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a, (1 + 3 + 5 + 7+ ... + 2003 + 2005) x (125125 x 127 - 127127 x125)
= (1 + 3 + 5 +...+2003 + 2005) x (125 x 1001 x 127 - 127x1001 x 125)
= (1 + 3 + 5 +...+ 2003 + 2005) x 0
= 0
Đề đọc khó hiểu quá bạn. Bạn nên viết lại để mọi người hỗ trợ tốt hơn.
\(5\dfrac{3}{4}:3+2\dfrac{1}{4}\times\dfrac{1}{3}-\dfrac{1}{8}=\left(5\dfrac{3}{4}+2\dfrac{1}{4}\right)\times\dfrac{1}{3}-\dfrac{1}{8}\\ =8\times\dfrac{1}{3}-\dfrac{1}{8}=\dfrac{8}{3}-\dfrac{1}{8}=\dfrac{61}{24}\\ ----\\ \dfrac{3}{5}:\dfrac{5}{6}:\dfrac{6}{7}:\dfrac{7}{8}+\dfrac{2}{5}+\dfrac{23}{35}\\ =\dfrac{3}{5}\times\dfrac{6}{5}\times\dfrac{7}{6}\times\dfrac{8}{7}+\dfrac{2}{5}+\dfrac{23}{35}=\dfrac{18}{25}\times\dfrac{4}{3}+\dfrac{2}{5}+\dfrac{23}{35}=\dfrac{353}{175}\)
6/7 * 5/8*7/3*7/6*8/5
= 6*5*7*7*8 /7*8*3*6*5
=1*1*7*1*1/1*1*3*1*1
= 7/3
$1-2+3-4+5-6+.....+97-98+99-100+101$
$= (1-2) + (3-4) + (5-6) + ...... + (99-100) + 101
$= (-50) + 101$
$= 51$
\(\frac{4}{3}:\frac{5}{4}:\frac{6}{5}:\frac{7}{6}:\frac{8}{7}:\frac{9}{8}\)
=\(\frac{4}{3}\times\frac{4}{5}\times\frac{5}{6}\times\frac{6}{7}\times\frac{7}{8}\times\frac{8}{9}\)
=\(\frac{16}{27}\)
Ta có : \(\frac{3}{5}:\frac{5}{6}:\frac{6}{7}:\frac{7}{8}+\frac{2}{5}+\frac{23}{25}\)
= \(\frac{3}{5}.\frac{6}{5}.\frac{7}{6}.\frac{8}{7}+\frac{2}{5}+\frac{23}{25}\)
= \(\frac{3.6.7.8}{5.5.6.7}+\frac{10}{25}+\frac{23}{25}\)
= \(\frac{24}{25}+\frac{10}{25}+\frac{23}{25}\)
= \(\frac{57}{25}\)
Lưu ý : dấu chấm là dấu nhân nhá
\(12,1+12,1\cdot101+99\cdot12,1-101\cdot12,1\)
\(=12,1\cdot\left(1+101+99-101\right)\)
\(=12,1\cdot\left(1+99\right)\)
\(=12,1\cdot100\)
\(=1210\)
`12,1 + 12,1 x 101 + 99 x 12,1 - 101 x 12,1`
`= 1 x 12,1 + 12,1 x 101 + 99 x 12,1 - 101 x 12,1`
`= 12,1 x (1 + 101 + 99 - 101)`
`= 12,1 x 100`
`= 1 210`
\(\dfrac{6}{35}+\dfrac{6}{63}+...+\dfrac{6}{399}+\dfrac{6}{483}\)
\(=\dfrac{6}{3\cdot5}+\dfrac{6}{5\cdot7}+...+\dfrac{6}{19\cdot21}+\dfrac{6}{21\cdot23}\)
\(=3\left(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{19\cdot21}+\dfrac{2}{21\cdot23}\right)\)
\(=3\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{19}-\dfrac{1}{21}+\dfrac{1}{21}-\dfrac{1}{23}\right)\)
\(=3\left(\dfrac{1}{3}-\dfrac{1}{23}\right)=3\cdot\dfrac{20}{69}=\dfrac{20}{23}\)
20/23