Đáp án:
`A_9=1/(x-2)`
`A_10=(x^2+3x)/(x^2+6x+8)`
`A_11=(a+1)/(a-4)`
Giải thích các bước giải:
`A_9=1/(x+2)+2/(x-2)-(2x)/(x^2-4)(x\ne±2)`
`=1/(x+2)+2/(x-2)-(2x)/[(x+2)(x-2)]`
`=(x-2)/[(x+2)(x-2)]+[2(x+2)]/[(x+2)(x-2)]-(2x)/[(x+2)(x-2)]`
`=[x-2+2(x+2)-2x]/[(x+2)(x-2)]`
`=(x-2+2x+4-2x)/[(x+2)(x-2)]`
`=[(x+2x-2x)+(-2+4)]/[(x+2)(x-2)]`
`=(x+2)/[(x+2)(x-2)]`
`=1/(x-2)`
`A_10=(x^2+5x+6)/(x^2+x-12):(x^2+4x+4)/(x^2-3x)(x\ne{-4;-2;0;3})`
`=(x^2+2x+3x+6)/(x^2+4x-3x-12):[(x+2)^2]/[x(x-3)]`
`=[x(x+2)+3(x+2)]/[x(x+4)-3(x+4)].[x(x-3)]/[(x+2)^2]`
`=[(x+2)(x+3)]/[(x+4)(x-3)].[x(x-3)]/[(x+2)^2]`
`=[x(x+3)]/[(x+2)(x+4)]`
`=(x^2+3x)/(x^2+4x+2x+8)`
`=(x^2+3x)/(x^2+6x+8)`
`A_11=(2/(a^2-5a+4)+3/(a^2-16)):5/(a^2+3a-4)(x\ne{1;4;-4})`
`=[2/(a^2-4a-a+4)+3/[(a+4)(a-4)]].(a^2+3a-4)/5`
`=[2/[a(a-4)-(a-4)]+3/[(a+4)(a-4)]].(a^2+4a-a-4)/5`
`=[2/[(a-4)(a-1)]+3/[(a+4)(a-4)]].[a(a+4)-(a+4)]/5`
`=[[2(a+4)]/[(a+4)(a-4)(a-1)]+[3(a-1)]/[(a+4)(a-4)(a-1)]].[(a+4)(a-1)]/5`
`=[2(a+4)+3(a-1)]/[(a+4)(a-4)(a-1)].[(a+4)(a-1)]/5`
`=(2a+8+3a-3)/[(a+4)(a-4)(a-1)].[(a+4)(a-1)]/5`
`=(5a+5)/[(a+4)(a-4)(a-1)].[(a+4)(a-1)]/5`
`=[5(a+1)]/[(a+4)(a-4)(a-1)].[(a+4)(a-1)]/5`
`=(a+1)/(a-4)`
