\(\frac{x}{8}< \frac{3}{5}< \frac{x+1}{8}\Rightarrow\frac{5x}{40}< \frac{24}{40}< \frac{5x+5}{40}\Rightarrow5x< 24< 5x+5\)

+)5x+1=24=>5x=23=>x=23/5 (loại)

+)5x+2=24=>5x=22=>x=22/5 (loại)

+)5x+3=24=>5x=21=>x=21/5 (loại)

+)5x+4=24=>5x=20=>x=4 (nhận)

Vậy x=4

Ta có: \(\frac{4}{7}< \frac{x}{10}< \frac{5}{7}\)

Ta lại có: \(\frac{8}{14}< \frac{x}{14}< \frac{x}{10}< \frac{10}{14}\)

\(\Rightarrow\frac{8}{14}< \frac{9}{14}< \frac{9}{10}< \frac{10}{14}\)

Vậy giá trị của x là 9.

Ta có:

\(\frac{4}{7}< \frac{x}{10}< \frac{5}{7}\)

\(\Rightarrow\frac{40}{70}< \frac{7x}{70}< \frac{50}{70}\)

\(\Rightarrow40< 7x< 50\)

Vì \(7x⋮7\)

\(\Rightarrow7x\in\left\{42;49\right\}\)

\(\Rightarrow x\in\left\{6;7\right\}\)

Vậy: x = 6 hoặc x = 7

 x = 6 hoặc 7

Bài 1:

\(A=\left(\frac{-5}{11}+\frac{7}{22}-\frac{4}{33}-\frac{5}{44}\right):\left(38\frac{1}{122}-39\frac{7}{22}\right)\)

\(=\frac{-49}{132}:\left(-\frac{879}{671}\right)=\frac{2989}{105408}\)

Bài 2:

\(\frac{4}{5}-\left(\frac{-1}{8}\right)=\frac{7}{8}-x\)

<=>  \(\frac{7}{8}-x=\frac{27}{40}\)

<=>  \(x=\frac{7}{8}-\frac{27}{40}=\frac{1}{5}\)

Vậy...

bài 2 mình tính sai, sửa

.......

<=>  \(\frac{7}{8}-x=\frac{37}{40}\)

<=>  \(x=\frac{7}{8}-\frac{37}{40}=\frac{-1}{20}\)

Vậy....

\(\frac{4}{5}-\frac{-1}{8}=\frac{7}{8}-x\)

\(\Leftrightarrow x=\frac{7}{8}-\frac{4}{5}-\frac{1}{8}\)

\(\Leftrightarrow x=\frac{-1}{20}\)

a) \(y+2\frac{3}{4}=5\frac{2}{3}\)

\(y+\frac{11}{4}=\frac{17}{3}\)

y = 35/12

b) \(y-1\frac{4}{5}=3\frac{2}{7}\)

y - 9/5 = 23/7

y = 178/35

\(a,y+2\frac{3}{4}=5\frac{2}{3}\)

\(\Rightarrow y+\frac{11}{4}=\frac{17}{3}\)

\(\Rightarrow y=\frac{17}{3}-\frac{11}{4}\)

\(\Rightarrow y=\frac{35}{12}\)

\(b,y-1\frac{4}{5}=3\frac{2}{7}\)

\(\Rightarrow y=3\frac{2}{7}-1\frac{4}{5}\)

\(\Rightarrow y=\frac{52}{35}\)

a) \(y+2\frac{3}{4}=5\frac{2}{3}\)

  \(y+\frac{11}{4}=\frac{17}{3}\)

 \(y=\frac{17}{3}-\frac{11}{4}\)

\(y=\frac{35}{12}\)

b) \(y-1\frac{4}{5}=3\frac{2}{7}\)

   \(y-\frac{9}{5}=\frac{23}{7}\)

\(y=\frac{23}{7}+\frac{9}{5}\)

\(y=\frac{178}{35}\)

x = 1 nha bạn mình đangtìm lời giải

a) \(\left(\frac{2}{3}x-\frac{4}{9}\right)\left(\frac{1}{2}-\frac{3}{7}:x\right)=0\)

\(\Rightarrow\orbr{\begin{cases}\frac{2}{3}x-\frac{4}{9}=0\\\frac{1}{2}-\frac{3}{7}:x=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=\frac{6}{7}\end{cases}}\)

Vậy \(x\in\left\{\frac{2}{3};\frac{6}{7}\right\}\)

b) 

 \(\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+5}{324}+\frac{x+349}{5}=0\)

        \(\frac{x+2}{327}+1+\frac{x+3}{326}+1+\frac{x+4}{325}+1+\frac{x+5}{324}+1+\frac{x+329}{5}+4=4\)

         \(\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}+\frac{x+329}{5}=0\)

          \(\left(x+329\right)\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)=0\)

Mà \(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\ne0\)

\(\Rightarrow x+329=0\)

\(\Rightarrow x=-329\)

Vậy \(x=-329\)

Ta có: \(\frac{x+1}{2014}+\frac{x+2}{2013}+\frac{x+3}{2012}=\frac{x+4}{2011}+\frac{x+5}{2010}+\frac{x+6}{2009}\)

\(\Rightarrow\frac{x+1}{2014}+1+\frac{x+2}{2013}+1+\frac{x+3}{2012}+1=\frac{x+4}{2011}+1+\frac{x+5}{2010}+1+\frac{x+6}{2009}+1\)

\(\Rightarrow\frac{2015+x}{2014}+\frac{2015+x}{2013}+\frac{2015+x}{2012}=\frac{2015+x}{2011}+\frac{2015+x}{2010}+\frac{2015+x}{2009}\)

\(\Rightarrow\left(2015+x\right)\left(\frac{1}{2014}+\frac{1}{2013}+\frac{1}{2012}-\frac{1}{2011}-\frac{1}{2010}-\frac{1}{2009}\right)=0\)

=> 2015 + x = 0

=> x = -2015

Các bạn check lại ở dáp án của Ngọc Vĩ nhé!

1/1/3+3/1/4+4/1/5

=4/3+13/4+21/5

=?