The following formulas are divided into: (1) 1 / A, 3 / 4A & sup2; B, 1 / 6ab & sup2; C; (2) 1 / 2x-2,1 / (x-1) & sup2;

The following formulas are divided into: (1) 1 / A, 3 / 4A & sup2; B, 1 / 6ab & sup2; C; (2) 1 / 2x-2,1 / (x-1) & sup2;

(1) (12b \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\/ 2 (x-1) (x-1) &

AB / (a + b) ^ 2 and B / A ^ 2-B ^ 2 x / xy-y ^ 2 and Y / x ^ 2-xy

ab/(a+b)^2=ab (a-b)/(a+b)²(a-b)
b/a^2-b^2=b(a+b)/(a+b)²(a-b)
x/xy-y^2=x²/xy(x-y)
y/x^2-xy =y²/xy(x-y)

Why is the original formula = (1 / x) (Y / y) + (1 / y) (x / x) = Y / XY + X / xy = (x + y) / XY where = (1 / x) (Y / y) + (1 / y) (x / x)?

Multiply up and down. Give me an example. 1 / 6 = 1 / 6 times 1 = 1 / 6 times 6, because 6 / 6 = 1

General score: (1) x / x ^ 2 + 2x + 1, (2) X-1 / x ^ 2 + X, (3) 1 / x ^ 2-1

x/(x^2+2x+1)
=x/(x+1)^2
=x^2(x-1)/[x(x+1)^2(x-1)]
=(x^3-x^2)/[x(x+1)^2(x-1)]
(x-1)/(x^2+x)
=(x-1)/[x(x+1)]
=(x-1)^2(x+1)//[x(x+1)^2(x-1)]
=(x^3-x^2-x+1)/[x(x+1)^2(x-1)]
1/(x^2-1)
=1/[(x+1)(x-1)]
=x(x+1)/[x(x+1)^2(x-1)]
=(x^2+x)/[x(x+1)^2(x-1)]