Tính nhanh: M =1/x*(x+1)+1/(x+1)*(x+2)+1/(x+2)*(x+3)+1/(x+3)*(x+4)+1/(x+4)*(x+5)+1/x+5 31/07/2021 Bởi Kennedy Tính nhanh: M =1/x*(x+1)+1/(x+1)*(x+2)+1/(x+2)*(x+3)+1/(x+3)*(x+4)+1/(x+4)*(x+5)+1/x+5
$ĐKXĐ : x \neq -1,-2,-3,-4,-5$ Ta có : $M = \dfrac{1}{x.(x+1)}+\dfrac{1}{(x+1).(x+2)} + \dfrac{1}{(x+2).(x+3)} + \dfrac{1}{(x+3).(x+4)}+\dfrac{1}{(x+4).(x+5)} + \dfrac{1}{x+5}$ $ = \dfrac{(x+1)-x}{x.(x+1)} + \dfrac{(x+2)-(x+1)}{(x+1).(x+2)} + \dfrac{(x+3)-(x+2)}{(x+2).(x+3)} + \dfrac{(x+4)-(x+3)}{(x+3).(x+4)}+\dfrac{(x+5)-(x+4)}{(x+4).(x+5)} + \dfrac{1}{x+5}$ $ = \dfrac{1}{x}-\dfrac{1}{x+1}+\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+5}$ $ = \dfrac{1}{x}$ Vậy $M = \dfrac{1}{x}$ Bình luận
$ĐKXĐ : x \neq -1,-2,-3,-4,-5$
Ta có : $M = \dfrac{1}{x.(x+1)}+\dfrac{1}{(x+1).(x+2)} + \dfrac{1}{(x+2).(x+3)} + \dfrac{1}{(x+3).(x+4)}+\dfrac{1}{(x+4).(x+5)} + \dfrac{1}{x+5}$
$ = \dfrac{(x+1)-x}{x.(x+1)} + \dfrac{(x+2)-(x+1)}{(x+1).(x+2)} + \dfrac{(x+3)-(x+2)}{(x+2).(x+3)} + \dfrac{(x+4)-(x+3)}{(x+3).(x+4)}+\dfrac{(x+5)-(x+4)}{(x+4).(x+5)} + \dfrac{1}{x+5}$
$ = \dfrac{1}{x}-\dfrac{1}{x+1}+\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+5}$
$ = \dfrac{1}{x}$
Vậy $M = \dfrac{1}{x}$