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

\(a;x:\left(9\dfrac{1}{2}-\dfrac{3}{2}\right)=\dfrac{\dfrac{2}{5}+\dfrac{4}{9}-\dfrac{5}{11}}{\dfrac{8}{5}+\dfrac{16}{9}-\dfrac{20}{11}}\\ x:8=\dfrac{1}{4}\Rightarrow x=\dfrac{1}{4}\cdot8=2\)

\(b;\left|2x-\dfrac{1}{3}\right|-\left(-2^2\right)=4\left(\dfrac{1}{-2}\right)^3\\ \left|2x-\dfrac{1}{3}\right|+4=-\dfrac{1}{2}\\ \left|2x-\dfrac{1}{3}\right|=-\dfrac{1}{2}-4=-\dfrac{9}{2}\\ \Rightarrow\left[{}\begin{matrix}2x-\dfrac{1}{3}=-\dfrac{9}{2}\Rightarrow x=-\dfrac{25}{12}\\2x-\dfrac{1}{3}=\dfrac{9}{2}\Rightarrow x=\dfrac{29}{12}\end{matrix}\right.\)

a: \(A=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}\)

\(=\sqrt{a}-\sqrt{b}-\sqrt{a}-\sqrt{b}=-2\sqrt{b}\)

b: \(B=\dfrac{2\sqrt{x}-x-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{x+\sqrt{x}+1}{x-1}\)

\(=\dfrac{-2x+\sqrt{x}-1}{\sqrt{x}-1}\cdot\dfrac{1}{x-1}\)

c: \(C=\dfrac{x-9-x+3\sqrt{x}}{x-9}:\left(\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}-2}{\sqrt{x}+3}+\dfrac{x-9}{x+\sqrt{x}-6}\right)\)

\(=\dfrac{3\left(\sqrt{x}-3\right)}{x-9}:\dfrac{9-x+x-4\sqrt{x}+4+x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)

\(=\dfrac{3}{\sqrt{x}+3}\cdot\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{x-4\sqrt{x}+4}\)

\(=\dfrac{3}{\sqrt{x}-2}\)

a) (2x+3)-(5x-17)=9

2x+3-5x+17=9

-3x+20=9

-3x=9-20=-11

=>x=11/3

b) \(\left(\frac{1}{5}+\frac{4}{5}x\right)-\left(\frac{2}{5}x+\frac{1}{9}\right)=\frac{3}{7}\)

=> \(\frac{1}{5}+\frac{4}{5}x-\frac{2}{5}x-\frac{1}{9}=\frac{3}{7}\)

=> \(\left(\frac{1}{5}-\frac{1}{9}\right)+\left(\frac{4}{5}x-\frac{2}{5}x\right)=\frac{3}{7}\)

\(\frac{4}{45}+\frac{2}{5}x=\frac{3}{7}\)

\(\frac{2}{5}x=\frac{3}{7}-\frac{4}{45}=\frac{107}{315}\)

x=107/315:2/5=107/126

dễ tí nữa tôi lấy nk khác của tôi làm thì và kb nhé

a) = 2x-5x+3+17=9\(\Rightarrow\)-3x+20=9\(\Rightarrow\)-3x=-11\(\Rightarrow\)x=11/3

b) = \(\frac{1}{5}-\frac{1}{9}+\frac{4}{5}x-\frac{2}{5}x=\frac{3}{7}\Rightarrow\frac{4}{45}+\frac{2}{5}x=\frac{3}{7}\Rightarrow\frac{2}{5}x=\frac{107}{315}\Rightarrow x=\frac{107}{126}\)

bạn ơi trả lời được câu này kông

( x + 1 ) + ( x - 3 ) + ( x + 5 ) + ............ + ( x +9) = 35