(7x6=5+2+6x7) =

ta gọi \(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{90}\)là A

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)

\(\Leftrightarrow1.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{9}-\frac{1}{10}\right)\)

\(\Rightarrow A=1-\frac{1}{10}=\frac{9}{10}\)

ta gọi B là biểu thức thứ2

\(B=\frac{2.2}{3}\times\frac{3.3}{2.4}\times\frac{4.4}{3.5}\times...\times\frac{10.10}{9.11}\)

\(\Rightarrow\)2 x \(\frac{10}{11}\)\(=\frac{20}{11}\)

\(\Rightarrow\)\(x+\frac{9}{10}=\frac{20}{11}+\frac{9}{110}\)

\(\Rightarrow x=1\)

mk nghĩ vậy bạn ạ, mk mong nó đúng

a) $\frac{1}{6} \times \frac{2}{3} = \frac{{1 \times 2}}{{6 \times 3}} = \frac{2}{{18}} = \frac{1}{9}$                            

b) $\frac{6}{5} \times \frac{3}{8} = \frac{{6 \times 3}}{{5 \times 8}} = \frac{{18}}{{40}} = \frac{9}{{20}}$

c) $\frac{4}{3} \times \frac{8}{9} = \frac{{4 \times 8}}{{3 \times 9}} = \frac{{32}}{{27}}$        

d) $\frac{5}{{12}} \times \frac{{12}}{5} = \frac{{5 \times 12}}{{12 \times 5}} = \frac{{60}}{{60}} = 1$

\(A=2\frac{1}{2}\)

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

\(\Leftrightarrow\frac{3}{5}.x=\frac{4}{5}+\frac{1}{5}\)

\(\Leftrightarrow\frac{3}{5}.x=1\)

\(\Leftrightarrow x=1:\frac{3}{5}\)

\(\Leftrightarrow x=\frac{5}{3}\)

Vậy : \(x=\frac{5}{3}\)

b) \(\frac{4}{7}+\frac{5}{7}:x=1\)

\(\Leftrightarrow\frac{5}{7}:x=1-\frac{4}{7}\)

\(\Leftrightarrow\frac{5}{7}:x=\frac{3}{7}\)

\(\Leftrightarrow x=\frac{5}{7}:\frac{3}{7}\)

\(\Leftrightarrow x=\frac{5}{3}\)

Vậy : \(x=\frac{5}{3}\)

c) \(-\frac{12}{7}.\left(\frac{3}{4}-x\right).\frac{1}{4}=-1\)

\(\Leftrightarrow\frac{-12.1}{7.4}.\left(\frac{3}{4}-x\right)=-1\)

\(\Leftrightarrow-\frac{3}{7}.\left(\frac{3}{4}-x\right)=-1\)

\(\Leftrightarrow\frac{3}{4}-x=-1:\left(-\frac{3}{7}\right)\)

\(\Leftrightarrow\frac{3}{4}-x=\frac{7}{3}\)

\(\Leftrightarrow x=\frac{3}{4}-\frac{7}{3}=-\frac{19}{12}\)

Vậy : \(x=-\frac{19}{12}\)

d) \(x:\frac{17}{8}=-\frac{2}{5}.-\frac{9}{17}+3\)

\(\Leftrightarrow x:\frac{17}{8}=\frac{273}{85}\)

\(\Leftrightarrow x=\frac{273}{85}.\frac{17}{8}\)

\(\Leftrightarrow x=\frac{273}{40}\)

Vậy : \(x=\frac{273}{40}\)

\(\)

tính giá trị biểu thức chứ còn cái gì nữa

a, \(A=\frac{22}{27}\)

b,\(B=\frac{1}{57}\)

C,\(C=\frac{1}{50}\)

d, \(D=0\)

\(\frac{6:\frac{3}{5}-1\frac{1}{6}\times\frac{6}{7}}{4\frac{1}{5}\times\frac{10}{11}+5\frac{2}{11}}\)

\(=\frac{\frac{6}{1}:\frac{3}{5}-\frac{7}{6}\times\frac{6}{7}}{\frac{21}{5}\times\frac{10}{11}\times\frac{57}{11}}\)

\(=\frac{\frac{6}{1}\times\frac{5}{3}-1}{\frac{210}{55}+\frac{57}{11}}\)

\(=\frac{\frac{30}{3}-1}{\frac{42}{11}+\frac{57}{11}}\)

\(=\frac{10-1}{\frac{99}{11}}\)

\(=\frac{9}{9}\)

\(=1\)

\(6:\frac{3}{5}-1\frac{1}{6}\)X \(\frac{6}{7}\)                                                     \(4\frac{1}{5}\)X \(\frac{10}{11}+5\frac{2}{11}\)

\(=\frac{33}{5}-\frac{7}{6}\)X \(\frac{6}{7}\)                                             \(=\)      \(\frac{21}{5}\)X  \(\frac{10}{11}+\frac{57}{11}\) 

\(=\frac{33}{5}-1\)                                                              \(=\frac{42}{11}+\frac{57}{11}\)

\(=\frac{28}{5}\)                                                                       \(=\frac{99}{11}=9\)