Đáp án:

 

Giải thích các bước giải:

 `a,1/[a-b]+[a+b]/[a^2 +ab+b^2]-[3ab]/[a^3 -b^3]`

`=[a^2 +ab+b^2 +(a+b)(a-b)-3ab]/[(a-b)(a^2 +ab+b^2 )]`

`=[a^2 +a^2 -b^2 +b^2 -2ab]/[(a-b)(a^2 +ab+b^2)]`

`=[2a^2 -2ab]/[(a-b)(a^2 +ab+b^2 )]`

`=[2a]/[a^2 +ab+b^2]`

`b,1/[x^2 +3x+2]+1/[x^2 +5x+6]+1/[x^2 +7x+12] +1/[x^2 +9x+20]`

`=1/[(x+1)(x+2)]+1/[(x+2)(x+3)]+1/[(x+3)(x+4)]+1/[(x+4)(x+5)]`

`=1/[x+1]-1/[x+2]+1/[x+2]-1/[x+3]+1/[x+3]-1/[x+4]+1/[x+4]-1/[x+5]`

`=1/[x+1]-1/[x+5]`

`=[x+5-x-1]/[(x+1)(x+5)]`

`= 4/[(x+1)(x+5)]`

(Bài tham khảo)

Gửi bạn:

$ĐKXĐ:a,b \neq 0$

$a,$ $\dfrac{1}{a-b}+\dfrac{a+b}{a^2+ab+b^2}-\dfrac{3ab}{a^3-b^3}$ 

$=\dfrac{1}{a-b}+\dfrac{a+b}{a^2+ab+b^2}-\dfrac{3ab}{(a-b)(a^2+ab+b^2)}$ 

$=\dfrac{a^2+ab+b^2}{(a-b)(a^2+ab+b^2)}+\dfrac{(a+b)(a-b)}{(a-b)(a^2+ab+b^2)}-\dfrac{3ab}{(a-b)(a^2+ab+b^2)}$ 

$=\dfrac{a^2+ab+b^2}{(a-b)(a^2+ab+b^2)}+\dfrac{a^2-b^2}{(a-b)(a^2+ab+b^2)}-\dfrac{3ab}{(a-b)(a^2+ab+b^2)}$ 

$=\dfrac{a^2+ab+b^2+a^2-b^2-3ab}{(a-b)(a^2+ab+b^2)}$ 

$=\dfrac{2a^2-2ab}{(a-b)(a^2+ab+b^2)}$ 

$=\dfrac{2a(a-b)}{(a-b)(a^2+ab+b^2)}$ 

$=\dfrac{2a}{a^2+ab+b^2}$ 

$b,$ $\dfrac{1}{x^2+3x+2}+\dfrac{1}{x^2+5x+6}+\dfrac{1}{x^2+7x+12}+\dfrac{1}{x^2+9x+20}$ 

$=\dfrac{1}{x^2+x+2x+2}+\dfrac{1}{x^2+2x+3x+6}+\dfrac{1}{x^2+3x+4x+12}+\dfrac{1}{x^2+5x+4x+20}$ 

$=\dfrac{1}{(x+1)(x+2)}+\dfrac{1}{(x+2)(x+3)}+\dfrac{1}{(x+3)(x+4)}+\dfrac{1}{(x+4)(x+5)}$ 

$=\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+5}$ 

$=\dfrac{1}{x+1}-\dfrac{1}{x+5}$ 

$=\dfrac{x+5}{(x+1)(x+5)}-\dfrac{x+1}{(x+5)(x+1)}$ 

$=\dfrac{4}{(x+1)(x+5)}$