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a) \(\left(56\times27+56\times35\right)\div62=56\times\left(27+35\right)\div62=56\times62\div62=56\)
b) \(\frac{0,18\times1230+0,9\times4567\times2+3\times5310\times6}{1+4+7+10+....+52+55-514}\)
\(=\frac{0,18\times1230+\left(0,9\times2\right)\times4567+\left(3\times6\right)\times5310}{1+4+5+.....+52+55-514}\)
\(=\frac{0,18\times1230+0,18\times4567+0,18\times5310}{1+4+7+...+52+55-514}\)
\(=\frac{0,18\times\left(1230+4567+5310\right)}{\left(55+1\right)\times55\div2-514}\)
\(=\frac{0,18\times11107}{971}=\frac{1999,26}{971}\)
mình ra cũng giống bạn Forever _ Alone nhé!!!Chẳng qua mình không biết viết phân số
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a, \(\frac{6}{7}.\frac{16}{15}.\frac{7}{6}.\frac{21}{32}=\frac{6}{7}.\frac{7}{6}.\frac{16}{15}.\frac{21}{32}\)=\(1.\frac{16}{15}.\frac{21}{32}=\frac{7}{5.2}=\frac{7}{10}\)
Phần b T2
c,\(\frac{7}{4}.\frac{11}{21}+\frac{11}{21}.\frac{5}{4}=\frac{11}{21}.\left(\frac{7}{4}+\frac{5}{4}\right)\)=\(\frac{11}{21}.3=\frac{11}{7}\)
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9/1-1/90-1/72-1/56-1/42-1/30-1/20-1/12-1/6-1/2=0/4
Giải :
ta có
\(\frac{9}{10}-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{90}\right)\)
=\(\frac{9}{10}-\left(\frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{9\times10}\right)\)
=\(\frac{9}{10}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
=\(\frac{9}{10}-\left[1+\left(\frac{-1}{2}+\frac{1}{2}\right)+\left(\frac{-1}{3}+\frac{1}{3}\right)+...+\left(\frac{-1}{9}+\frac{1}{9}\right)-\frac{1}{10}\right]\)
=\(\frac{9}{10}-\left(1-\frac{1}{10}\right)\)
=\(\frac{9}{10}-1+\frac{1}{10}=0\) (Mong online math ks cho mình nhé)
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\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{56}=\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}...\frac{1}{7x8}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{7}-\frac{1}{8}\)\(-\frac{1}{8}=\frac{1}{2}-\frac{1}{8}=\frac{3}{8}\)
b,
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Câu 1:\(\frac{12}{23}\times\frac{9}{28}+\frac{11}{23}\times\frac{9}{28}\)
\(=\frac{9}{28}\times\left(\frac{12}{23}+\frac{11}{23}\right)\)
\(=\frac{9}{28}\times\frac{23}{23}\)
\(=\frac{9}{28}\times1\)
\(=\frac{9}{28}\)
Câu 2:
\(3+\frac{1}{9}+\frac{3}{9}+2+\frac{5}{9}+3+2\)
\(=\left(3+2+2+3\right)+\left(\frac{1}{9}+\frac{3}{9}+\frac{5}{9}\right)\)
\(=10+\frac{9}{9}\)
\(=10+1\)
\(=11\)
\(\frac{12}{23}x\frac{9}{28}+\frac{11}{23}x\frac{9}{28}=\)\(\left(\frac{12}{23}+\frac{11}{23}\right)x\frac{9}{28}\)\(=\) \(1x\frac{9}{28}=\frac{9}{28}\)
\(3+\frac{1}{9}+\frac{3}{9}+2+\frac{5}{9}+3+2\)\(=\) \(\left(\frac{1}{9}+\frac{3}{9}+\frac{5}{9}\right)+\left(3+2+3+2\right)\)\(=\) \(1+3+2+3+2=\)\(11\)
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a) $\frac{2}{3} = \frac{{2 \times 6}}{{3 \times 6}} = \frac{{12}}{{18}}$
Ta có $\frac{{12}}{{18}} > \frac{{11}}{{18}}$ nên $\frac{2}{3} > \frac{{11}}{{18}}$
b) $\frac{{36}}{{63}} = \frac{{36:9}}{{63:9}} = \frac{4}{7}$
Ta có $\frac{4}{7} < \frac{5}{7}$ nên $\frac{{36}}{{63}}$ < $\frac{5}{7}$
c)
$\frac{{55}}{{110}} = \frac{{55:55}}{{110:55}} = \frac{1}{2}$ ; $\frac{4}{8} = \frac{1}{2}$
Vậy $\frac{{55}}{{110}}$ = $\frac{4}{8}$
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Số ngày thứ 2 số trẻ em tiêm chủng nhiều hơn ngày thứ nhất là :
11 / 23 - 8/23 = 3/23 ( phần )
Đáp số : 3/23 phần
2 ) tính
\(\frac{3}{4}-\frac{2}{3}=\frac{9}{12}-\frac{8}{12}=\frac{1}{12}\)
\(\frac{7}{5}-\frac{3}{7}=\frac{49}{35}-\frac{15}{35}=\frac{34}{35}\)
\(\frac{4}{3}-\frac{3}{5}=\frac{20}{15}-\frac{9}{15}=\frac{11}{15}\)
\(\frac{11}{3}-\frac{2}{5}=\frac{55}{15}-\frac{6}{15}=\frac{49}{15}\)
3 tính
\(\frac{8}{9}-\frac{1}{3}=\frac{24}{27}-\frac{9}{27}=\frac{15}{27}\)
\(\frac{4}{5}-\frac{6}{25}=\frac{20}{25}-\frac{6}{25}=\frac{14}{25}\)
Hãy tk và kp với mk nha
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\(=\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{11.12}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{11}-\frac{1}{12}\)
\(=\frac{1}{2}-\frac{1}{12}\)
\(=\frac{5}{12}\)
bn sẽ tinh theo kieeuranhaan 2 nha xin lỗi mik làm bi này rùi nhưng mik quên mik có sacks xem lại
a/\(\frac{64}{3}\)
b/\(\frac{449}{63}\)
c/\(\frac{13}{55}\)
d/\(\frac{399}{50}\)
e/\(\frac{9073}{253}\)
Học Tốt!!