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.

x = -2 nha bạn . 

\(\frac{x+\frac{2\left(3-x\right)}{5}}{14}-\frac{5x-4\left(x-1\right)}{24}=\frac{7x+2+\frac{9-3x}{5}}{12}+\frac{2}{3}\)

\(\Leftrightarrow\frac{x}{840}-\frac{17}{210}=\frac{8x}{15}+\frac{19}{60}+\frac{2}{3}\)

\(\Leftrightarrow\frac{x}{840}.840-\frac{17}{210}.840=\frac{8x}{15}.840+\frac{19}{60}.840+\frac{2}{3}.840\)

\(\Leftrightarrow x-68=448x+226+560\)

\(\Leftrightarrow x-68=448x+826\)

\(\Leftrightarrow x=448x+826+68\)

\(\Leftrightarrow x=448x+894\)

\(\Leftrightarrow-447x=894\)

=> x = -2

a)

\(\begin{array}{l}x:{\left( {\frac{{ - 1}}{2}} \right)^3} =  - \frac{1}{2}\\x =  - \frac{1}{2}.{\left( {\frac{{ - 1}}{2}} \right)^3}\\x = {\left( {\frac{{ - 1}}{2}} \right)^4}\\x = \frac{1}{{16}}\end{array}\)              

Vậy \(x = \frac{1}{{16}}\).

 b)

\(\begin{array}{l}x.{\left( {\frac{3}{5}} \right)^7} = {\left( {\frac{3}{5}} \right)^9}\\x = {\left( {\frac{3}{5}} \right)^9}:{\left( {\frac{3}{5}} \right)^7}\\x = {\left( {\frac{3}{5}} \right)^2}\\x = \frac{9}{{25}}\end{array}\)

Vậy \(x = \frac{9}{{25}}\).

c)

\(\begin{array}{l}{\left( {\frac{{ - 2}}{3}} \right)^{11}}:x = {\left( {\frac{{ - 2}}{3}} \right)^9}\\x = {\left( {\frac{{ - 2}}{3}} \right)^{11}}:{\left( {\frac{{ - 2}}{3}} \right)^9}\\x = {\left( {\frac{{ - 2}}{3}} \right)^2}\\x = \frac{4}{9}.\end{array}\)         

Vậy \(x = \frac{4}{9}\).

d)

\(\begin{array}{l}x.{\left( {0,25} \right)^6} = {\left( {\frac{1}{4}} \right)^8}\\x.{\left( {\frac{1}{4}} \right)^6} = {\left( {\frac{1}{4}} \right)^8}\\x = {\left( {\frac{1}{4}} \right)^8}:{\left( {\frac{1}{4}} \right)^6}\\x = {\left( {\frac{1}{4}} \right)^2}\\x = \frac{1}{{16}}\end{array}\)

Vậy \(x = \frac{1}{{16}}\).

\(a,\left(\frac{6^3-10.5^3}{6^2.3^3-15^2.5^2}.|x-2|\right):10=\left(1-\frac{1}{2}\right)....\left(1-\frac{1}{10}\right)\)

\(=\frac{1.2.3.4...9}{1.2.....10}=\frac{1}{10}\Leftrightarrow\frac{6^3-10.5^3}{6^2.3^3-15^2.5^2}.|x-2|=1\)

\(\Leftrightarrow\frac{6^2.6-2.5^4}{6^2.3^2-3^2.5^4}.|x-2|=1\Leftrightarrow|x-2|.\frac{2}{3}=1\Leftrightarrow|x-2|=\frac{3}{2}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{7}{2}\end{cases}}\)

\(\left(\frac{6^3-10,5^3}{6^2.3^3-15^2.5^2}.\left|x-2\right|\right):10=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).....\left(1-\frac{1}{9}\right).\left(1-\frac{1}{10}\right)\)

\(=\frac{1.2.3.4...9}{1.2.....10}=\frac{1}{10}\)

\(\Leftrightarrow\frac{6^3-10,5^3}{6^2.3^3-15^2.5^2}.\left|x-2\right|=1\)

\(\Leftrightarrow\frac{6^2.6-2.5^4}{6^2.3^2-3^2.5^4}.\left|x-2\right|=1\)

\(\Leftrightarrow\left|x-2\right|.\frac{2}{3}=1\Leftrightarrow\left|x-2\right|=\frac{3}{2}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{7}{2}\end{cases}}\)

Mình làm tiếp câu b nha !

b,                                               Bài giải

\(\frac{x-2018}{2}+\frac{x-2020}{4}=\frac{x-2040}{8}+\frac{x-2030}{14}\)

\(\left(\frac{x-2018}{2}+1\right)+\left(\frac{x-2020}{4}+1\right)=\left(\frac{x-2040}{8}+1\right)+\left(\frac{x-2030}{14}+1\right)\)

\(\frac{x-2016}{2}+\frac{x-2016}{4}=\frac{x-2032}{8}+\frac{x-2016}{14}\)

\(\left(x-2016\right)\left(\frac{1}{2}+\frac{1}{4}\right)=\frac{x-2016}{8}-2+\frac{x-2016}{14}\)

\(\left(x-2016\right)\cdot\frac{3}{4}=\left(x-2016\right)\left(\frac{1}{8}+\frac{1}{14}\right)-2\)

\(\left(x-2016\right)\cdot\frac{3}{4}=\left(x-2016\right)\cdot\frac{11}{56}-2\)

\(\left(x-2016\right)\cdot\frac{3}{4}-\left(x-2016\right)\cdot\frac{11}{56}=-2\)

\(\left(x-2016\right)\left(\frac{3}{4}-\frac{11}{56}\right)=-2\)

\(\left(x-2016\right)\cdot\frac{31}{56}=-2\)

\(x-2016=-2\text{ : }\frac{31}{56}\)

\(x-2016=-\frac{112}{31}\)

\(x=-\frac{112}{31}+2016\)

\(x=\frac{62384}{31}\)

\(\frac{2^{4-x}}{16^5}=32^6\)

=> \(\frac{2^{4-x}}{\left(2^4\right)^5}=\left(2^5\right)^6\)

=> \(\frac{2^{4-x}}{2^{20}}=2^{30}\)

=> \(2^{4-x}=2^{30}.2^{20}\)

=> \(2^{4-x}=2^{50}\)

=> 4  - x = 50

=> x = 4 - 50 = -46

\(\frac{3^{2x+3}}{9^3}=9^{14}\)

=> \(\frac{3^{2x+3}}{\left(3^2\right)^3}=\left(3^2\right)^{14}\)

=> \(\frac{3^{2x+3}}{3^6}=3^{28}\)

=> \(3^{2x+3}=3^{28}.3^6\)

=> \(3^{2x+3}=3^{34}\)

=> 2x + 3 = 34

=> 2x = 34 - 3

=> 2x = 31

=> x = 31/2

\(\frac{2}{\left(x+2\right)\left(x+4\right)}+\frac{4}{\left(x+4\right)\left(x+8\right)}+\frac{6}{\left(x+8\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+4\right)}\)

\(\Rightarrow\frac{1}{x+2}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+8}-\frac{1}{x+8}-\frac{1}{x+16}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)

\(\Rightarrow\frac{1}{x+2}-\frac{1}{x+16}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)

\(\Rightarrow\frac{\left(x+16\right)-\left(x+2\right)}{\left(x+2\right)\left(x+16\right)}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)

\(\Rightarrow x+16-x-2=x\)

\(\Rightarrow x=14\)