1. Molecule = 2x5-5x4-x3+x2-6x = (x2-3x+1) * (2x3+x2)-6x =-6x.
Molecules are calculated by short division
Denominator = 3 times
So the original formula =-2.
2.a^3+b^3=(a+b)(a^2-ab+b^2)
I write (4x 2+4xy+y 2) = (2x+y) 2 alone.
[(2x-y)/(8x^3+y^3)]=(2x-y)/[(2x+y)(4x^2-2xy+y^2)]
[(4x^2-2xy+y^2)/(4x^2-y^2)]=(4x^2-2xy+y^2)/[(2x+y)(2x-y)]
After finishing, it is the original formula = 1.
There seems to be a lack of conditions, but I offer some ideas.
The numerator and denominator are divided by x 2 at the same time
Then consider (y/x) as a whole.
Bring the plus and minus 3 into 4, 1. Molecule = 2x5-5x4-x3+x2-6x = (x2-3x+1) * (2x3+x2)-6x =-6x.
Denominator = 3 times
So the original formula =-2.
2.a^3+b^3=(a+b)(a^2-ab+b^2)
(4x^2+4xy+y^2)=(2x+y)^2
[(2x-y)/(8x^3+y^3)]=(2x-y)/[(2x+y)(4x^2-2xy+y^2)]
[(4x^2-2xy+y^2)/(4x^2-y^2)]=(4x^2-2xy+y^2)/[(2x+y)(2x-y)]
Finishing, the original formula = 1, 1, score calculation problem
1. If x 2-3x+ 1 = 0, find the score (2x 5-5x 4-x 3+x 2-6x)/(x 2+ 1).
2.(4x^2+4xy+y^2)*[(2x-y)/(8x^3+y^3)]*[(4x^2-2xy+y^2)/(4x^2-y^2)]
* The title says that this problem needs a 3+ (or negative) Y 3 = ...
So what is a 3+ (or negative) y 3?
3. Given the absolute value of y/x is 3, find the fraction [(x 2+2xy+y 2)/(x 2y 2)] * [(x-y)/(x-2y)].
I hope that a friend who knows the answer can give me some advice, and I am very grateful.