[ Toán 6 ] Dãy phân số

[thieukhang61]

[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
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Tính tổng sau:
\[\frac{1}{{10}} + \frac{2}{9} + \frac{3}{8} + ... + \frac{9}{2} + 10\]
Tính A, biết:
\[A = \frac{{2008 + \frac{{2007}}{2} + \frac{{2006}}{3} + \frac{{2005}}{4} + ... + \frac{2}{{2007}} + \frac{1}{{2008}}}}{{\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + ... + \frac{1}{{2008}} + \frac{1}{{2009}}}}\]

 

[hiennguyenthu082]

Tính tổng sau:
\[\frac{1}{{10}} + \frac{2}{9} + \frac{3}{8} + ... + \frac{9}{2} + 10\]

Ta có :

$A=\dfrac{1}{10}+\dfrac{2}{9}+\dfrac{3}{8}+...+ \dfrac{9}{2} +10$

$A=\dfrac{11-10}{10}+\dfrac{11-9}{9}+\dfrac{11-8}{8}+...+\dfrac{11-2}{2}+\dfrac{11-1}{1}$

$A=(\dfrac{11}{10}+\dfrac{11}{9}+\dfrac{11}{8}+...+\dfrac{11}{2}+\dfrac{11}{1})-(\dfrac{10}{10}+\dfrac{9}{9}+\dfrac{8}{8}+...+ \dfrac{2}{2} +\dfrac{1}{1})$

$A=11+11(\dfrac{1}{10}+\dfrac{1}{9}+\dfrac{1}{8}+...+\dfrac{1}{2})-10$

$A=1+11(\dfrac{1}{10}+\dfrac{1}{9} + \dfrac{1}{8} +...+\dfrac{1}{2})$

$A=11(\dfrac{1}{11}+\dfrac{1}{10}+\dfrac{1}{9}+ \dfrac{1}{8} +...+\dfrac{1}{2})$

 

[thieukhang61]

Ta có :

$A=\dfrac{1}{10}+\dfrac{2}{9}+\dfrac{3}{8}+...+ \dfrac{9}{2} +10$

$A=\dfrac{11-10}{10}+\dfrac{11-9}{9}+\dfrac{11-8}{8}+...+\dfrac{11-2}{2}+\dfrac{11-1}{1}$

$A=(\dfrac{11}{10}+\dfrac{11}{9}+\dfrac{11}{8}+...+\dfrac{11}{2}+\dfrac{11}{1})-(\dfrac{10}{10}+\dfrac{9}{9}+\dfrac{8}{8}+...+ \dfrac{2}{2} +\dfrac{1}{1})$

$A=11+11(\dfrac{1}{10}+\dfrac{1}{9}+\dfrac{1}{8}+...+\dfrac{1}{2}+1)-10$

$A=1+11(\dfrac{1}{10}+\dfrac{1}{9} + \dfrac{1}{8} +...+\dfrac{1}{2}+1)$

$A=11(\dfrac{1}{11}+\dfrac{1}{10}+\dfrac{1}{9}+ \dfrac{1}{8} +...+\dfrac{1}{2}+1)$


Bạn đưa ra kết quả cụ thể được không.

 

[huy14112]

Ta có :

$A=\dfrac{1}{10}+\dfrac{2}{9}+\dfrac{3}{8}+...+ \dfrac{9}{2} +10$

$A=\dfrac{11-10}{10}+\dfrac{11-9}{9}+\dfrac{11-8}{8}+...+\dfrac{11-2}{2}+\dfrac{11-1}{1}$

$A=(\dfrac{11}{10}+\dfrac{11}{9}+\dfrac{11}{8}+...+\dfrac{11}{2}+\dfrac{11}{1})-(\dfrac{10}{10}+\dfrac{9}{9}+\dfrac{8}{8}+...+ \dfrac{2}{2} +\dfrac{1}{1})$

$A=11+11(\dfrac{1}{10}+\dfrac{1}{9}+\dfrac{1}{8}+...+\dfrac{1}{2}+1)-10$

$A=1+11(\dfrac{1}{10}+\dfrac{1}{9} + \dfrac{1}{8} +...+\dfrac{1}{2}+1)$

$A=11(\dfrac{1}{11}+\dfrac{1}{10}+\dfrac{1}{9}+ \dfrac{1}{8} +...+\dfrac{1}{2}+1)$

Em này làm sai mà dài quá.

$A+10=\dfrac{1}{10}+1+\dfrac{2}{9}+1+\dfrac{3}{8}+1+...+ \dfrac{9}{2}+1 +10+1$

$A+10=\dfrac{11}{10}+\dfrac{11}{9}+\dfrac{11}{8}+...+\dfrac{11}{2}+\dfrac{11}{1}$

$A+10=11(\dfrac{1}{10}+\dfrac{1}{9} + \dfrac{1}{8} +...+\dfrac{1}{2}+1)$

$A=11(\dfrac{1}{10}+\dfrac{1}{9} + \dfrac{1}{8} +...+\dfrac{1}{2}+1)-10$

 

[hiennguyenthu082]

Tính A, biết:
\[A = \dfrac{{2008 + \dfrac{{2007}}{2} + \dfrac{{2006}}{3} + \dfrac{{2005}}{4} + ... + \dfrac{2}{{2007}} + \dfrac{1}{{2008}}}}{{\dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{4} + ... + \dfrac{1}{{2008}} + \dfrac{1}{{2009}}}}\]

Đặt

$A= \dfrac{B}{C} = \dfrac{{2008 + \dfrac{{2007}}{2} + \dfrac{{2006}}{3} + \dfrac{{2005}}{4} + ... + \dfrac{2}{{2007}} + \dfrac{1}{{2008}}}}{{\dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{4} + ... + \dfrac{1}{{2008}} + \dfrac{1}{{2009}}}}$

\Rightarrow $B=\dfrac{2009-1}{1}+\dfrac{2009-2}{2}+\dfrac{2009-3}{3}+\dfrac{2009-4}{4}...+ \dfrac{2009-2007}{2007} +\dfrac{2009-2008}{2008}$

$B=(\dfrac{2009}{1}+\dfrac{2009}{2}+\dfrac{2009}{3}+\dfrac{2009}{4}...+\dfrac{2009}{2007}+\dfrac{2009}{2008})-(\dfrac{1}{1}+\dfrac{2}{2}+\dfrac{3}{3}+\dfrac{4}{4}...+\dfrac{2007}{2007}+\dfrac{2008}{2008})$

$B=2009+2009(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}...+\dfrac{1}{2007}+\dfrac{1}{2008})-2008$

$B=1+2009(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}...+\dfrac{1}{2007}+\dfrac{1}{2008})$

$B=2009(\dfrac{1}{2}+\dfrac{1}{3} + \dfrac{1}{4} +...+\dfrac{1}{2008}+\dfrac{1}{2009})$

\Rightarrow $\dfrac{B}{C}=\dfrac{2009}{1}=2009$

Vậy $A=2009$

 

[pandahieu]

Tính tổng sau:
\[\frac{1}{{10}} + \frac{2}{9} + \frac{3}{8} + ... + \frac{9}{2} + 10\]
Tính A, biết:
\[A = \frac{{2008 + \frac{{2007}}{2} + \frac{{2006}}{3} + \frac{{2005}}{4} + ... + \frac{2}{{2007}} + \frac{1}{{2008}}}}{{\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + ... + \frac{1}{{2008}} + \frac{1}{{2009}}}}\]

$\boxed{1}$
Ta có: $A=\dfrac{1}{10}+\dfrac{2}{9}+\dfrac{3}{8}+...+ \dfrac{9}{2} +10$
$A=\dfrac{11-10}{10}+\dfrac{11-9}{9}+\dfrac{11-8}{8}+...+\dfrac{11-2}{2}+\dfrac{11-1}{1}$
$A=(\dfrac{11}{10}+\dfrac{11}{9}+\dfrac{11}{8}+...+\dfrac{11}{2}+\dfrac{11}{1})-(\dfrac{10}{10}+\dfrac{9}{9}+\dfrac{8}{8}+...+ \dfrac{2}{2} +\dfrac{1}{1})$
$A=11+11(\dfrac{1}{10}+\dfrac{1}{9}+\dfrac{1}{8}+...+\dfrac{1}{2}+1)-10$
$A=1+11(\dfrac{1}{10}+\dfrac{1}{9} + \dfrac{1}{8} +...+\dfrac{1}{2}+1)$
$A=11(\dfrac{1}{11}+\dfrac{1}{10}+\dfrac{1}{9}+ \dfrac{1}{8} +...+\dfrac{1}{2}+1)$

$\boxed{2}$
Đặt tử là $A$, mẫu là $B$ thì ta có :
$A+2008=(2008+1)+(\frac{2007}{2}+1)+...+(\frac{1}{2008}+1)=2009(1+B-\frac{1}{2009})$
$\Rightarrow \frac{A}{B}=2009$