How to find ∫ 1 / √ x-x & # 178; DX?

How to find ∫ 1 / √ x-x & # 178; DX?

x-x²=1/4-1/4+x-x²=1/4-(x-1/2)^2
The original integral is ∫ 1 / √ x-x & # 178; = ∫ D (2x-1) / √ [1 - (2x-1) ^ 2]
=arcsin(2x-1)+c
Solving equation X-1% x = 19.8
x-40%x=15
x+25%x=2.5
50%x+x=31.5
x-15%x=8.5
x-1%x=19.80.99x=19.8x=19.8/0.99x=20x-40%x=150.6x=15x=15/0.6x=25x+25%x=2.51.25x=2.5x=2.5/1.25x=250%x+x=31.51.5x=31.5x=31.5/1.5x=21x-15%x=8.50.85x=8.5x=8.5/0.85x=10
15-(1-x+x²)+(1-x+x²-x³),
First remove the brackets, the original formula = 15-1 + x-x ＾ 2 + 1-x + X ＾ 2-x ＾ 3
Reduction = 15-x ＾ 3
Or is it too simple?
Solution equation: 19% x = 17% (x + 30)
0.19x=0.17x+0.17×30
0.19x-0.17x=5.1
0.02x=5.1
x=5.1÷0.02
x=255
0.19x=0.17x+0.17×30
0.19x-0.17x=5.1
0.02x=5.1
x=5.1÷0.02
x=255
19%x=17%(x+30)
19%x-17%x=17%×30
2%x=510%
2x=510
x=255
2%x=17%*30
x=17*15
Simplification: 189; sin: 178; X [(1 / 2 / Tan x) - 2 / Tan x] + 2 / 2 root triple cos2x
1/2*sin²x*[1/tan(x/2)-tan(x/2)]+√3/2*cos2x=1/2*sin²x*[cos(x/2)/sin(x/2)-sin(x/2)/cos(x/2)]+√3/2*cos2x=1/2*sin²x*[cos²(x/2)-sin²(x/2)]/[sin(x/2)cos(x/2)]+√3/2*cos2x=sin²...
3x + X + 5 = 17.8
4x=17.8-5=12.8
x=12.8/4=3.2
x=3.2
On the two equations X & # 178; - X-2 = 0 and (x + 1) 1 = (x + a) 2 have the same solution, then a =? Ask the great God to help me
Solving equation 2.5 * 9-x = 17.7
solve equations:
2.5*9-x=17.7
22.5-x=17.7
x=22.5-17.7
x=4.8
----------------------
Please click "evaluation" in the upper right corner, and then you can choose "satisfied, the problem has been solved perfectly"!
2.5*9-x=17.7
22.5-x = 17.7 (transfer)
x=22.5-17.7
x=4.8
tan²X-3tanX+2=0,0°≤X≤360°,X=?
Such as the title
tan²x-3tanx+2=0
(tanx-1)(tanx-2)=0
TaNx = 1 or TaNx = 2
0°≤x≤360°
X = 45 ° or x = 225 ° or x = arctan2 or x = arctan2 + 180 degree
30% X × 24-17 / 2 = 36.5 solution
30% X × 24-17 / 2 = 36.5
30% X × 24-17 / 2 = 36.5
Multiply both sides by 10
72x-85=365
72x=450
x=6.25
7.2x-8.5=36.5
7.2x=45
x=6.25
30% X × 24-17 / 2 = 36.5
0.72x=36.5+17/2
0.72x=45
x=62.5