Đáp án+Giải thích các bước giải:
Bài 1:
`a,ĐKXĐ:x neq 2;5`
`A=1/(x-2)+(x^2-x-2)/(x^2-7x+10)-(2x-4)/(x-5)`
`=(x-5+x^2-x-2-(x-2)(2x-4))/(x^2-7x+10)`
`=(x^2-7-2(x^2-4x+4))/(x^2-7x+10)`
`=(x^2-7-2x^2+8x-8)/(x^2-7x+10)`
`=(-x^2+8x-15)/(x^2-7x+10)`
`=(-(x^2-8x+15))/(x^2-7x+10)`
`=(-(x-3)(x-5))/(x-2)(x-5)`
`=(3-x)/(x-2)`
`b,A in Z`
`=>3-x vdots x-2`
`=>-(x-2)+1 vdots x-2`
`=>1 vdots x-2`
`=>x-2 in Ư(1)={1,-1}`
`+)x-2=1=>x=3`
`+)x-2=-1=>x=1`
Vậy `x in {1,3}` thì `A in Z`
Bài 2:
`a,A=((1-x^3)/(1-x)-x):(1-x^2)/(1-x-x^2+x^3)`
`ĐK:x ne 1`
`A=(x^2+x+1-x):(1-x^2)/(1-x-x^2(1-x))`
`=(x^2+1):(1-x^2)/((1-x)(1-x^2))`
`=(x^2+1):1/(1-x)`
`=-(x-1)(x^2+1)`
`b,x=-1 2/3`
`=>x=-5/3`
`=>A=-(-5/3-1)(25/9+1)`
`=-(-8/3). 34/9`
`=272/27`
`c,A<0`
`<=>-(x-1)>0` do `x^2+1>0`
`<=>x-1<0`
`<=>x<1`