\(a,\frac{0,75-0,6+\frac{3}{7}+\frac{3}{13}}{2,75-2,2+\frac{11}{7}+\frac{11}{13}}\)

\(=\frac{\frac{3}{4}-\frac{3}{5}+\frac{3}{7}+\frac{3}{13}}{\frac{11}{4}-\frac{11}{5}+\frac{11}{7}+\frac{11}{13}}\)

\(=\frac{3\left[\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right]}{11\left[\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right]}=\frac{3}{11}\)

Câu b tương tự

#)Giải :

b)\(\frac{5}{9}:\left(\frac{1}{3}+\frac{1}{4}\right)+\frac{5}{9}:\left(\frac{1}{9}+\frac{2}{3}\right)\)

\(=\frac{5}{9}:\frac{1}{3}+\frac{5}{9}:\frac{1}{4}+\frac{5}{9}:\frac{1}{9}+\frac{5}{9}:\frac{2}{3}\)

\(=\frac{5}{9}:\left(\frac{1}{3}+\frac{1}{4}+\frac{1}{9}+\frac{2}{3}\right)\)

\(=\frac{5}{9}:\frac{49}{36}\)

\(=\frac{20}{49}\)

\(a,\frac{21}{36}.\frac{5}{2}-\frac{7}{12}.\frac{2}{7}+\left(2018-2019\right)^0\)

=\(\frac{7}{12}.\frac{5}{2}-\frac{7}{12}.\frac{2}{7}+\left(-1\right)\)

= \(\frac{7}{12}.\left(\frac{5}{2}+\frac{2}{7}\right)+\left(-1\right)\)

=\(\frac{7}{12}.\frac{39}{14}+\left(-1\right)\)

=\(\frac{13}{8}+\left(-1\right)\)

= \(\frac{5}{8}\)

\(b,-12\frac{1}{3}-\frac{5}{7}+7\frac{1}{3}+1\frac{5}{7}+1^{2019}\)

=\(-\frac{37}{3}+\frac{-5}{7}+\frac{22}{3}+\frac{12}{7}+1\)

=\(\left(\frac{-37+22}{3}\right)+\left(\frac{-5+12}{7}\right)+=1\)

= \(-5+1+1\)

=\(-3\)

Bằng 86/41

= \(\frac{83}{41}\)

hello

#)Giải :

a)\(\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{24.25}\)

\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{24}-\frac{1}{25}\)

\(=\frac{1}{5}-\frac{1}{25}\)

\(=\frac{4}{25}\)

b)\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)

\(=1-\frac{1}{101}\)

\(=\frac{100}{101}\)

a) \(\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{24.25}\)

= \(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{24}-\frac{1}{25}\)

= \(\frac{1}{5}-\frac{1}{25}\)

= \(\frac{4}{25}\)

b) \(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\)

= \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\)

= \(1-\frac{1}{101}\)

= \(\frac{100}{101}\)

c) \(5\frac{2}{7}.\frac{8}{11}+5\frac{2}{7}.\frac{5}{11}-5\frac{2}{7}.\frac{2}{11}\)

= \(5\frac{2}{7}.\left(\frac{8}{11}+\frac{5}{11}-\frac{2}{11}\right)\)

= \(5\frac{2}{7}\)

= \(\frac{37}{7}\)

a)\(\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{24.25}\)

\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{24}-\frac{1}{25}\)

\(=\frac{1}{5}-\frac{1}{25}\)

\(=\frac{5}{25}-\frac{1}{25}=\frac{4}{25}\)

\(\frac{ }{ }\)

\(1)\frac{1}{2}x-\frac{3}{5}=\frac{-4}{5}\)

\(\Rightarrow\frac{1}{2}x=\frac{-4}{5}+\frac{3}{5}\)

\(\Rightarrow\frac{1}{2}x=\frac{-1}{5}\)

\(\Rightarrow x=\frac{-1}{5}:\frac{1}{2}=\frac{-1}{5}\cdot\frac{2}{1}=\frac{-2}{5}\)

\(\Leftrightarrow x=\frac{-2}{5}\)

\(2)3\frac{1}{5}-2\frac{1}{3}x=-1\frac{3}{5}+1\frac{7}{10}\)

\(\Rightarrow\frac{16}{5}-\frac{7}{3}x=-\frac{8}{5}+\frac{17}{10}\)

\(\Rightarrow\frac{7}{3}x=\frac{16}{5}-\frac{-8}{5}+\frac{17}{10}\)

\(\Rightarrow\frac{7}{3}x=\frac{16}{5}+\frac{8}{5}+\frac{17}{10}\)

\(\Rightarrow\frac{7}{3}x=\frac{24}{5}+\frac{17}{10}\)

\(\Rightarrow\frac{7}{3}x=\frac{48}{10}+\frac{17}{10}\)

Đến đây tìm được rồi nhé

3,4, áp dụng bài 1,2 rồi làm :v

1. 1/2x - 3/5 = -4/5

=> 1/2x = -4/5 + 3/5

=> 1/2x = -1/5

=> x = -1/5 : 1/2

=> x = -2/5

i don't know i mới học lớp 5

bn eie mik lớp 6 nha bn

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