[Toán 7]Chứng Minh

[t.linh2111]

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Chứng minh rằng:
[TEX]\frac{1}{1.2}[/TEX]+[TEX]\frac{1}{3.4}[/TEX]+[TEX]\frac{1}{4.5}[/TEX]+...+[TEX]\frac{1}{49.50}[/TEX]=[TEX]\frac{1}{26}[/TEX]+[TEX]\frac{1}{27}[/TEX]+[TEX]\frac{1}{28}[/TEX]+...+[TEX]\frac{1}{50}[/TEX]

 

[hiennguyenthu082]

Chứng minh rằng:
[TEX]\frac{1}{1.2}[/TEX]+[TEX]\frac{1}{3.4}[/TEX]+[TEX]\frac{1}{4.5}[/TEX]+...+[TEX]\frac{1}{49.50}[/TEX]=[TEX]\frac{1}{26}[/TEX]+[TEX]\frac{1}{27}[/TEX]+[TEX]\frac{1}{28}[/TEX]+...+[TEX]\frac{1}{50}[/TEX]

Ta có : $\dfrac{1}{26}+\dfrac{1}{27}+ \dfrac{1}{28} +...+\dfrac{1}{49}+\dfrac{1}{50}-(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+ \dfrac{1}{4.5}+...+\dfrac{1}{49.50})$

$=\dfrac{1}{26}+\dfrac{1}{27}+ \dfrac{1}{28} +...+\dfrac{1}{49}+\dfrac{1}{50}-(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50})$

$=\dfrac{1}{26}+\dfrac{1}{27}+ \dfrac{1}{28} +...+\dfrac{1}{49}+\dfrac{1}{50}+(\dfrac{1}{26}-\dfrac{1}{27}+\dfrac{1}{28}-...-\dfrac{1}{49}+\dfrac{1}{50})-
(\dfrac{1}{13}-\dfrac{1}{14}+\dfrac{1}{15}-...-\dfrac{1}{24}+\dfrac{1}{25})-(\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{9}-...-\dfrac{1}{12})+(\dfrac{1}{6}-\dfrac{1}{5}+\dfrac{1}{4})+(\dfrac{1}{2}-1)$

$=\dfrac{1}{7}+\dfrac{1}{8}+...+\dfrac{1}{12}- (\dfrac{1}{7}- \dfrac{1}{8}+\dfrac{1}{9}-...-\dfrac{1}{12})(\dfrac{1}{6}-\dfrac{1}{5}+\dfrac{1}{4})+(\dfrac{1}{2}-1)$

$=\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+(\dfrac{1}{6}-\dfrac{1}{5}+\dfrac{1}{4})+(\dfrac{1}{2}-1)$

$=\dfrac{1}{2}+\dfrac{1}{2}-1$

$=0$

Vậy $\dfrac{1}{26}+\dfrac{1}{27}+ \dfrac{1}{28} +...+\dfrac{1}{49}+\dfrac{1}{50}=(\dfrac{1}{1.2}+ \dfrac{1}{2.3}+\dfrac{1}{3.4}+ \dfrac{1}{4.5}+...+\dfrac{1}{49.50})$

 

[tayhd20022001]

Chứng minh rằng:
$\dfrac{1}{1.2}$+$\dfrac{1}{2.3}$+$\dfrac{1}{3.4}$ + ... +$\dfrac{1}{49.50}$=$\dfrac{1}{26}$+$\dfrac{1}{27}$+$\dfrac{1}{28}$+...+$\dfrac{1}{50}$
Ta có :
A:
=> $\dfrac{1}{1.2}$+$\dfrac{1}{2.3}$+$\dfrac{1}{3.4}$ + ... +$\dfrac{1}{49.50}$.
=>1-$\dfrac{1}{2}$+$\dfrac{1}{2}$-$\dfrac{1}{3}$+$\dfrac{1}{3}$-$\dfrac{1}{4}$+....+$\dfrac{1}{49}$-$\dfrac{1}{50}$
=>1-$\dfrac{1}{50}$
=>$\dfrac{49}{50}$
B:
=>$\dfrac{1}{26}$+$\dfrac{1}{27}$+$\dfrac{1}{28}$+...+$\dfrac{1}{50}$
=> 1+$\dfrac{1}{2}$+$\dfrac{1}{3}$+...+$\dfrac{1}{50}$-(1+$\dfrac{1}{2}$+$\dfrac{1}{3}$+...+$\dfrac{1}{25}$)
=>1+$\dfrac{1}{2}$+$\dfrac{1}{3}$+...+$\dfrac{1}{50}$-2(1+$\dfrac{1}{2}$+$\dfrac{1}{4}$+...+$\dfrac{1}{50}$
=>1-$\dfrac{1}{2}$+$\dfrac{1}{2}$-$\dfrac{1}{3}$+$\dfrac{1}{3}$-$\dfrac{1}{4}$+....+$\dfrac{1}{49}$-$\dfrac{1}{50}$
=>$\dfrac{49}{50}$
=> Vậy :$\dfrac{1}{1.2}$+$\dfrac{1}{2.3}$+$\dfrac{1}{3.4}$ + ... +$\dfrac{1}{49.50}$=$\dfrac{1}{26}$+$\dfrac{1}{27}$+$\dfrac{1}{28}$+...+$\dfrac{1}{50}$