Đáp án :
Giải thích các bước giải :
Tính ( hợp lí ) :
$\frac{70}{3}$ × ( $\frac{39}{40}$ + $\frac{39}{42}$ ) - $\frac{246}{7}$ ÷ ( $\frac{41}{56}$ + $\frac{41}{72}$
= $\frac{70}{3}$ × ( $\frac{13}{10}$ + $\frac{13}{14}$ ) - $\frac{246}{7}$ ÷ ( $\frac{41}{7 × 8}$ + $\frac{41}{8 × 9}$ )
= $\frac{70}{3}$ × ( 1 + $\frac{3}{10}$ + 1 - $\frac{1}{14}$ ) - $\frac{246}{7}$ ÷ ( $\frac{40 + 1}{7 × 8}$ + $\frac{40 + 1}{8 × 9}$ )
= $\frac{70}{3}$ × ( ( 1 + 1 ) + ( $\frac{3}{10}$ - $\frac{1}{14}$ ) ) - $\frac{246}{7}$ ÷ ( $\frac{5}{7}$ + $\frac{1}{7 × 8}$ + $\frac{5}{9}$ + $\frac{1}{8 × 9}$ )
= $\frac{70}{3}$ × ( 2 + $\frac{8}{35}$ ) - $\frac{246}{7}$ ÷ ( ( $\frac{5}{7}$ + $\frac{5}{9}$ ) + ( $\frac{1}{7 × 8}$ + $\frac{1}{8 × 9}$ ) )
= $\frac{70}{3}$ × $\frac{78}{35}$ - $\frac{246}{7}$ ÷ ( ( $\frac{5}{7}$ + $\frac{5}{9}$ ) + ( $\frac{1}{7}$ - $\frac{1}{8}$ + $\frac{1}{8}$ - $\frac{1}{9}$ ) )
= $\frac{35 × 2 × 26 × 3}{3 × 35}$ - $\frac{246}{7}$ ÷ ( $\frac{5}{7}$ + $\frac{5}{9}$ + $\frac{1}{7}$ - $\frac{1}{9}$ )
= 52 - $\frac{246}{7}$ ÷ ( ( $\frac{5}{7}$ + $\frac{1}{7}$ ) + ( $\frac{5}{9}$ - $\frac{1}{9}$ ) )
= 52 - $\frac{246}{7}$ ÷ ( $\frac{6}{7}$ + $\frac{4}{9}$ )
= 52 - $\frac{246}{7}$ ÷ $\frac{82}{63}$
= 52 - $\frac{246}{7}$ ÷ $\frac{82 × 3 × 9 × 7}{7 × 82}$
= 52 - 27
= 25
$\frac{57}{20}$ - $\frac{26}{15}$ + $\frac{139}{20}$ ÷ 3
= $\frac{57}{20}$ - $\frac{26}{15}$ + $\frac{139}{60}$
= $\frac{171}{60}$ - $\frac{104}{60}$ + $\frac{139}{60}$
= $\frac{103}{30}$
( 1 - $\frac{1}{2}$ ) × ( 1 - $\frac{1}{3}$ ) × ( 1 - $\frac{1}{4}$ ) × ...... ( 1 - $\frac{1}{2004}$ )
= $\frac{1}{2}$ × $\frac{2}{3}$ × $\frac{3}{4}$ × ..... × $\frac{2001}{2002}$ × $\frac{2002}{2003}$ × $\frac{2003}{2004}$
= $\frac{1 × 2 × 3 × ..... × 2001 × 2002 × 2004}{2 × 3 × 4 × .... × 2002 × 2003 × 2004}$
= $\frac{1}{2004}$
No copy .
Chúc bn học tốt !
Cho mink ctlhn nha .
