a)

\(\begin{array}{l}\frac{2}{9}:x + \frac{5}{6} = 0,5\\\frac{2}{9}:x = \frac{1}{2} - \frac{5}{6}\\\frac{2}{9}:x = \frac{3}{6} - \frac{5}{6}\\\frac{2}{9}:x = \frac{{ - 2}}{6}\\x = \frac{2}{9}:\frac{{ - 2}}{6}\\x = \frac{2}{9}.\frac{{ - 6}}{2}\\x = \frac{{ - 2}}{3}\end{array}\)                        

Vậy \(x = \frac{{ - 2}}{3}\).

b)

\(\begin{array}{l}\frac{3}{4} - \left( {x - \frac{2}{3}} \right) = 1\frac{1}{3}\\x - \frac{2}{3} = \frac{3}{4} - 1\frac{1}{3}\\x - \frac{2}{3} = \frac{3}{4} - \frac{4}{3}\\x - \frac{2}{3} = \frac{9}{{12}} - \frac{{16}}{{12}}\\x - \frac{2}{3} = \frac{{ - 7}}{{12}}\\x = \frac{{ - 7}}{{12}} + \frac{2}{3}\\x = \frac{{ - 7}}{{12}} + \frac{8}{{12}}\\x = \frac{1}{12}\end{array}\)

Vậy\(x = \frac{1}{12}\).

c)

\(\begin{array}{l}1\frac{1}{4}:\left( {x - \frac{2}{3}} \right) = 0,75\\\frac{5}{4}:\left( {x - \frac{2}{3}} \right) = \frac{3}{4}\\x - \frac{2}{3} = \frac{5}{4}:\frac{3}{4}\\x - \frac{2}{3} = \frac{5}{4}.\frac{4}{3}\\x - \frac{2}{3} = \frac{5}{3}\\x = \frac{5}{3} + \frac{2}{3}\\x = \frac{7}{3}\end{array}\)               

Vậy \(x = \frac{7}{3}\).

d)

\(\begin{array}{l}\left( { - \frac{5}{6}x + \frac{5}{4}} \right):\frac{3}{2} = \frac{4}{3}\\ - \frac{5}{6}x + \frac{5}{4} = \frac{4}{3}.\frac{3}{2}\\ - \frac{5}{6}x + \frac{5}{4} = 2\\ - \frac{5}{6}x = 2 - \frac{5}{4}\\ - \frac{5}{6}x = \frac{8}{4} - \frac{5}{4}\\ - \frac{5}{6}x = \frac{3}{4}\\x = \frac{3}{4}:\left( { - \frac{5}{6}} \right)\\x = \frac{3}{4}.\frac{{ - 6}}{5}\\x = \frac{{ - 9}}{{10}}\end{array}\)

Vậy \(x = \frac{{ - 9}}{{10}}\).

a)

\(\begin{array}{l}x.\frac{{14}}{{27}} = \frac{{ - 7}}{9}\\x = \frac{{ - 7}}{9}:\frac{{14}}{{27}}\\x = \frac{{ - 7}}{9}.\frac{{27}}{{14}}\\x = \frac{{ - 3}}{2}\end{array}\)                

Vậy \(x = \frac{{ - 3}}{2}\).

b)

\(\begin{array}{l}\left( {\frac{{ - 5}}{9}} \right):x = \frac{2}{3}\\x = \left( {\frac{{ - 5}}{9}} \right):\frac{2}{3}\\x = \left( {\frac{{ - 5}}{9}} \right).\frac{3}{2}\\x = \frac{{ - 5}}{6}\end{array}\)

Vậy \(x = \frac{{ - 5}}{6}\).

c)

\(\begin{array}{l}\frac{2}{5}:x = \frac{1}{{16}}:0,125\\\frac{2}{5}:x = \frac{1}{{16}}:\frac{1}{8}\\\frac{2}{5}:x = \frac{1}{{16}}.8\\\frac{2}{5}:x = \frac{1}{2}\\x = \frac{2}{5}:\frac{1}{2}\\x = \frac{2}{5}.2\\x = \frac{4}{5}\end{array}\)      

Vậy \(x = \frac{4}{5}\)

d)

\(\begin{array}{l} - \frac{5}{{12}}x = \frac{2}{3} - \frac{1}{2}\\ - \frac{5}{{12}}x = \frac{4}{6} - \frac{3}{6}\\ - \frac{5}{{12}}x = \frac{1}{6}\\x = \frac{1}{6}:\left( { - \frac{5}{{12}}} \right)\\x = \frac{1}{6}.\frac{{ - 12}}{5}\\x = \frac{{ - 2}}{5}\end{array}\)

Vậy \(x = \frac{{ - 2}}{5}\).

Chú ý: Khi trình bày lời giải bài tìm x, sau khi tính xong, ta phải kết luận.

a)

\(\begin{array}{l}x - \left( { - \frac{7}{9}} \right) =  - \frac{5}{6}\\x + \frac{7}{9} =  - \frac{5}{6}\\x =  - \frac{5}{6} - \frac{7}{9}\\x =  - \frac{{15}}{{18}} - \frac{{14}}{{18}}\\x = \frac{{ - 29}}{{18}}\end{array}\)

Vậy \(x = \frac{{ - 29}}{{18}}\).

b)

\(\begin{array}{l}\frac{{15}}{{ - 4}} - x = 0,3\\x = \frac{{15}}{{ - 4}} - 0,3\\x =  - 3,75 - 0,3\\x =  - 4,05\end{array}\)

Vậy \(x =  - 4,05\).

B=\(\frac{5}{9}\)x{\(\frac{7}{13}\)+\(\frac{9}{13}\)-\(\frac{3}{13}\)}

B=\(\frac{5}{9}\)x 1

B=\(\frac{5}{9}\)

Vậy B=\(\frac{5}{9}\)

Chúc bạn hok tốt

bằng 5/9 , chúc bạn học tốt !

B = \(\frac{5}{9}.\frac{7}{13}+\frac{5}{9}.\frac{9}{13}-\frac{5}{9}.\frac{3}{13}\)

Ta thấy có phân số \(\frac{5}{9}\) là thừa số chung

\(\Leftrightarrow\frac{5}{9}\left(\frac{7}{13}+\frac{9}{13}-\frac{3}{13}\right)=\frac{5}{9}\left(\frac{7+9+3}{13}\right)\)

\(=\frac{5}{9}.\frac{19}{13}\left(\frac{5.19}{9.13}\right)\)

\(=\frac{95}{117}\)

a) \(x-\frac{5}{7}=\frac{1}{9}\Rightarrow x=\frac{1}{9}+\frac{5}{7}\Rightarrow x=\frac{52}{63}\)

b) \(\frac{-3}{7}-x=\frac{4}{5}+\frac{-2}{3}\Rightarrow\frac{-3}{7}-x=\frac{2}{15}\Rightarrow x=\frac{-3}{7}-\frac{2}{15}\Rightarrow x=\frac{-59}{105}\)

c) \(x-\frac{1}{5}=\frac{2}{7}.\frac{-11}{5}\Rightarrow x-\frac{1}{5}=\frac{-22}{35}\Rightarrow x=\frac{-22}{35}+\frac{1}{5}\Rightarrow x=\frac{-3}{7}\)

d) \(\frac{x}{182}=\frac{-6}{14}.\frac{35}{91}\Rightarrow\frac{x}{182}=\frac{-15}{91}\Rightarrow x=\frac{\left(-15\right).182}{91}\Rightarrow x=-30\)

a) Ta có : \(\frac{x+5}{5}+\frac{x+5}{7}+\frac{x+5}{9}=\frac{x+5}{11}+\frac{x+5}{13}\)

\(\Rightarrow\frac{x+5}{5}+\frac{x+5}{7}+\frac{x+5}{9}-\left(\frac{x+5}{11}+\frac{x+5}{13}\right)=0\)

\(\Rightarrow\frac{x+5}{5}+\frac{x+5}{7}+\frac{x+5}{9}-\frac{x+5}{11}-\frac{x+5}{13}=0\)

\(\Rightarrow\left(x+5\right)\left(\frac{1}{5}+\frac{1}{7}+\frac{1}{9}-\frac{1}{11}-\frac{1}{13}\right)=0\)

Do \(\frac{1}{5}+\frac{1}{7}+\frac{1}{9}-\frac{1}{11}-\frac{1}{13}\ne0\)

\(\Rightarrow x+5=0\Rightarrow x=-5\)

Vậy x = -5

b) Ta có : \(\frac{x+2}{100}+\frac{x+3}{99}+\frac{x+4}{98}=\frac{x+5}{97}+\frac{x+6}{96}+\frac{x+7}{95}\)

\(\Rightarrow\frac{x+2}{100}+\frac{x+3}{99}+\frac{x+4}{98}+3=\frac{x+5}{97}+\frac{x+6}{96}+\frac{x+7}{95}+3\)

\(\Rightarrow\frac{x+2}{100}+1+\frac{x+3}{99}+1+\frac{x+4}{98}+1=\frac{x+5}{97}+1+\frac{x+6}{96}+1+\frac{x+7}{95}+1\)

\(\Rightarrow\frac{x+102}{100}+\frac{x+102}{99}+\frac{x+102}{98}=\frac{x+102}{97}+\frac{x+102}{96}+\frac{x+102}{95}\)

\(\Rightarrow\frac{x+102}{100}+\frac{x+102}{99}+\frac{x+102}{98}-\left(\frac{x+102}{97}+\frac{x+102}{96}+\frac{x+102}{95}\right)=0\)

\(\Rightarrow\frac{x+102}{100}+\frac{x+102}{99}+\frac{x+102}{98}-\frac{x+102}{97}-\frac{x+102}{96}-\frac{x+102}{95}\)

\(\Rightarrow\left(x+102\right)\left(\frac{1}{100}+\frac{1}{99}+\frac{1}{98}-\frac{1}{97}-\frac{1}{96}-\frac{1}{95}\right)=0\)

Do \(\frac{1}{100}+\frac{1}{99}+\frac{1}{98}-\frac{1}{97}-\frac{1}{96}-\frac{1}{95}\ne0\)

\(\Rightarrow x+102=0\Rightarrow x=-102\)

Vậy x = -102

c) Ta có : (x + 2) - (x + 3) = x + 2 - x - 3

                                      = x - x + 2 - 3

                                      = -1

mà (x + 2) - (x + 3) > 0 => không tồn tại x sao cho (x + 2) - (x + 3) > 0

d) Ta có : \(\left(x-5\right)\left(x+\frac{7}{3}\right)\ge0\)

\(\Rightarrow\orbr{\begin{cases}x\ge5\\x\ge\frac{-7}{3}\end{cases}}\)

\(\Rightarrow x\ge\frac{-7}{3}\)

Vậy \(x\ge\frac{-7}{3}\)

\(a,12,3\cdot4,7+5,3\cdot12,3-1,4\cdot7,5+7,5\cdot5,4\)

\(=12,3\cdot\left(4,7+5,3\right)-7,5\cdot\left(1,4+5,4\right)\)

\(=12,3\cdot10-7,5\cdot6,8\)

\(=123-51\)

\(=72.\)

\(b,3\cdot x-1=\frac{3}{4}:\frac{15}{16}\)

\(\Leftrightarrow3\cdot x-1=\frac{3}{4}\cdot\frac{16}{15}\)

\(\Leftrightarrow3\cdot x-1=\frac{4}{5}\)

\(\Leftrightarrow3\cdot x=\frac{4}{5}+1\)

\(\Leftrightarrow3\cdot x=\frac{9}{5}\)

\(\Leftrightarrow x=\frac{9}{5}\div3\)

\(\Leftrightarrow x=\frac{9}{5}\cdot\frac{1}{3}\)

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

a) \(12,3\times4,7+5,3\times12,3-1,4\times7,5+7,5\times5,4\)

\(=12,3\times\left(4,7+5,3\right)+7,5\times\left(5,4-1,4\right)\)

\(=12,3\times10+7,5\times4\)

\(=123+30\)

\(=153\)

b) \(3.x-1=\frac{3}{4}:\frac{15}{16}\)

\(\Rightarrow3.x-1=\frac{4}{5}\)

\(\Rightarrow3.x=\frac{9}{5}\Rightarrow x=\frac{3}{5}\)

c,  Trong phép nhân 1 x 3 x 5 x...x 19 có chứa thừa số 5 nên tích là một số chia hết cho 5

=>chữ số tận cùng của tích là 0 hoặc 5

Vì các thừa số là số lẻ nên tích là số lẻ nên chữ số tận cùng của tích là 5

Trong phép nhân 1 x 2 x 3 x...x 8 x 9 có chứa tích 2 x 5 = 10 nên phép nhân 1 x 2 x 3 x...x 9 có chữ số tận cùng là 0

=> 1 x 3 x 5 x 7 x 9 x ....x 13 x 15 x 17 x 19 + 1 x 2 x 3 x ....x 8 x 9 = ...5 + ....0 = ....5

\(a,2\frac{x}{7}=\frac{75}{35}\)

\(\frac{75}{35}=2\frac{5}{35}=2\frac{1}{7}\)

\(\Rightarrow x=1\)

\(b,47:4=...\left(+3\right)\)

     \(\Rightarrow\left(47-3\right):4=11\)

      \(\Rightarrow x=11\)

\(c,\frac{112}{5}=\frac{336}{15}=22\left(+6\right)\)

     \(\Rightarrow x=22\frac{x=6}{15}\)

\(d,7,5x=\frac{75x}{100}:\left(9-6\frac{13}{21}\right)=2\frac{13}{25}\)

     \(\frac{75x}{100}:\left(9-\frac{139}{21}\right)=\frac{63}{25}\)

       \(\frac{75x}{100}:\frac{50}{21}=\frac{63}{25}\)

           \(\Rightarrow\frac{63^3}{25^{\left(1\right)}}\cdot\frac{50^{\left(2\right)}}{21^1}=6=\frac{6}{1}\)

             \(\Rightarrow\frac{6}{1}=\frac{600}{100}\)

Con d sai đề