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.

\(\left(X+5\right)-\left(X-9\right)=X+2\)\(2\)

\(=>X+5-X+9=X+2\)

\(=>\left(X-X\right)+\left(5+9\right)=X+2\)

 \(=>0+14=X+2\)

\(=>14=X+2\)

\(=>X=12\)

(x+5)-(x-9)=x+2

x+5-x+9=x+2

x-x+14=x+2

12=x(cũng bớt mới về đi 2 đơn vị)

hoặc x=14-2suy ra x=12

làm cau 2 đi mk tích

\(\Leftrightarrow\frac{\frac{x-3+5}{5}}{4}=\frac{\frac{4x-3}{6}}{6}\Leftrightarrow\frac{x+2}{20}=\frac{4x-3}{36}\Leftrightarrow36x+72=80x-60\Leftrightarrow44x=132\Rightarrow x=2\)

\(\Leftrightarrow\frac{\frac{10x+x+2}{2}}{9}-\frac{\frac{x+3+75}{5}}{12}=x-2\)\(\Leftrightarrow\frac{11x+2}{18}-\frac{x+78}{60}=x-2\)\(\Leftrightarrow\left(\frac{11}{18}-\frac{1}{60}-1\right)x+\left(\frac{2}{18}-\frac{78}{60}+2\right)=0\).Giải típ nha, ko có Casio nên mk ko bấm

\(\Leftrightarrow5x+\frac{x+2}{18}-x-\frac{x+3+75}{60}+2=0\Leftrightarrow\left(5+\frac{1}{18}-1-\frac{1}{60}\right)x+\left(\frac{2}{18}-\frac{78}{60}+2\right)=0\).Giải típ nha

= \(x^8.\frac{1}{10}.\frac{2}{9}.\frac{3}{8}.\frac{4}{7}.\frac{5}{6}.\frac{6}{5}.\frac{7}{4}.\frac{8}{3}.\frac{9}{2}\)

= \(x^8.\frac{1}{10}.\left(\frac{2}{9}.\frac{9}{2}\right).\left(\frac{3}{8}.\frac{8}{3}\right).\left(\frac{4}{7}.\frac{7}{4}\right).\left(\frac{5}{6}.\frac{6}{5}\right)\)

= \(x^8.\frac{1}{10}.1.1.1.1\)

= \(x^8.\frac{1}{10}\)

Mk ko pik co dung ko nua

=4/10

\(\frac{x^8}{10}\)

d: =>4x+6=15x-12

=>4x-15x=-12-6=-18

=>-11x=-18

hay x=18/11

e: =>\(45x+27=12+24x\)

=>21x=-15

hay x=-5/7

f: =>35x-5=96-6x

=>41x=101

hay x=101/41

g: =>3(x-3)=90-5(1-2x)

=>3x-9=90-5+10x

=>3x-9=10x+85

=>-7x=94

hay x=-94/7

\(\frac{x-3}{5}-\frac{2x-1}{10}=\frac{x+1}{2}+\frac{1}{4}\)

\(< =>\frac{\left(x-3\right).4}{20}-\frac{\left(2x-1\right).2}{20}=\frac{\left(x+1\right).10}{20}+\frac{5}{20}\)

\(< =>4x-12-4x+2=10x+10+5\)

\(< =>10x=-10-10-5=-25\)

\(< =>x=-\frac{25}{10}=-\frac{5}{2}\)

\(\frac{x+3}{2}-\frac{2x-1}{3}-1=\frac{x+5}{5}\)

\(< =>\frac{\left(x+3\right).15}{30}-\frac{\left(2x-1\right).10}{30}-\frac{30}{30}=\frac{\left(x+5\right).5}{30}\)\(< =>15x+45-20x+10-30=5x+25\)

\(< =>-5x+25=5x+25< =>10x=0< =>x=0\)

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

\(< =>\frac{\left(x-1\right).9}{36}-\frac{\left(5-2x\right).4}{36}=\frac{3x.36}{36}-\frac{2.12}{36}\)

\(< =>\left(x-1\right).9-\left(5-2x\right).4=108x-24\)

\(< =>9x-9-20+8x=108x-24\)

\(< =>108x-17x=-29+24\)

\(< =>91x=-5< =>x=-\frac{5}{91}\)

a) Đk: x \(\ne\)-2

Ta có: \(\frac{2}{x+2}-\frac{2x^2+16}{x^2+8}=\frac{5}{x^2-2x+4}\)

<=> \(\frac{2\left(x^2-2x+4\right)-\left(2x^2+16\right)}{\left(x+2\right)\left(x^2-2x+4\right)}=\frac{5\left(x+2\right)}{\left(x+2\right)\left(x^2-2x+4\right)}\)

<=> 2x² - 4x + 8 - 2x² - 16 = 5x + 10

<=> -4x - 8 = 5x + 10

<=> -4x - 5x = 10 + 8

<=> -9x = 18

<=> x = -2 (ktm)

=> pt vô nghiệm

b) Đk: x \(\ne\)2; x \(\ne\)-3

Ta có: \(\frac{1}{x-2}-\frac{6}{x+3}=\frac{5}{6-x^2-x}\)

<=> \(\frac{x+3}{\left(x-2\right)\left(x+3\right)}-\frac{6\left(x-2\right)}{\left(x-2\right)\left(x+3\right)}=-\frac{5}{\left(x-2\right)\left(x+3\right)}\)

<=> x + 3 - 6x + 12 = -5

<=> -5x = -5 - 15

<=> -5x = -20

<=> x = 4 

vậy S = {4}

c) Đk: x \(\ne\)8; x \(\ne\)9; x \(\ne\)10; x \(\ne\)11

Ta có: \(\frac{8}{x-8}+\frac{11}{x-11}=\frac{9}{x-9}+\frac{10}{x-10}\)

<=> \(\left(\frac{8}{x-8}+1\right)+\left(\frac{11}{x-11}+1\right)=\left(\frac{9}{x-9}+1\right)+\left(\frac{10}{x-10}+1\right)\)

<=> \(\frac{x}{x-8}+\frac{x}{x-11}-\frac{x}{x-9}-\frac{x}{x-10}=0\)

<=> \(x\left(\frac{1}{x-8}+\frac{1}{x-11}-\frac{1}{x-9}-\frac{1}{x-10}\right)=0\)

<=> x = 0 (vì \(\frac{1}{x-8}+\frac{1}{x-11}-\frac{1}{x-9}-\frac{1}{x-10}\ne0\)

Vậy S = {0}

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