1、(1)计算{(3-x)⼀(x-2)}⼀(x-2-(13-4x)⼀(x-2)) (2) {1+4⼀(a-2)(a-4+4⼀a)-3}⼀(1⼀a눀-1)

2026年09月23日 22:47
有2个网友回答
网友(1):

 (1){(3-x)/(x-2)}/(x-2-(13-4x)/(x-2))
  ={(3-x)/(x-2)}/{(x-2)*(x-2)-(13-4x)/(x-2)
  ={(3-x)/(x-2)}/{((x²-4x+4)-(13-4x))/(x-2}
  ={(3-x)/(x-2)}/{(x²-4x+4-13+4x)/(x-2}
  ={(3-x)/(x-2)}/{(x²-9)/(x-2)}
  ={(3-x)/(x-2)}/{(x-3)*(x+3)/(x-2)}
  =(3-x)/(x-2)*(x-2)/(x-3)/(x+3)
  =-(x-3)/(x-3)/(x+3)
  =-1/(x+3)
 (2) {1+4/(a-2)(a-4+4/a)-3}/(1/a²-1)
  = {1+4/(a-2)((a-2)-2/a(a-2))-3}/{(1/a-1)(1/a+1)}
  =(1+4-2/a-3)/{(1/a-1)(1/a+1)}
  =(2-2/a)/{(1/a-1)(1/a+1)}
  =2(1-1/a)/{-(1-1/a)(1/a+1)}
  =-2/(1/a+1)

网友(2):

(1){(3-x)/(x-2)}/(x-2-(13-4x)/(x-2))
={(3-x)/(x-2)}/(((x-2)^2-(13-4x))/(x-2))
=(3-x)/((x-2)^2-13+4x)
=(3-x)/(x^2-4x+4-13+4x)
=(3-x)/(x^2-9)
=(3-x)/((x+3)(x-3))
=-1/(x+3)