The solution equation is: X × 1 / 3 × 4 / 9 = 1 / 6 8 × x = 6 / 25 △ 1 / 5

The solution equation is: X × 1 / 3 × 4 / 9 = 1 / 6 8 × x = 6 / 25 △ 1 / 5

Ⅹ×1/3×4/9＝1/6
4/27x=1/6
x=1/6÷4/27
x=27/24
x=9/8
8×Ⅹ＝6/25÷1/5
8x=6/5
x=6/40
x=3/20

Solving equation 80 + 80 × 25% = x

80(1+25%)=X

(1-25%) x = 12 (solving equation,) 6x-4.8 = 5.28 x-80% x = 42

（1—25%）X=12
0.75X=12
X=12÷0.75
X=16
6X—4.8=5.28
6X=5.28+4.8
6X=10.8
X=10.8÷6
X=1.8
X—80%X=42
0.2X=42
X=42÷0.2
X=210

Let the sequence {an} satisfy the following relation A1 = 2A (a is a constant that is not zero) an = 2A - A ^ 2 / an-1 sequence BN = 1 / (an-a)
It is proved that {BN} is an arithmetic sequence

(1)
b1=1/(a1-a)=1/a
Bn-bn-1 = 1 / (an-a) - 1 / (an-1-a)
=1/a (n》2)
So BN is an arithmetic sequence
（2）
bn = n/a
bn=1/an-a
So an = (n + 1) n / A

The floor of a sitting room is paved with square bricks. If the side length is 6 decimeters, 80 pieces are needed. If the side length is 8 decimeters, how many pieces are needed?

6x6x80 (8x8) = 45 pieces

Simple calculation: 13 / 20 × 3 / 8 + 13 / 20 △ 1 and 3 / 5

13 of 20 × 3 of 8 + 13 of 20 △ 1 and 3 of 5
=(13/20)×(3/8+5/8)
=(13/20)×1
=13/20

LIM (1-x) x to the power of 1 / 1, and below Lim is x → 0

x->0 lim(1-x)^(1/x)
=lim[1+(-x)]^[-(-1/x)]
=e^(-1)
=1/e

As shown in the figure, ⊙ o is the inscribed circle of equilateral triangle ABC with side length 2, then the area of the shadow part in the figure is______ ．

Since the inner center of an equilateral triangle is its outer center, we can get ad = 12ab = 1, ∠ OAB = 12 ∠ cab = 30 °; in RT △ oad, Tan 30 ° = odad, i.e. 33 = od1, we can get 0d = 33. The area of the shaded part in the graph is equal to s △ abc-s ⊙ o = 34 × 22 - π (33) 2 = 3 − 13 π

Find the range of the following functions (expressed in intervals) y = x & sup2; - 3x + 4

[7/4,+∞)

1. Y = sin, X is the inverse function of [π / 2,3 π / 2]
2, find the function y = arcsinx, X belongs to [0,1] de inverse function

1)y=sinx,x∈[π/2,3π/2]
-1≤y≤1
x=arcsiny
y=arcsinx -1≤x≤1
2)y=arcsinx,x∈[0,1],y∈[k∏,∏/2+2k∏]
x=siny
y=sinx,x∈[k∏,∏/2+2k∏]