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a) \(\frac{51}{3}-\frac{22}{3}=\frac{51-22}{3}=\frac{29}{3}\)
b) \(\frac{5}{12}+\frac{5}{6}-\frac{3}{4}=\frac{5}{12}+\frac{10}{12}-\frac{9}{12}=\frac{5+10-9}{12}=\frac{6}{12}=\frac{1}{2}\)
c) \(1-\left(\frac{1}{5}+\frac{1}{2}\right)=\frac{10}{10}-\frac{2}{10}-\frac{5}{10}=\frac{10-5-2}{10}=\frac{3}{10}\)
d) \(\frac{111}{4}-\left(\frac{25}{7}+\frac{51}{4}\right)=\frac{777}{28}-\frac{60}{28}-\frac{357}{28}=\frac{360}{28}=\frac{90}{7}\)
e) \(\left(\frac{85}{11}+\frac{35}{7}\right)-\frac{35}{11}=\left(\frac{85}{11}-\frac{35}{11}\right)+\frac{35}{7}=\frac{50}{11}-\frac{35}{7}=\frac{350}{77}-\frac{385}{77}=-\frac{35}{77}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
A=(6 : 35−116 x 67) : (415 x 1011+5211)A=(6 : 35−116 x 67) : (415 x 1011+5211)
A=(6 : 35−76 x 67) : (215 x 1011+5711)A=(6 : 35−76 x 67) : (215 x 1011+5711)
A=(10−1) : (4211+5711)A=(10−1) : (4211+5711)
A=9 : 9A=9 : 9
A=1
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Bài giải
\(\frac{2}{3}+\frac{3}{4}+\frac{4}{5}=\frac{40}{60}+\frac{45}{60}+\frac{48}{60}=\frac{133}{60}\)
\(\frac{8}{5}+\frac{7}{6}+\frac{10}{9}+\frac{1}{2}=\frac{144}{90}+\frac{105}{90}+\frac{100}{90}+\frac{45}{90}=\frac{394}{90}\)
\(\frac{15}{17}-\frac{11}{13}+\frac{3}{26}=\frac{390}{442}+\frac{374}{442}+\frac{51}{442}=\frac{815}{442}\)
\(\frac{9}{12}\text{ x }\frac{4}{3}\text{ : }\frac{8}{5}=\frac{9}{12}\text{ x }\frac{4}{3}\text{ x }\frac{5}{8}=\frac{9\text{ x }4\text{ x }5}{12\text{ x }3\text{ x }8}=\frac{5}{8}\)
\(\frac{4}{5}\text{ x }\frac{15}{8}\text{ : }\frac{5}{7}=\frac{4}{5}\text{ x }\frac{15}{8}\text{ x }\frac{7}{5}=\frac{4\text{ x }15\text{ x }7}{5\text{ x }8\text{ x }5}=\frac{21}{10}\)
\(\frac{2}{3}+\frac{3}{4}+\frac{4}{5}=\frac{40}{60}+\frac{45}{60}+\frac{48}{60}=\frac{133}{60}\)
\(\frac{8}{5}+\frac{7}{6}+\frac{10}{9}+\frac{1}{2}=\frac{144}{90}+\frac{105}{90}+\frac{100}{90}+\frac{45}{90}=\frac{197}{45}\)
\(\frac{15}{17}-\frac{11}{13}+\frac{1}{26}=\frac{390}{442}+\frac{374}{442}+\frac{51}{442}=\frac{815}{442}\)
\(\frac{9}{12}\times\frac{4}{3}:\frac{8}{5}=1:\frac{8}{5}=\frac{5}{8}\)
\(\frac{4}{5}\times\frac{15}{8}:\frac{5}{7}=\frac{3}{2}:\frac{5}{7}=\frac{21}{10}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\frac{1.3+3.5+5.7+7.9}{3.6+9.10+15.14+21.18}\)
= \(\frac{1.3+3.5+5.7+7.9}{1.3.2.3+3.5.2.3+5.7.2.3+7.9.2.3}\)
= \(\frac{1.3+3.5+5.7+7.9}{1.3.6+3.5.6+5.7.6+7.9.6}\)
= \(\frac{1.3+3.5+5.7+7.9}{6.\left(1.3+3.5+5.7+7.9\right)}=\frac{1}{6}\)
Dấu "." là dấu nhân cấp 2
b) \(\frac{1.2+2.3+3.4+4.5}{3.6+6.9+9.12+12.15}\)
= \(\frac{1.2+2.3+3.4+4.5}{1.2.3.3+2.3.3.3+3.4.3.3+4.5.3.3}\)
= \(\frac{1.2+2.3+3.4+4.5}{1.2.9+2.3.9+3.4.9+4.5.9}\)
= \(\frac{1.2+2.3+3.4+4.5}{9.\left(1.2+2.3+3.4+4.5\right)}=\frac{1}{9}\)
Dấu "." là dấu nhân cấp 2
c) \(\frac{0,3+\frac{3}{7}+\frac{3}{11}}{0,4+\frac{4}{7}+\frac{4}{11}}\)= \(\frac{\frac{3}{10}+\frac{3}{7}+\frac{3}{11}}{\frac{4}{10}+\frac{4}{7}+\frac{4}{11}}\)= \(\frac{3.\left(\frac{1}{10}+\frac{1}{7}+\frac{1}{11}\right)}{4.\left(\frac{1}{10}+\frac{1}{7}+\frac{1}{11}\right)}=\frac{3}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)\(\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{23.27}=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{23}-\frac{1}{27}=\frac{1}{3}-\frac{1}{27}=\frac{8}{27}\)
b)\(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{6.7}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{6}-\frac{1}{7}=\frac{1}{2}-\frac{1}{7}=\frac{5}{14}\)
c)\(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{11.13}+\frac{2}{1.2}+\frac{2}{2.3}+...+\frac{2}{9.10}=\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}\right)+2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=\frac{1}{3}-\frac{1}{13}+2\left(1-\frac{1}{10}\right)=\frac{10}{39}+\frac{9}{5}=\frac{401}{195}\)
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\(\frac{1.3.5+2.6.10+4.12.20+7.21.35}{1.5.7+2.10.14+4.20.28+7.35.49}\)
\(=\frac{1.3.5\left(1+2+4+7\right)}{1.5.7\left(1+2+7+7\right)}=\frac{1.3.5}{1.5.7}=\frac{15}{35}=\frac{3}{7}\)
\(\frac{1\cdot3\cdot5+2\cdot6\cdot10+4\cdot12\cdot20+7\cdot21\cdot35}{1\cdot5\cdot7+2\cdot10\cdot14+4\cdot20\cdot28+7\cdot35\cdot49}\)
\(=\)\(\frac{1\cdot3\cdot5\cdot\left(1+2+4+7\right)}{1\cdot5\cdot7\cdot\left(1+2+7+7\right)}\)
\(=\frac{1\cdot3\cdot5}{1\cdot5\cdot7}\)\(=\frac{15}{35}=\frac{3}{7}\)
bn vội quá viết nhầm lun kìa
hj hj chúc bn làm bài tốt nha
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\(1-2-3+4+5-6-7+8+...+97-98-99+100+101\)
\(=\left(1-2-3+4\right)+\left(5-6-7+8\right)+...+\left(97-98-99+100\right)+101\)
\(=0+0+...+0+101\)
\(=101\)
1/6 + 3/6 + 1 + 2/6 = 2/3 +1+2/6 = 5/3 +2/6 = 12/6 = 2 15/7 + 2/7 = 17/7
1/7+ 3/5 = 26/35 5/6 + 2/9 = 19/18
2/9 + 3/6 = 13/16 2/10 + 3/5 = 8/10
8/11 - 5/11 = 3/11 15/35 - 7/35 = 8/35
3/5 - 1/3 = 4/15 4 - 7/3 = 5/3
tk mình nha mn
THank nhìu lém >.<