It is known that curve C1: y = X3 (x ≥ 0) and curve C2: y = - 2x3 + 3x (x ≥ 0) intersect at O and a, and the straight line x = t (0

It is known that curve C1: y = X3 (x ≥ 0) and curve C2: y = - 2x3 + 3x (x ≥ 0) intersect at O and a, and the straight line x = t (0

1. Curve C1: y = X3 (x ≥ 0) and curve C2: y = - 2x3 + 3x (x ≥ 0) intersect at O and a
The simultaneous equations are y = X3
y=-2x3+3x
The solution is x = 0, x = 1
Then O and a coordinates are (0,0) (1,1)
Straight line x = t (0

The domain of the function y = arcsin (1-3x / x-1) is to help do the next procedure 3Q

Because the definition field of y = arcsinu is - 1 ≤ u ≤ 1 (because the value field of sin is greater than or equal to - 1 and less than or equal to 1)
So - 1 ≤ 1 - 3x / X - 1 ≤ 1, just solve this inequality

1. X-2 / 1 + (2Y + y to the second power), then x + (y to the second power) = ()
2. The original number of approximate number 1.8 is not less than () and less than ()
3. If a = 5, B = 2 and a + B = B + A, then B-A = ()

The first question should be this
If X-2 / 1 + (2Y + y) to the second power = 0, then x + (y to the second power) = (1 / 2)
2. The original number of approximate number 1.8 is not less than (1.75) and less than (1.85)
3. If a = 5, B = 2 and a + B = B + A, then B-A = (- 3)

Please help me solve the following math problem!
It is known that a, B and C are three sides of the triangle ABC, and satisfy the relation that the square of a + the square of C = the square of 2Ab + 2ac-2b. Try to explain that the triangle ABC is an equilateral triangle

a^2+c^2=2ab+2ac-2b^2
a^2+b^2 -2ab + b^2+c^2 -2ac=0
(a-b)^2 + (b-c)^2 =0
a=b
b=c
The triangle ABC is an equilateral triangle