a)

\(\begin{array}{l}x + \left( { - \frac{1}{5}} \right) = \frac{{ - 4}}{{15}}\\x = \frac{{ - 4}}{{15}} + \frac{1}{5}\\x = \frac{{ - 4}}{{15}} + \frac{3}{{15}}\\x = \frac{{ - 1}}{{15}}\end{array}\)                 

Vậy \(x = \frac{{ - 1}}{{15}}\).

b)

\(\begin{array}{l}3,7 - x = \frac{7}{{10}}\\x = 3,7 - \frac{7}{{10}}\\x = \frac{{37}}{{10}} - \frac{7}{{10}}\\x=\frac{30}{10}\\x = 3\end{array}\)

Vậy \(x = 3\).

c)

\(\begin{array}{l}x.\frac{3}{2} = 2,4\\x.\frac{3}{2} = \frac{{12}}{5}\\x = \frac{{12}}{5}:\frac{3}{2}\\x = \frac{{12}}{5}.\frac{2}{3}\\x = \frac{8}{5}\end{array}\)

Vậy \(x = \frac{8}{5}\)       

d)

\(\begin{array}{l}3,2:x =  - \frac{6}{{11}}\\\frac{{16}}{5}:x =  - \frac{6}{{11}}\\x = \frac{{16}}{5}:\left( { - \frac{6}{{11}}} \right)\\x = \frac{{16}}{5}.\frac{{ - 11}}{6}\\x = \frac{{ - 88}}{{15}}\end{array}\)

Vậy \(x = \frac{{ - 88}}{{15}}\).

giúp mk vs ạ!!

ở câu b mk sót ở cuối là =0 nha

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 \(x = \frac{{ - 9}}{{10}}\).

xin lỗi bn mik mới học lớp 6 thôi

a/ đề \(=\frac{\sqrt{x}}{\sqrt{x}-5}-\frac{10\sqrt{x}}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}-\frac{5}{\sqrt{x}+5}\)

       \(=\frac{\sqrt{x}\left(\sqrt{x}+5\right)-10\sqrt{x}-5\left(\sqrt{x}-5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\)

       \(=\frac{x+5\sqrt{x}-10\sqrt{x}-5\sqrt{x}+25}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}=\frac{x-10\sqrt{x}+25}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\)

          \(=\frac{\left(\sqrt{x}-5\right)^2}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}=\frac{\sqrt{x}-5}{\sqrt{x}+5}\)

b/ đề \(=\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{3}{\sqrt{x}+1}-\frac{6\sqrt{x}-4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)

            \(=\frac{\sqrt{x}\left(\sqrt{x}+1\right)+3\left(\sqrt{x}-1\right)-\left(6\sqrt{x}-4\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)

            \(=\frac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\frac{x-2\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)

              \(=\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)

c/ đề \(=\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}+1}{\sqrt{x}-3}-\frac{3-11\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

          \(=\frac{2\sqrt{x}\left(\sqrt{x}-3\right)+\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)-\left(3-11\sqrt{x}\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

             \(=\frac{2x-6\sqrt{x}+x+4\sqrt{x}+3-3+11\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\frac{3x+9\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

               \(=\frac{3\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\frac{3\sqrt{x}}{\sqrt{x}-3}\)

Rút gọn nhé

      \(\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+5}{324}+\frac{x+349}{5}=0\)

\(\Leftrightarrow\)\(\frac{x+2}{327}+1+\frac{x+3}{326}+1+\frac{x+4}{325}+1+\frac{x+5}{324}+1 +\frac{x+349}{5}-4=0\)

\(\Leftrightarrow\)\(\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}+\frac{x+329}{5}=0\)

\(\Leftrightarrow\)\(\left(x+329\right)\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)=0\)

\(\Leftrightarrow\)\(x+329=0\)   (vì  1/327 + 1/326 + 1/325 + 1/324 + 1/5  khác  0  )

\(\Leftrightarrow\)\(x=-329\)

Bài 1 : 

\(\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+5}{324}+\frac{x+349}{5}=0\)

\(\Leftrightarrow\)\(\left(\frac{x+2}{327}+1\right)+\left(\frac{x+3}{326}+1\right)+\left(\frac{x+4}{325}+1\right)+\left(\frac{x+5}{324}+1\right)+\left(\frac{x+349}{5}-4\right)=0\)

\(\Leftrightarrow\)\(\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}+\frac{x+329}{5}=0\)

\(\Leftrightarrow\)\(\left(x+329\right)\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)=0\)

Vì \(\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)\ne0\)

\(\Rightarrow\)\(x+329=0\)

\(\Rightarrow\)\(x=-329\)

Vậy \(x=-329\)

 a) \(1\frac{2}{7} = 1 + \frac{2}{7} = \frac{9}{2}\)

\(\begin{array}{l}x:1\frac{2}{7} =  - 3,5\\x:\frac{9}{7} =  - \frac{7}{2}\\x =  - \frac{7}{2}.\frac{9}{7}\\x =  - \frac{9}{2}\end{array}\)

b) \(0,4.x - \frac{1}{5}.x = \frac{3}{4}\)

\(\begin{array}{l}\frac{2}{5}.x - \frac{1}{5}.x = \frac{3}{4}\\\left( {\frac{2}{5} - \frac{1}{5}} \right).x = \frac{3}{4}\\\frac{1}{5}.x = \frac{3}{4}\\x = \frac{3}{4}:\frac{1}{5}\\x = \frac{3}{4}.5\\x = \frac{{15}}{4}\end{array}\)