do các phân số ở hàng số thứ 2  đã tối giản nên x=0=>7x=0 =>tổng các phân số sau đều tối giản

\(a)\frac{1}{7}x-\frac{1}{2}x+\frac{5}{7}x=-\frac{1}{2}\)

\(\Rightarrow\left(\frac{1}{7}-\frac{1}{2}+\frac{5}{7}\right)x=-\frac{1}{2}\)

\(\Rightarrow\left(\frac{2}{14}-\frac{7}{14}+\frac{10}{14}\right)x=-\frac{1}{2}\)

\(\Rightarrow\frac{5}{14}x=-\frac{1}{2}\)

\(\Rightarrow x=-\frac{1}{2}:\frac{5}{14}\)

\(\Rightarrow x=-\frac{1}{2}.\frac{14}{5}\)

\(\Rightarrow x=-\frac{7}{5}\)

\(b)(\frac{2}{11.13}+\frac{2}{13.15}+...+\frac{2}{49.51})x=-\frac{1}{3}\)

\(\Rightarrow\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...+\frac{1}{49}-\frac{1}{51}\right)x=-\frac{1}{3}\)

\(\Rightarrow\left(\frac{1}{11}-\frac{1}{51}\right)x=-\frac{1}{3}\)

\(\Rightarrow\left(\frac{51}{561}-\frac{11}{561}\right)x=-\frac{1}{3}\)

\(\Rightarrow\frac{40}{561}x=-\frac{1}{3}\)

\(\Rightarrow x=-\frac{1}{3}:\frac{40}{561}\)

\(\Rightarrow x=-\frac{1}{3}.\frac{561}{40}\)

\(\Rightarrow x=-\frac{187}{40}\)

Chúc bạn học tốt !!! 

Ta có : 

\(\frac{7x+2}{5x+7}=\frac{7x-1}{5x+1}\)

\(=>\left(7x+2\right)\left(5x+1\right)=\left(7x-1\right)\left(5x+7\right)\)

\(=>35x^2+7x+10x+2=35x^2+49x-5x-7\)

\(=>35x^2+17x+2=35x^2+44x-7\)

\(=>17x+2=44x-7\)

\(=>44x-17x=2+7\)

\(=>27x=9\)

\(=>x=\frac{9}{27}=\frac{1}{3}\)

A

\(1,\Rightarrow3^{x-3}=\left(3^2\right)^8:\left(3^3\right)^5=3^{16}:3^{15}=3^1\\ \Rightarrow x-3=1\\ \Rightarrow x=4\\ 2,\Rightarrow7^x\left(1+7^2\right)=350\\ \Rightarrow7^x=\dfrac{350}{50}=7=7^1\\ \Rightarrow x=1\)

\(3,\Rightarrow2^{2+2x+2}-2^{2x}=240\\ \Rightarrow2^{2x}\left(2^4-1\right)=240\\ \Rightarrow2^{2x}=\dfrac{240}{15}=16=2^4\\ \Rightarrow2x=4\Rightarrow x=2\)

Đề bài????

= cái j

7x.(2+x)-7x.(x+3)=14

7x.(2+x-x-3)=14

7x.(-1)=14

7x=14:(-1)

7x=-14

x=-14:7

x=2

Vậy x=2

tk nha!

Tìm x €Z biết

4.(2x+9)+(-8.x+3) -(x-13)=0

\(7\frac{x}{2.5}+7\frac{x}{5.8}+.....+7.\frac{x}{17.20}=\frac{21}{10}\)

\(7\left(\frac{x}{2.5}+\frac{x}{5.8}+...+\frac{x}{17.20}\right)=\frac{21}{10}\)

\(\frac{x}{2.5}+\frac{x}{5.8}+...+\frac{x}{17.20}=\frac{21}{70}\)

\(\frac{x.3}{2.5.3}+\frac{x.3}{5.8.3}+...+\frac{x.3}{17.20.3}=\frac{21}{70}\)

\(x.\frac{1}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{17.20}\right)=\frac{21}{70}\)

\(x.\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{20}\right)=\frac{21}{70}\)

\(x.\frac{1}{3}.\frac{9}{20}=\frac{21}{70}\)

=> \(x=2\)

\(x=\frac{7x}{2}\)\(-\frac{7x}{5}+\)\(\frac{7x}{5}\)\(-\frac{7x}{8}\)\(+\frac{7x}{8}\)\(-\frac{7x}{11}\)\(+\frac{7x}{11}\)\(-\frac{7x}{14}\)\(+\frac{7x}{14}\)\(-\frac{7x}{17}+\)\(\frac{7x}{17}\)\(-\frac{7x}{20}\)\(=\frac{21}{10}\)

\(x=\frac{7x}{2}\)\(-\frac{7x}{20}\)\(=\frac{21}{10}\)

\(x=\frac{7x.10}{20}\)\(+\frac{7x}{20}\)\(=\frac{21}{10}\)

\(x=\frac{7x.10+7x}{20}\)\(=\frac{21}{10}\)

\(x=\frac{7x.\left(10+2\right)}{20.2}\)\(=\frac{7x.12}{40}\)\(=\frac{21}{10}\)

\(=>\frac{7x.12:4}{40:4}=\)\(\frac{21}{10}\)

\(=>x=1\)

1. 

2|x-6|+7x-2=|x-6|+7x

2|x-6| - |x-6|=7x-(7x-2)

     |x-6|       =    2

=>x-6 = +2

*x-6=2                                                                                                          *x-6 = -2

x     =2+6                                                                                                       x     = (-2)+6

x      =8                                                                                                          x      =    4

2.

|x-5|-7(x+4)=5-7x

|x-5|-7x-28 =5-7x

|x-5|-28       =5-7x+7x

|x-5|-28       =      5

|x-5|            =   5+28

|x-5|            =      33

=>x-5          =    +33

*x-5=33                                                                      *x-5=-33

 x    =38                                                                      x   = -28

3.

3|x+4|-2(x-1)=7-2x

3|x+4|-2x+2 =7-2x

3|x+4|-2       =7-2x+2x

3|x+4|-2       =7

3|x+4|          =7+2

3|x+4|          =  9

  |x+4|          =9:3

  |x+4|          =  3

=>x+4          = +3

*x+4=3                                   *x+4=-3

 x     =-1                                   x    = -7