[Toán 6] Bài tập tính nhanh

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[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
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$A = \dfrac{(\dfrac{-3333}{1212} - \dfrac{-3333}{2020} - \dfrac{-3333}{3030} - \dfrac{-3333}{4242} - \dfrac{-3333}{5656})}{\dfrac{5}{808}}$

$B = (\dfrac{3^2}{10} + \dfrac{3^2}{40} + \dfrac{3^2}{88}) + (\dfrac{-3}{5} - \dfrac{3}{20} - \dfrac{3}{44} - \dfrac{3}{77} + \dfrac{-3}{119} - \dfrac{3}{170})$

 

[tayhd20022001]

$B$=( $\dfrac{3^2}{10}$ + $\dfrac{3^2}{40}$ + $\dfrac{3^2}{88}$)$ + $( $\dfrac{-3}{5}$ -$\dfrac{3}{20}$ - $\dfrac{3}{44}$ - $\dfrac{3}{77}$ + $\dfrac{-3}{119}$ - $\dfrac{3}{170}$ )
\Rightarrow $B$=( $\dfrac{9}{10}$ + $\dfrac{9}{40}$ + $\dfrac{9}{88}$)$ + $3$ .$( $\dfrac{-1}{5}$ -$\dfrac{1}{20}$ - $\dfrac{1}{44}$ - $\dfrac{1}{77}$ + $\dfrac{-1}{119}$ - $\dfrac{1}{170}$ )
\Rightarrow $B$=3.3.( $\dfrac{1}{10}$ + $\dfrac{1}{40}$ + $\dfrac{1}{88}$)$ + $3$ .$( $\dfrac{-1}{5}$ -$\dfrac{1}{20}$ - $\dfrac{1}{44}$ - $\dfrac{1}{77}$ + $\dfrac{-1}{119}$ - $\dfrac{1}{170}$ )
\Rightarrow $B$=3.3.( $\dfrac{1}{2.5}$ + $\dfrac{1}{5.8}$ + $\dfrac{1}{8.11}$)$ + $3$ .$( $\dfrac{-1}{5}$ -$\dfrac{1}{20}$ - $\dfrac{1}{44}$ - $\dfrac{1}{77}$ + $\dfrac{-1}{119}$ - $\dfrac{1}{170}$ )
Vậy \Rightarrow $\dfrac{1}{2}$-$\dfrac{1}{5}$=$\dfrac{3}{10}$
\Rightarrow B sẽ phải giảm đi 3 lần .
Từ đó ta có :

\Rightarrow $B$=3.( $\dfrac{1}{2}$-$\dfrac{1}{5}$ + $\dfrac{1}{5}$-$\dfrac{1}{8}$ + $\dfrac{1}{8}$-$\dfrac{1}{11}$)$ + $3$ .$( $\dfrac{-1}{5}$ -$\dfrac{1}{20}$ - $\dfrac{1}{44}$ - $\dfrac{1}{77}$ + $\dfrac{-1}{119}$ - $\dfrac{1}{170}$ )
\Rightarrow $B$=3.( $\dfrac{1}{2}$-$\dfrac{1}{11}$)$ + $3$ .$( $\dfrac{-1}{5}$ -$\dfrac{1}{20}$ - $\dfrac{1}{44}$ - $\dfrac{1}{77}$ + $\dfrac{-1}{119}$ - $\dfrac{1}{170}$ )
\Rightarrow $B$= $\dfrac{27}{22}$ + $3$ .( $\dfrac{-1}{5}$ -$\dfrac{1}{20}$ - $\dfrac{1}{44}$ - $\dfrac{1}{77}$ + $\dfrac{-1}{119}$ - $\dfrac{1}{170}$ )
\Rightarrow $B$= $\dfrac{27}{22}$ + 3.$\dfrac{-3}{10}$
\Rightarrow $B$= $\dfrac{27}{22}$ + $\dfrac{-9}{10}$
\Rightarrow $B$= $\dfrac{18}{55}$

 

[tayhd20022001]

A= $\dfrac{\dfrac{-3333}{1212}-\dfrac{-3333}{2020}-\dfrac{-3333}{3030}-\dfrac{-3333}{4242}-\dfrac{-3333}{5656}}{\dfrac{5}{808}}$
A= $\dfrac{-3333.(\dfrac{1}{1212}-\dfrac{1}{2020}-\dfrac{1}{3030}-\dfrac{1}{4242}-\dfrac{1}{5656})}{\dfrac{5}{808}}$
A= $\dfrac{-3333.(\dfrac{1}{12.101}-\dfrac{1}{20.101}-\dfrac{1}{30.101}-\dfrac{1}{42.101}-\dfrac{1}{56.101})}{\dfrac{5}{8.101}}$
A= -3333.($\dfrac{1}{12.101}$-$\dfrac{1}{20.101}$-$\dfrac{1}{30.101}$-$\dfrac{1}{42.101}$-$\dfrac{1}{56.101}$):$\dfrac{5}{8.101}$
A= -3333.($\dfrac{1}{12}$.$\dfrac{1}{101}$-$\dfrac{1}{20}$.$\dfrac{1}{101}$-...-$\dfrac{1}{56}$.$\dfrac{1}{101}$):$\dfrac{5}{8.101}$
A= -3333.$\dfrac{1}{101}$.($\dfrac{1}{12}$-$\dfrac{1}{20}$-$\dfrac{1}{30}$-...-$\dfrac{1}{56}$):$\dfrac{5}{8}$:$\dfrac{5}{101}$.
A= -3333.($\dfrac{1}{3.4}$-$\dfrac{1}{4.5}$-$\dfrac{1}{5.6}$-...-$\dfrac{1}{7.8}$):$\dfrac{5}{8}$:$\dfrac{1}{5}$.
A= -3333.($\dfrac{1}{3}$-$\dfrac{1}{4}$-$\dfrac{1}{4}$-$\dfrac{1}{5}$...-$\dfrac{1}{7}$-$\dfrac{1}{8}$):$\dfrac{1}{8}$
A= -3333.($\dfrac{1}{3}$-$\dfrac{1}{8}$).8
A= -3333.$\dfrac{5}{24}$.8
A= -3333.$\dfrac{5}{3}$
\Rightarrow A=-5555
Đáp số : A=-5555

 

[huy14112]

$A = \dfrac{(\dfrac{-3333}{1212} - \dfrac{-3333}{2020} - \dfrac{-3333}{3030} - \dfrac{-3333}{4242} - \dfrac{-3333}{5656})}{\dfrac{5}{808}}$

$A = \dfrac{\dfrac{-3333}{101}(\dfrac{1}{12} - \dfrac{1}{20} - \dfrac{1}{30} - \dfrac{1}{42} - \dfrac{1}{56})}{\dfrac{5}{808}}$

$A = \dfrac{\dfrac{-3333}{101}}{\dfrac{5}{808}}.(\dfrac{1}{12} - \dfrac{1}{20} - \dfrac{1}{30} - \dfrac{1}{42} - \dfrac{1}{56})$

$A = \dfrac{(-3333).8}{5}. [\dfrac{1}{12} -( \dfrac{1}{20} + \dfrac{1}{30} +\dfrac{1}{42} + \dfrac{1}{56})]$

$A = \dfrac{(-3333).8}{5}. [\dfrac{1}{12} -( \dfrac{1}{4.5} + \dfrac{1}{5.6} + \dfrac{1}{6.7} + \dfrac{1}{7.8})]$


$A=\dfrac{(-3333).8}{5}.(\dfrac{1}{12}-(\dfrac{1}{4}-\dfrac{1}{8})$

$A=\dfrac{(-3333).8}{5}.(\dfrac{1}{12}-\dfrac{1}{8})$

$A=\dfrac{(-3333).8}{5}.(-\dfrac{1}{24})=\dfrac{1111}{5}$