters. (9 points)
(3) When 0
When 15 ≤ A
When a≥20, two rectangular chicken farms with different lengths and widths can be enclosed; (12)
Eight,
23.( 1) Drawing (2 points)
(2) proof: from the meaning of the question: △ Abd △ Abe, △ ACD △ ACF.
∴∠DAB=∠EAB, ∠DAC=∠FAC, and ∠ BAC = 45,
∴∠EAF=90。
Are you ∵AD⊥BC?
∴∠E=∠ADB=90 ∠F=∠ADC=90。
AE = AD，AF=AD。
∴AE=AF.
∴ Quadrilateral AEGF is a square. (7 points)
(3) solution: let AD=x, then AE = eg = gf = X.
BD = 2，DC=3
∴BE=2，CF=3
∴BG=x-2,CG=x-3.
At Rt△BGC, BG2+CG2=BC2.
∴( x-2)2+(x-3)2=52。
Simplified, x2-5x-6=0
The solution is x 1=6, x2=- 1 (excluding), so AD=x=6. (12)