Đáp án:
`13)A_13=(x-29)/[(x-1)^2]`
`14)A_14=(x^3+2x^2-3x+2)/[2(x+1)(x-1)(x-2)]`
`15)A_15=(x-1)/2`
`16)A_16=(x-4)/(x-2)`
`17)A_17=1/(x+5)`
Giải thích các bước giải:
`13)A_13=(3x-31)/[(x-1)^2]+(2+2x)/(1-x^2)`
`=(3x-31)/[(x-1)^2]-(2x+2)/(x^2-1)`
`=(3x-31)/[(x-1)^2]-(2x+2)/[(x+1)(x-1)]`
`=[(3x-31)(x+1)]/[(x+1)(x-1)^2]-[(2x+2)(x-1)]/[(x+1)(x-1)^2]`
`=[(3x-31)(x+1)-(2x+2)(x-1)]/[(x+1)(x-1)^2]`
`=[3x^2+3x-31x-31-(2x^2-2x+2x-2)]/[(x+1)(x-1)^2]`
`=(3x^2+3x-31x-31-2x^2+2x-2x+2)/[(x+1)(x-1)^2]`
`=[(3x^2-2x^2)+(3x-31x+2x-2x)+(-31+2)]/[(x+1)(x-1)^2]`
`=(x^2-28x-29)/[(x+1)(x-1)^2]`
`=(x^2-29x+x-29)/[(x+1)(x-1)^2]`
`=[x(x-29)+(x-29)]/[(x+1)(x-1)^2]`
`=[(x-29)(x+1)]/[(x+1)(x-1)^2]`
`=(x-29)/[(x-1)^2]`
`14)A_14=x/(x-2)+(x^2+1)/(2-2x^2)`
`=x/(x-2)-(x^2+1)/(2x^2-2)`
`=x/(x-2)-(x^2+1)/[2(x^2-1)]`
`=x/(x-2)-(x^2+1)/[2(x+1)(x-1)]`
`=[2x(x+1)(x-1)]/[2(x+1)(x-1)(x-2)]-[(x^2+1)(x-2)]/[2(x+1)(x-1)(x-2)]`
`=[2x(x+1)(x-1)-(x^2+1)(x-2)]/[2(x+1)(x-1)(x-2)]`
`=[2x(x^2-1)-(x^3-2x^2+x-2)]/[2(x+1)(x-1)(x-2)]`
`=(2x^3-2x-x^3+2x^2-x+2)/[2(x+1)(x-1)(x-2)]`
`=[(2x^3-x^3)+2x^2-(2x+x)+2]/[2(x+1)(x-1)(x-2)]`
`=(x^3+2x^2-3x+2)/[2(x+1)(x-1)(x-2)]`
`15)A_15=(x^2+2x)/(2x+10)+(x-5)/x+(50-5x)/[2x(x+5)]`
`=(x^2+2x)/[2(x+5)]+(x-5)/x+(50-5x)/[2x(x+5)]`
`=[x(x^2+2x)]/[2x(x+5)]+[2(x+5)(x-5)]/[2x(x+5)]+(50-5x)/[2x(x+5)]`
`=[x(x^2+2x)+2(x+5)(x-5)+50-5x]/[2x(x+5)]`
`=[x^3+2x^2+2(x^2-25)+50-5x]/[2x(x+5)]`
`=(x^3+2x^2+2x^2-50+50-5x)/[2x(x+5)]`
`=[x^3+(2x^2+2x^2)-5x+(-50+50)]/[2x(x+5)]`
`=(x^3+4x^2-5x)/[2x(x+5)]`
`=(x^3+5x^2-x^2-5x)/[2x(x+5)]`
`=[x^2(x+5)-x(x+5)]/[2x(x+5)]`
`=[(x+5)(x^2-x)]/[2x(x+5)]`
`=[x(x+5)(x-1)]/[2x(x+5)]`
`=(x-1)/2`
`16)A_16=(x+2)/(x+3)-5/(x^2+x-6)+1/(2-x)`
`=(x+2)/(x+3)-5/(x^2+3x-2x-6)-1/(x-2)`
`=(x+2)/(x+3)-5/[x(x+3)-2(x+3)]-1/(x-2)`
`=(x+2)/(x+3)-5/[(x-2)(x+3)]-1/(x-2)`
`=[(x+2)(x-2)]/[(x-2)(x+3)]-5/[(x-2)(x+3)]-(x+3)/[(x-2)(x+3)]`
`=[(x+2)(x-2)-5-(x+3)]/[(x-2)(x+3)]`
`=(x^2-4-5-x-3)/[(x-2)(x+3)]`
`=(x^2-x-12)/[(x-2)(x+3)]`
`=(x^2-4x+3x-12)/[(x-2)(x+3)]`
`=[x(x-4)+3(x-4)]/[(x-2)(x+3)]`
`=[(x-4)(x+3)]/[(x-2)(x+3)]`
`=(x-4)/(x-2)`
`17)A_17=1/(x+5)+2/(x-5)-(2x+10)/(x^2-25)`
`=1/(x+5)+2/(x-5)-(2x+10)/[(x+5)(x-5)]`
`=(x-5)/[(x+5)(x-5)]+[2(x+5)]/[(x+5)(x-5)]-(2x+10)/[(x+5)(x-5)]`
`=[x-5+2(x+5)-(2x+10)]/[(x+5)(x-5)]`
`=(x-5+2x+10-2x-10)/[(x+5)(x-5)]`
`=[(x+2x-2x)+(-5+10-10)]/[(x+5)(x-5)]`
`=(x-5)/[(x+5)(x-5)]`
`=1/(x+5)`