Answer:
i.e answer A.
Step-by-step explanation:
This question involves knowing the following power/exponent rule:
![\sqrt[n]{x^m} = x^\frac{m}{n} \\\\so \sqrt[7]{x^2} = x^\frac{2}{7} \\\\and \\\\ \sqrt[5]{y^3} = y^\frac{3}{5} \\](https://tex.z-dn.net/?f=%5Csqrt%5Bn%5D%7Bx%5Em%7D%20%3D%20x%5E%5Cfrac%7Bm%7D%7Bn%7D%20%5C%5C%5C%5Cso%20%5Csqrt%5B7%5D%7Bx%5E2%7D%20%3D%20x%5E%5Cfrac%7B2%7D%7B7%7D%20%5C%5C%5C%5Cand%20%20%5C%5C%5C%5C%20%5Csqrt%5B5%5D%7By%5E3%7D%20%3D%20y%5E%5Cfrac%7B3%7D%7B5%7D%20%5C%5C)
Next, when a power is on the bottom of a fraction, if we want to move it to the top, this makes the power become negative.
so the y-term, when moved to the top of the fraction, becomes:

So the answer is: 
a(x+1)(x-1)+b(x-2)(x+1)+(x+1)^2=9x^2-x-10
(x+1)(a(x-1)+b(x-2)+(x+1))=(x+1)(9x-10)
a(x-1)+b(x-2)+x+1=9x-10
Now this equation is much simpler!
(a+b)x-a-2b+x+1=9x-10
(a+b)x-a-2b=8x-11
(a+b-8)x-a-2b-11=0
a+b-8=(a+2b-11)/x
I can't solve it 3 variables and 1 equations means infinite answers so yea.
The distance between any two points is:
d^2=(x2-x1)^2+(y2-y1)^2
d^2=(6--2)^2+(4--4)^2
d^2=8^2+8^2
d^2=64+64
d^2=128
d=√128 units