Giúp mình
$\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+…+\frac{1}{200}}{\frac{1}{199}+\frac{2}{198}+\frac{3}{197}+…+\frac{198}{2}+\frac{199}{1}}$
Giúp mình
$\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+…+\frac{1}{200}}{\frac{1}{199}+\frac{2}{198}+\frac{3}{197}+…+\frac{198}{2}+\frac{199}{1}}$
*Lời giải :
`(1/2 + 1/3 + 1/4 + … + 1/200)/(1/199 + 2/198 + 3/197 + … + 198/2 + 199/1)`
`= (1/2 + 1/3 + 1/4 + … + 1/200)/( (1/199 + 1) + (2/198 + 1) + (3/197 + 1) + … + (198/2 + 1) + 1 )`
`= (1/2 + 1/3 + 1/4 + … + 1/200)/(200/199 + 200/198 + 200/197 + .. + 200/2 + 1)`
`= (1/2 + 1/3 + 1/4 + … + 1/200)/(200 . (1/2 + 1/3 + 1/4 + … + 1/200) )`
`= 1/200`
Đáp án + Giải thích các bước giải:
Xét MS : `1/199+2/198+3/197+…+198/2+199/1`
`= (1/199+1)+(2/198+1)+(3/197+1)+…+(198/2+1)+1`
`= 200/199+200/198+200/197+…+200/2+200/200`
`= 200 . (1/2+….+1/197+1/198+1/199+1/200)`
`to`
$\dfrac{ \dfrac{1}{2}+\dfrac{1}{3}+…+\dfrac{1}{200} }{ \dfrac{1}{199}+\dfrac{2}{198}+\dfrac{3}{197}+…+\dfrac{198}{2}+\dfrac{199}{1}}$
$=\dfrac{ \dfrac{1}{2}+\dfrac{1}{3}+…+\dfrac{1}{200} }{200 . (\dfrac{1}{2}+…+\dfrac{1}{199}+\dfrac{1}{200})}=\dfrac{1}{200}$