Question

数学分式计算

Answer 1

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

2、=1/m(m-3)+1/(m-3)(m-6)+1/(m-6)(m-9)
=1/3[3/m(m-3)+3/(m-3)(m-6)+3/(m-6)(m-9)]
=1/3[1/(m-3)-1/m+1/(m-6)-1/(m-3)+1/(m-9)-1/(m-6)]
=1/3[1/(m-9)-1/m]
=1/3×9/m(m-9)
=3/(m²-9m）

Answer 2

（1）1/(x-1)+1/{(x-1)(x-2)]+1/{(x-2)(x-3)]
={(x-2)(x-3)]/{(x-1)(x-2)(x-3)]+(x-3)/{(x-1)(x-2)(x-3)]+(x-1)/{(x-1)(x-2)(x-3)]
={(x-2)(x-3)+(x-3)+(x-1)]/{(x-1)(x-2)(x-3)]
=(x²-5x+6+x-3+x-1)/{(x-1)(x-2)(x-3)]
=(x²-3x+2)/{(x-1)(x-2)(x-3)]
=(x-1)(x-2)/{(x-1)(x-2)(x-3)]
=1/(x-3)