How to solve these problems 1/a+3-6/9-a^2 (a/a-b-b/b-a) ·ab/a+b (1/x+1-x+3/x^2-1)/1/x-1 [2/3a-2/a+b(a+b/3a-a-b)]/a-b/a

How to solve these problems 1/a+3-6/9-a^2 (a/a-b-b/b-a) ·ab/a+b (1/x+1-x+3/x^2-1)/1/x-1 [2/3a-2/a+b(a+b/3a-a-b)]/a-b/a


In the first question, I didn't understand the original formula 2 = (A / A-B + B / a-b) AB / A + B = (a + B / a-b) AB / A + B = AB / a-b3 = (x-1 / X & # 178; - 1 - (x + 3) / X & # 178; - 1 / X-1 = - 4 / x + 14, the original formula = 2 / 3A & # 178; - 2 / A & # 178; + B + B & # 178; / 3A & # 178; - B-B & # 178; / A-B / A



Ask some junior high school calculation questions
1 [the fourth power of X, the second power of y-2x (the second power of x-3xy)] divided by (- the third power of 2x)
2 [(a + b) to the second power + (a-b) to the second power] (2a to the second power - 2b to the second power)
3 (1 / 2) - 1 power + i-3i + (the root of 2-3 is 1.732) - 0 power + (- 1)
4(2x-y-1/2)(2x-y+1/2)


1 [the 4th power of X, the 2nd power of y-2x (the 2nd power of x-3xy)] divided by (- 2x 3rd power) = (x ^ 4y-2x ^ 4 + 6x ^ 3Y ^ 2) divided by (- 2x 3rd power) = - 1 / 2XY + x-3y ^ 2 [(a + B) 2nd power + (a-b) 2nd power] (2a 2nd power - 2b 2nd power) = (A & # 178; + 2Ab + B & # 178; + A & # 178; - 2Ab