2.c. set out; ⊙ When the central coordinate of O2 is (4,0), it is tangent.

3, D. MP<m is 0.

When p

4、d . 75 X = 1+sin 30/cos 30 = 1+√3/3

5. There is no map

6. B.M. Make a straight line perpendicular to the upper bottom and the lower bottom, and divide the upper bottom and the lower bottom equally with vertical lines.

Then from the isosceles right triangle at the bottom, the midline of the hypotenuse = half of the hypotenuse, it is concluded that the height = half of the upper and lower bottoms = midline.

7. This question is incomplete

8. If b. 1:3∠A = 90, this is the answer.

∠ B = 90-60 = 30, AC/BC =1/2 (the right-angled side of 30 = half of the hypotenuse).

△ABC∽△ACD

△ACD/S△ABC=( 1/2)？ = 1/4

△ACD/△ABD= 1/3

9、 D.xy/y-x y \u( y-x)/x = xy/(y-x)

10、C. 1/(2a- 1) x/(x？ -ax+ 1) = 1，(x？ -ax+ 1)/x = 1x+ 1/x = a+ 1x？ + 1/x？ =(a+ 1)？ -2=a？ +2a- 1

(x^4-a？ x？ + 1)/x？ =x？ + 1/x？ -a？ =a？ +2a- 1-a？ =2a- 1

∴x？ /(x^4-a？ x？ + 1)= 1/(2a- 1)

1 1 、( a+b)(a-b)+4(b- 1)

=a？ -B? +4b-4

=a？ -(b-2)？

=(a+b-2)(a-b+2)

12、2x？ +x+m+ 1/4=0

(m+ 1/4)/2>0 m>- 1/4

1? -4×2×(m+ 1/4)>0

1-( 8m+2 ) >0

m< - 1/8

So- 1/4 < m

13, and the sector area is πR? /4 is a quarter of the circular area, and the arc length of the sector is 2πR/4=πR/2.

Radius of cone bottom R/4 height =√(R? -R？ / 16)=(√ 15/4)R

14、ab、

15、