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

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

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

Ta có : \(\frac{x-1991}{9}+\frac{x-1993}{7}+\frac{x-1995}{5}+\frac{x-1997}{3}+\frac{x-1999}{1}\)\(=\frac{x-9}{1991}+\frac{x-7}{1993}+\frac{x-5}{1995}+\frac{x-3}{1997}+\frac{x-1}{1999}\)
\(\Rightarrow\left(\frac{x-1991}{9}-1\right)+\left(\frac{x-1993}{7}-1\right)+\left(\frac{x-1995}{5}-1\right)+\left(\frac{x-1997}{3}-1\right)+\left(\frac{x-1999}{1}-1\right)\)
\(=\left(\frac{x-9}{1991}-1\right)+\left(\frac{x-7}{1993}-1\right)+\left(\frac{x-5}{1995}-1\right)+\left(\frac{x-3}{1997}-1\right)+\left(\frac{x-1}{1999}\right)\)
\(\Rightarrow\frac{x-2000}{9}+\frac{x-2000}{7}+\frac{x-2000}{5}+\frac{x-2000}{3}\)
\(=\frac{x-2000}{1991}+\frac{x-2000}{1993}+\frac{x-2000}{1995}+\frac{x-2000}{1997}+\frac{x-2000}{1999}\)
\(\Rightarrow\left(x-2000\right)\left(\frac{1}{9}+\frac{1}{7}+\frac{1}{5}+\frac{1}{3}\right)=\left(x-2000\right)\left(\frac{1}{1991}+\frac{1}{1993}+\frac{1}{1995}+\frac{1}{1997}+\frac{1}{1999}\right)\)
\(\Rightarrow\left(x-2000\right)\left(\frac{1}{9}+\frac{1}{7}+\frac{1}{5}+\frac{1}{3}\right)-\left(x-2000\right)\left(\frac{1}{1991}+\frac{1}{1993}+\frac{1}{1995}+\frac{1}{1997}+\frac{1}{1999}\right)=0\)
\(\Rightarrow\left(x-2000\right)\left[\left(\frac{1}{9}+\frac{1}{7}+\frac{1}{5}+\frac{1}{3}\right)-\left(\frac{1}{1991}+\frac{1}{1993}+\frac{1}{1995}+\frac{1}{1997}+\frac{1}{1999}\right)\right]=0\)
Vì \(\left(\frac{1}{9}+\frac{1}{7}+\frac{1}{5}+\frac{1}{3}\right)-\left(\frac{1}{1991}+\frac{1}{1993}+\frac{1}{1995}+\frac{1}{1997}+\frac{1}{1999}\right)\ne0\)
=> x - 2000 = 0
=> x = 2000

a. \(\frac{7}{9}:\left(2+\frac{3}{4}x\right)+\frac{5}{9}=\frac{23}{27}\)
\(\Rightarrow\frac{7}{9}:\left(2+\frac{3}{4}x\right)=\frac{23}{27}-\frac{5}{9}\)
\(\Rightarrow\frac{7}{9}:\left(2+\frac{3}{4}x\right)=\frac{8}{27}\)
\(\Rightarrow2+\frac{3}{4}x=\frac{7}{9}:\frac{8}{27}\)
\(\Rightarrow2+\frac{3}{4}x=\frac{21}{8}\)
\(\Rightarrow\frac{3}{4}x=\frac{21}{8}-2\)
\(\Rightarrow\frac{3}{4}x=\frac{5}{8}\)
\(\Rightarrow x=\frac{5}{8}:\frac{3}{4}\)
\(\Rightarrow x=\frac{5}{6}\)
b. \(\frac{-2}{3}x+\frac{1}{5}=\frac{3}{10}\)
\(\Rightarrow\frac{-2}{3}x=\frac{3}{10}-\frac{1}{5}\)
\(\Rightarrow\frac{-2}{3}x=\frac{1}{10}\)
\(\Rightarrow x=\frac{1}{10}:\left(\frac{-2}{3}\right)\)
\(\Rightarrow x=\frac{-3}{20}\)
c. \(\left|x\right|-\frac{3}{4}=\frac{5}{3}\)
\(\Rightarrow\left|x\right|=\frac{5}{3}+\frac{3}{4}\)
\(\Rightarrow\left|x\right|=\frac{29}{12}\)
\(\Rightarrow x=\frac{29}{12}\) hoặc \(x=-\frac{29}{12}\)

\(\left(X+5\right)-\left(X-9\right)=X+2\)\(2\)
\(=>X+5-X+9=X+2\)
\(=>\left(X-X\right)+\left(5+9\right)=X+2\)
\(=>0+14=X+2\)
\(=>14=X+2\)
\(=>X=12\)
(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

\(x+\frac{3+\frac{3}{20}+\frac{3}{13}+\frac{3}{2013}}{5+\frac{5}{20}+\frac{5}{13}+\frac{5}{2013}}=\frac{9}{2}\)
\(x+\frac{3\times\left(1+\frac{1}{20}+\frac{1}{13}+\frac{1}{2013}\right)}{5\times\left(1+\frac{1}{20}+\frac{1}{13}+\frac{1}{2013}\right)}=\frac{9}{2}\)
\(x+\frac{3}{5}=\frac{9}{2}\)
\(x=\frac{39}{10}\)
\(x=3,9\)
-5/3 = x-3/9
Khi đó 3.(x-3) = (-5).9
3.(x-3) = -45
x-3 = (-45) : 3
x-3 = -15
x = -15 + 3
x= -12.
-15=x-3
x=-15+3
x=-12