Ask the master for your advice, thank you, ask the god for help, thank you

1) Lianmu Shen connects AC1 and A1C to point E, connecting ED.

Because d and e are the midpoints of ab and ac1, respectively, in abc1, de bc1 (median theorem).

Because BC1 is not in planar A1CD and DE is in planar A1CD, BC1 is planar A1CD

2) A1B1C1 area = quarter of the root number three sterling silver.

v = root number three x quarter of the root to do the nai yan number three x one-third = 1 4

y=cos2x+sinx=1-2sin^2x+sinx= -2(sinx-1/4)^2+9/8

1<=sinx<=1, y=cos2x+sinx is in the range of [-2,0].

y=cosx(cosx+sinx)=cos^2x+sinx*cosx

1/2(2cos^2x-1)+1/2*2sinx*cosx+1/2

1/2cos2x+1/2sin2x+1/2

2/2sin(2x+π/4)+1/2

The value range of y=cosx(cosx+sinx) is [(1-2) 2,(1+ 2)2].

y=cos(2x-π/3)+2sin(x-π/4)sin(x+π/4)

cos(2x-π/3)-1/2*2[cos(x-π/4+x+π/4)-cos(x-π/4-x-π/4)]

cos(2x-π/3)-cos2x+cosπ/2

1/2cos2x+√3/2sin2x-cos2x

3/2sin2x-1/2cos2x

sin(2x-π/6)

The value range of y=cos(2x- 3)+2sin(x- 4)sin(x+4) is [-1,1].

Do you still need to be examined.

1) If the fruit is scattered, a b+b a-ab=-2, then the chong bu a+b=ab

2) Proof of: a 2 + b 2 - (ab) 2 = -2aba 2 + b 2 + 2 ab = (ab) 2

a>0,b>0

So a+b=ab

y=7-4sinxcosx+4cos^2x-4cos^4xcos^2x=(cos2x+1)/2

cos^4x=(cos2x+1)^2/4

sinxcosx=(sin2x)/2

So y=7-2sin2x+2cos2x+2-(cos2x+1) 27-2sin2x+2cos2x+2-(cos2x) 2-2cos2x-1

8-2sin2x-(cos2x)^2

8-2sin2x-[1-(sin2x)^2](sin2x)^2-2sin2x+7

sin2x-1)^2+6

1<=sin2x "Bino Waiter's imitation noise=1."

So sin 2x = 1 and y min = 6

sin2x=-1, y max=10

55.Let the full name be x km x = 140km The first small hail is rented when the god is missing;

56.Suppose the book has x pages, 45+ x=216 pages.

The rule is that both horizontal and vertical calculations are equal to 4