(1) 2 × (- 3) 3-4 × (- 3) + 15; (2) (12-59 + 712) × (- 36); (3) known 1 + 2 + 3 + +31 + 32 + 33 = 17 × 33, find 1-3 + 2-6 + 3-9 + 4-12 + +31-93 + 32-96 + 33-99

(1) 2 × (- 3) 3-4 × (- 3) + 15; (2) (12-59 + 712) × (- 36); (3) known 1 + 2 + 3 + +31 + 32 + 33 = 17 × 33, find 1-3 + 2-6 + 3-9 + 4-12 + +31-93 + 32-96 + 33-99

（1）2×（-3）3-4×（-3）+15=2×（-27）-4×（-3）+15=-54-（-12）+15=-54+12+15=-54+27=-（54-27）=-27；（2）（12-59+712）×（-36）=（-36）×12+（-36）×（-59）+（-36）×712=-18+20+（-21）=[-18+（-21）]+...

2.5 - (17 / 20 + 0.15) △ 0.6

=2.5-(85+15)/100÷0.6=5/2-1*10/6=15/6-10/6=5/6=0.83

Simple calculation: [2.5 - (17 / 20 + 0.15) △ 0.6] × 2 / 5
Please, hehe

I'll answer first and give you the process right away
Original formula = [2.5 - (17 / 20 + 3 / 20) × 5 / 3] × 2 / 5
=(5/2 -5/3) ×2/5
=5/2 × 2/5 - 5/3× 2/5
=1 - 2/3
=1/3

No matter why m is a real number, the line y = MX + m-2 always passes through the point (- 1, - 2), right?

Yes
Substituting (- 1, - 2) into
-2=-m+m-2
-2=-2

In known rectangle ABCD, ab = 3cm, ad = 9cm, fold the rectangle so that point B coincides with point D, and the crease is EF, then the area of △ Abe is ()
A 6cm² B 8cm² C 10cm² D 12cm²

B and D coincide, so be = ed, let ed = X. from Pythagorean theorem: (9 - x) ^ 2 + 3 ^ 2 = x ^ 2, we get x = 4
So the area is 6, choose a

Finding the extremum of the known function f (x) = 1 / (1-x) ^ 2 + AlN (x-1)
If the tangent slope of curve y = f (x) at point (2, f (2)) is - 1, find the value of real number a and its tangent equation

x-1>0,x>1
f(x)=1/(1-x)^2+aln(x-1)
f(x)=1/(1-x)^2+ln(x-1)^a
x=2,y=1

Plane rectangular coordinate system of mathematical problems
There is a kind of toy in the exhibition hall. In the plane rectangular coordinate system, the coordinates of three known points are A1 (1,1 A2 (2,0 A3 (1, negative 1). An electronic frog is located at the origin of the coordinate. For the first time, the electronic frog jumps from the origin to P1 with A1 as the symmetry center, for the second time, it jumps from P1 to P2 with A2 as the symmetry center, and for the third time, it jumps from P2 to P3 with A3 as the symmetry center, According to this rule, the electronic frog continues to jump with A1, A2, A3 as the symmetry center. When the electronic frog jumps for 2010 times, what is the coordinate p2010 of the landing point of the electronic frog?

The origin is a cycle of three times. Divide three by zero is the origin

If the result of (x to the second power + PX + 1) (x to the second power - 2x + Q) does not contain the terms of X to the third power and X to the third power, find the value of PQ?

(x to the second power + PX + 1) (x to the second power - 2x + Q)
=x^4+(p-2)x³+(q-2p+1)x²+(pq-2)x+q
If not, the coefficient is 0
So P-2 = 0, q-2p + 1 = 0
therefore
p=2
q=2p-1=3

Given that the edge length of cube abcd-a1b1c1d1 is 1, find the tangent of the angle between b1c1 and plane ab1c

This is what I do

The interval of zero point of F (x) = 2 to the x power + x-4

f(x)=2^x+x-4
f(2)=2^2+2-4=2>0
f(1)=2+1-4=-1