Giải thích các bước giải:
18/.
A18 = `1/(x+5)` = `2/(x-5)` - `(2x+10)/(x²-25)`
ĐKXĐ: x# 5 ; x # - 5
MTC: (x + 5)(x - 5)
A18 = `(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-5)/[(x+5)(x-5)]`
= `1/(x+5)`
Vậy A18 = `1/(x+5)`
19/.
A19 = `3/(x+3)` + `1/(x-3)` - `(18)/(9-x²)`
A19 = `3/(x+3)` + `1/(x-3)` + `(18)/(x²-9)`
ĐKXĐ: x # 3; x # - 3
MTC: (x + 3)(x - 3)
A19 = `3(x-3)/[(x+3)(x-3)]` + `(x+3)/[(x+3)(x-3)]` + `(18)/[(x+3)(x-3)]`
= `(3x-9+x+3+18)/[(x+3)(x-3)]`
= `(4x+12)/[(x+3)(x-3)]`
= `[4(x+3)]/[(x+3)(x-3)]`
= `4/(x-3)`
Vậy A19 = `4/(x-3)`
20/.
A20 = (`(x+1)/(2x-2)` + `3/(x²-1)` - `(x+3)/(2x+2)`) . `(4x²-4)/5`
Ta có: 2x -2 = 2(x-1); 2x + 2 = 2(x +1); x² - 1 = (x-1)(x+1)
ĐKXĐ: x # 1; x # -1
⇒ MTC: 2(x-1)(x+1)
A20 = (`(x+1)/(2x-2)` + `3/(x²-1)` - `(x+3)/(2x+2)`) - `(4x²-4)/5`
= (`[(x+1)(x-1)]/[2(x-1)(x+1)]` + `(3.2)/[2(x-1)(x+1)]` - `[(x+3)(x-1)]/[2(x-1)(x+1)]`). `(4x²-4)/5`
= (`[(x+1)²]/[2(x-1)(x+1)]` + `6/[2(x-1)(x+1)]` - `[(x+3)(x-1)]/[2(x-1)(x+1)]`). `[4(x²-1)]/5`
= (`[x²+2x+1+6-(x²-x+3x-3)]/[2(x-1)(x+1)]`). `[4(x²-1)]/5`
= (`(x²+2x+1+6-x²+x-3z+3)/[2(x-1)(x+1)]`). `[4(x²-1)]/5`
= `(10)/[2(x²-1)]`.`[4(x²-1)]/5`
= `[40(x²-1)]/[10(x²-1)]`
= 4
Vậy A20 = 4
21/.
A21 = (`(5x+2)/(x²-10)` + `(5x-2)/(x²+10)`).`(x²-100)/(x²+4)`
ĐKXĐ: x # ± √10
MTC: (x²-10)(x²+10)
A21 = (`[(5x+2)(x²+10)]/[(x²-10)(x²+10)]` + `[(5x-2)(x²-10)]/[(x²-10)(x²+10)]`).`(x²-100)/(x²+4)`
= (`(5x³+2x²+50x+20+5x³-2x²-50x+100)/[(x²-10)(x²+10)]`).`(x²-100)/(x²+4)`
= `(10x³+120)/[(x²-10)(x²+10)]`. `(x²-100)/(x²+4)`
= `[(10x³+120)(x²-100)]/[(x4-100)(x²+4)]`
= `(10x5+120x²-100x³-1200)/(x6-100x²+4x4-400)`
= `(10x5+120x²-100x³-1200)/(x6 -100x²+4x4-400)`
= `[10x²(x³+12)-100(x³+12)/[x²(x4-100)+4(x4-100)]`
= `[(x³+12)(10x²-100))/[(x4-100)(x²+4)`
22/.
A22 =`(x²+2x)/(2x+10)`+ `(x-5)/x` + `(50-5x)/[2x(x+5)]`
ĐKXĐ: x# 0; x # - 5
MTC: 2x(x+5)
A22 =`[x(x²+2x)]/[2x(x+5)]`+ `[2(x-5)(x+5)]/[2x(x+5)]` + `(50-5x)/[2x(x+5)]`
= `(x³+2x²+2x²-50+50-5x)/[2x(x+5)]`
= `(x³+4x²-5x)/[2x(x+5)]`
= `[x(x²+4x-5)]/[2x(x+5)]`
= `[x(x²+5x-x-5)]/[2x(x+5)]`
= `[x(x(x+5))-(x+5)]/[2x(x+5)]`
= `[x(x+5)(x-1)]/[2x(x+5)]`
= `(x-1)/2`
