=.8/5+((2*17*37+2*7*37+2*7*17)/(7*17*37))/((5*17*37+5*7*37+5*7*17)/(7*17*37))*x=16/5

=>8/5+((2*17*37+2*7*37+2*7*17)/(7*17*37))*((7*17*37)/(5*17*37+5*7*37+5*7*17))*x=16/5

=>8/5+(2(17*37+7*37+7*17))/(5(17*37+7*37+7*17))*x=16/5

=>8/5+(2/5)*x=16/5

=>(2/5)*x=16/5-8/5

=>(2/5)*x=8/5

=>x=(8/5)/(2/5)

=>x=4

Vậy x=4

ngothithuyhien  làm dài dòng quá !

8/5 + [(2/7 + 2/17 + 2/37)/(5/7 + 5/17 + 5/37)].x = 16/5

=> 8/5 + [(2/7 : 5/7) + (2/17 : 5/17) + (2/37 + 5/37)].x = 16/5

=> 8/5 + (2/5 + 2/5 + 2/5).x = 16/5

=> 8/5 + (2/5 . 3).x = 16/5

=> 8/5 + 1/1/5 .x = 16/5

=>          6/5 .x = 16/5 - 8/5

=>          6/5 .x = 1/3/5

=>                 x = 1/3/5 : 6/5 

=>                 x = 1/1/3

Có gì sai xin m.n giúp đỡ ^^ 

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

\(\Rightarrow\left(x+1\right)6=5\left(x+2\right)\)

\(\Leftrightarrow6x+6=5x+10\)

\(\Leftrightarrow6x-5x=10-6\)

\(\Rightarrow x=4\)

\(\frac{x+1}{2}\)= \(\frac{8}{x+1}\) 

x + 1 . x + 1 = 2 . 8

x . 2             = 16

x                  = 16 : 2

x                  = 8

\(\frac{-37}{8}+\frac{13}{8}< x< \frac{1}{4}+\left(-\frac{5}{4}\right)\)

\(\frac{-37+13}{8}< x< \frac{1-5}{4}\)

\(\frac{-24}{8}< x< \frac{-4}{4}\)

\(\Leftrightarrow-3< x< -1\)

=> x = -2 

a) \(x+\left(-7\right)=-20\)

\(\Rightarrow x=-20+7\)

\(\Rightarrow x=-13\)

Vậy \(x=-13\)

b) \(8-x=-12\)

\(\Rightarrow x=8-\left(-12\right)\)

\(\Rightarrow x=20\)

Vậy \(x=20\)

c) \(|x|-7=-6\)

\(\Rightarrow|x|=-6+7\)

\(\Rightarrow|x|=1\)

\(\Rightarrow\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)

Vậy \(x\in\left\{1;-1\right\}\)

d) \(5^2.2^2-7.|x|=65\)

\(\Rightarrow\left(5.2\right)^2-7.|x|=65\)

\(\Rightarrow10^2-7.|x|=65\)

\(\Rightarrow100-7.|x|=65\)

\(\Rightarrow7.|x|=35\)

\(\Rightarrow|x|=5\)

\(\Rightarrow\orbr{\begin{cases}x=5\\x=-5\end{cases}}\)

Vậy \(x\in\left\{5;-5\right\}\)

e) \(37-3.|x|=2^3-4\)

\(\Rightarrow37-3.|x|=8-4\)

\(\Rightarrow37-3.|x|=4\)

\(\Rightarrow3.|x|=33\)

\(\Rightarrow|x|=11\)

\(\Rightarrow\orbr{\begin{cases}x=11\\x=-11\end{cases}}\)

Vậy \(x\in\left\{11;-11\right\}\)

f) \(|x|+|-5|=|-37|\)

\(\Rightarrow|x|+5=37\)

\(\Rightarrow|x|=32\)

\(\Rightarrow\orbr{\begin{cases}x=32\\x=-32\end{cases}}\)

Vậy \(x\in\left\{32;-32\right\}\)

g)\(5.|x+9|=40\)

\(\Rightarrow|x+9|=8\)

\(\Rightarrow\orbr{\begin{cases}x+9=8\\x+9=-8\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-1\\x=-17\end{cases}}\)

Vậy \(x\in\left\{-1;-17\right\}\)

h) \(-\frac{5}{6}+\frac{8}{3}+\frac{-29}{6}\le x\le\frac{-1}{2}+2+\frac{5}{2}\)

\(\Rightarrow\frac{-5}{6}+\frac{16}{6}+\frac{-29}{6}\le x\le\frac{-1}{2}+\frac{4}{2}+\frac{5}{2}\)

\(\Rightarrow-3\le x\le4\)

Vậy \(-3\le x\le4\)

câu a

x+(-7)=-20

x=-20-(-7)

x=-13

\(\frac{x-1}{5}+\frac{x-1}{7}+...+\frac{x-1}{37}=0=>\left(x-1\right).\left(\frac{1}{5}+\frac{1}{7}+...+\frac{1}{37}\right)=0\)

mà \(\frac{1}{5}+\frac{1}{7}+...+\frac{1}{37}>0\)

=> x-1 = 0

=> x = 1

Vậy x=1

đúng nha

\(1\frac{3}{5}+\left(\frac{\frac{2}{7}+\frac{2}{17}+\frac{2}{37}}{\frac{5}{7}+\frac{5}{17}+\frac{5}{37}}\right)x=\frac{16}{5}\)

\(\Rightarrow\frac{2\left(\frac{1}{7}+\frac{1}{17}+\frac{1}{37}\right)}{5\left(\frac{1}{7}+\frac{1}{17}+\frac{1}{37}\right)}.x=\frac{16}{5}-\frac{8}{5}\)

\(\Rightarrow\frac{2}{5}.x=\frac{8}{5}\)

\(\Rightarrow x=\frac{8}{5}:\frac{2}{5}=4\)

vậy x=4

\(\frac{7}{5}+\left(\frac{2\left(\frac{1}{7}+\frac{1}{17}+\frac{1}{37}\right)}{5\left(\frac{1}{7}+\frac{1}{17}+\frac{1}{37}\right)}\right)\cdot x=\frac{16}{5}\)

\(\frac{2}{5}x=\frac{16}{5}-\frac{7}{5}\)

\(\frac{2}{5}x=\frac{9}{5}\)

 x      = \(\frac{9}{5}:\frac{2}{5}\)

x        = 9/2 

\(1\frac{2}{5}+\left(\frac{\frac{2}{7}+\frac{2}{17}+\frac{2}{37}}{\frac{5}{7}+\frac{5}{17}+\frac{5}{37}}\right).x=\frac{16}{5}\)

\(\frac{7}{5}+\left[\frac{2.\left(\frac{1}{7}+\frac{1}{17}+\frac{1}{37}\right)}{5.\left(\frac{1}{7}+\frac{1}{17}+\frac{1}{37}\right)}\right].x=\frac{16}{5}\)

\(\frac{7}{5}+\frac{2}{5}.x=\frac{16}{5}\)

\(\frac{2}{5}.x=\frac{16}{5}-\frac{7}{5}\)

\(\frac{2}{5}.x=\frac{9}{5}\)

\(x=\frac{9}{5}:\frac{2}{5}\)

\(x=\frac{9}{5}.\frac{5}{2}\)

\(x=\frac{9}{2}\)

\(\frac{7}{5}+\left(\frac{2\left(\frac{1}{7}+\frac{1}{17}+\frac{1}{37}\right)}{5\left(\frac{1}{7}+\frac{1}{17}+\frac{1}{37}\right)}\right).x=\frac{16}{5}\)

                                                \(\frac{2}{5}x=\frac{16}{5}-\frac{7}{5}\)

                                               \(\frac{2}{5}x=\frac{9}{5}\)

                                                    \(x=\frac{9}{5}\div\frac{2}{5}\)

                                                   \(x=\frac{9}{2}\)

theo đề bài ta có:

\(x+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}=\frac{-37}{45}\)

\(x+\left(\frac{1}{5}-\frac{1}{45}\right)=\frac{-37}{45}\)

\(x+\frac{8}{45}=\frac{-37}{45}\)

\(x=\frac{-37}{45}-\frac{8}{45}\)

\(x=\frac{-45}{45}=1\)

đặt A=4/5.9+4/9.13+4/13.17+...+4/41.45

=1/5-1/9+1/9-1/13+1/13-1/17+...+1/41-1/45

=1/5-1/45

=8/45

suy ra x+8/45=-37/45

suy ra x=-1

x+ 1/5.9 + ... + 1/41+ 1/45 = -37/45

x  + 1/5 - 1/9 + 1/9 -  ......+ 1/41 - 1/45 = -37/45

x+ 1/5 - 1/45 = -37 45

x+ 8/45 =-37/45

x            = -37 /45 - 8/45

x                = -1