1. a) \(\frac{3}{4}-\frac{-1}{2}+\frac{1}{3}=\frac{3}{4}+\frac{1}{2}+\frac{1}{3}=\frac{9}{12}+\frac{6}{12}+\frac{4}{12}=\frac{19}{12}\)

   b) \(5\frac{5}{27}+\frac{7}{23}+\frac{1}{2}-\frac{5}{27}+\frac{16}{23}\)

\(=\frac{140}{27}-\frac{5}{27}+\frac{7}{23}+\frac{16}{23}+\frac{1}{2}\)

\(=\frac{135}{27}+\frac{23}{23}+\frac{1}{2}\)

\(=5+1+0,5=6,5\)

2) a) 1/2 + 2/3x = 1/4

=> 2/3x            = 1/4 - 1/2

=> 2/3x            = -1/4

=> x                = -1/4 : 2/3

=> x                = -3/8

b) 3/5 + 2/5 : x = 3 1/2

=> 3/5 + 2/5 : x = 7/2

=>         2/5 : x  = 7/2 - 3/5

=>         2/5 : x  = 29/10

=>               x    = 2/5 : 29/10

=>               x    = 4/29

c) x+4/2004 + x+3/2005 = x+2/2006 + x+1/2007

=> x+4/2004 + 1 + x+3/2005 + 1 = x+2/2006 + 1 + x+1/2007 + 1

=>   x+2008/2004 + x+2008/2005 = x+2008/2006 + x+2008/2007

=>  x+2008/2004 + x+2008/2005 - x+2008/2006 - x+2008/2007 = 0

=> (x+2008). (1/2004 + 1/2005 - 1/2006 - 1/2007) = 0

Vì 1/2004 + 1/2005 - 1/2006 - 1/2007 khác 0

Nên x + 2008 = 0 <=> x = -2008

Vậy x = -2008

1,a,\(\frac{3}{4}-\frac{-1}{2}+\frac{1}{3}=\frac{3}{4}+\frac{2}{4}+\frac{1}{3}=\frac{5}{4}+\frac{1}{3}=\frac{15}{12}+\frac{4}{12}=\frac{19}{12}\)

  b, \(5\frac{5}{27}+\frac{7}{23}+\frac{1}{2}-\frac{5}{27}+\frac{16}{23}=\frac{140}{27}-\frac{5}{27}+\frac{7}{23}+\frac{16}{23}+\frac{1}{2}=\frac{135}{27}+\frac{23}{23}+\frac{1}{2}=5+1+\frac{1}{2}=\frac{13}{2}\)2,a,\(\frac{1}{2}+\frac{2}{3}.x=\frac{1}{4}\)

    <=>\(\frac{2}{3}.x=-\frac{1}{2}\)

   <=>\(x=-\frac{3}{4}\)

b,\(\frac{3}{5}+\frac{2}{5}\div x=3\frac{1}{2}\)

 <=>\(\frac{2}{5x}=\frac{29}{10}\)

 <=>\(x=\frac{29}{4}\)

c,\(\frac{x+4}{2004}+\frac{x+3}{2005}=\frac{x+2}{2006}+\frac{x+1}{2007}\)

<=> \(\frac{x+4}{2004}+1+\frac{x+3}{2005}+1=\frac{x+2}{2006}+1+\frac{x+1}{2007}+1\)

<=>\(\frac{x+2008}{2004}+\frac{x+2008}{2005}=\frac{x+2008}{2006}+\frac{x+2008}{2007}\)

<=>\(\left(x+2008\right)\left(\frac{1}{2004}+\frac{1}{2005}-\frac{1}{2006}-\frac{1}{2007}\right)\)=0

<=>x+2008=0 vì cái ngoặc còn lại\(\ne0\)

<=>x=-2008

 Vậy x=-2008

Bạn nhớ tk cho mình vì mình đã chăm chỉ làm hết bài bạn hỏi nha!

1.

a) \(\frac{3}{4}-\frac{-1}{2}+\frac{1}{3}=\frac{3}{4}+\frac{1}{2}+\frac{1}{3}=\frac{19}{12}\)

b) \(5\frac{5}{27}+\frac{7}{23}+\frac{1}{2}-\frac{5}{27}+\frac{16}{23}\)

\(=\frac{140}{27}+\frac{7}{23}+\frac{1}{2}-\frac{5}{27}+\frac{16}{23}\)

\(=5+1+\frac{1}{2}=\frac{13}{2}\)

2.

a) \(\frac{1}{2}+\frac{2}{3}x=\frac{1}{4}\)

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

\(x=-\frac{3}{8}\)

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

\(\frac{2}{5}x=\frac{7}{2}-\frac{3}{5}\)

\(\frac{2}{5}x=\frac{29}{10}\)

\(x=\frac{29}{4}\)

c) \(\frac{x+4}{2004}+\frac{x+3}{2005}=\frac{x+2}{2006}+\frac{x+1}{2007}\)

\(\left(\frac{x+4}{2004}+1\right)+\left(\frac{x+3}{2005}+1\right)=\left(\frac{x+2}{2006}+1\right)\left(\frac{x+1}{2007}+1\right)\)

\(\frac{x+2008}{2004}+\frac{x+2008}{2005}-\frac{x+2008}{2006}-\frac{x+2008}{2007}=0\)

\(\left(x+2008\right)\left(\frac{1}{2004}+\frac{1}{2005}+\frac{1}{2006}+\frac{1}{2007}\right)=0\)

\(x+2008=0\)

\(x=-2008\) ( vì \(\frac{1}{2004}+\frac{1}{2005}+\frac{1}{2006}+\frac{1}{2007}\ne0\))

Tất cả đều trừ 1 vào mỗi thừa số=>tử số là:x-29(cái cuối+3)

=> cả cái đó=0

=> x-29=0 

=>x=29

\(\frac{x-2}{27}+\frac{x-3}{26}+\frac{x-4}{25}+\frac{x-5}{24}+\frac{x-44}{5}=1\)

\(\Rightarrow\left(\frac{x-2}{27}-1\right)+\left(\frac{x-3}{26}-1\right)+\left(\frac{x-4}{25}-1\right)+\left(\frac{x-5}{24}-1\right)+\left(\frac{x-44}{5}+3\right)=0\)

\(\Rightarrow\frac{x-29}{27}+\frac{x-29}{26}+\frac{x-29}{25}+\frac{x-29}{24}+\frac{x-29}{5}=0\)

\(\Rightarrow\left(x-29\right)\left(\frac{1}{27}+\frac{1}{26}+\frac{1}{25}+\frac{1}{24}+\frac{1}{5}\right)=0\)

\(\Rightarrow x-29=0\)

\(\Rightarrow x=29\)

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.

pt \(\Leftrightarrow\left(\frac{x-2}{27}-1\right)+\left(\frac{x-3}{26}-1\right)+\left(\frac{x-4}{25}-1\right)+\left(\frac{x-5}{24}-1\right)+\left(\frac{x-44}{5}+3\right)=0\)

\(\Leftrightarrow\frac{x-29}{27}+\frac{x-29}{26}+\frac{x-29}{25}+\frac{x-29}{24}+\frac{x-29}{5}=0\)

\(\Leftrightarrow\left(x-29\right)\left(\frac{1}{27}+\frac{1}{26}+\frac{1}{25}+\frac{1}{24}+\frac{1}{5}\right)=0\)

Mà \(\frac{1}{27}+\frac{1}{26}+\frac{1}{25}+\frac{1}{24}+\frac{1}{5}\ne0\)

\(\Rightarrow x-29=0\Leftrightarrow x=29\)

\(\left(\frac{x}{-5}+1\frac{1}{2}\right):\frac{28}{75}-1,4.\frac{15}{49}=\left|-\frac{2}{3}\right|.\left(-\frac{3}{2}\right)^3\)

\(\left(\frac{x}{-5}+\frac{3}{2}\right).\frac{75}{28}-\frac{14}{10}.\frac{15}{49}=\frac{2}{3}.\frac{-27}{8}\)

\(\left(\frac{-x}{5}+\frac{3}{2}\right).\frac{75}{28}-\frac{3}{7}=\frac{-9}{4}\)

\(\left(\frac{-x}{5}+\frac{3}{2}\right).\frac{75}{28}=\frac{-9}{4}+\frac{3}{7}\)

\(\left(\frac{-x}{5}+\frac{3}{2}\right).\frac{75}{28}=\frac{-63}{28}+\frac{12}{28}\)

\(\left(\frac{-x}{5}+\frac{3}{2}\right).\frac{75}{28}=\frac{-51}{28}\)

\(\frac{-x}{5}+\frac{3}{2}=\frac{-51}{28}:\frac{75}{28}\)

\(\frac{-x}{5}+\frac{3}{2}=\frac{-51}{28}.\frac{28}{75}\)

\(\frac{-x}{5}+\frac{3}{2}=\frac{-17}{25}\)

\(\frac{-x}{5}=\frac{-17}{25}-\frac{3}{2}\)

\(\frac{-x}{5}=\frac{-34}{50}-\frac{75}{50}\)

\(\frac{-x}{5}=\frac{-109}{50}\)

\(\frac{-10x}{50}=\frac{-109}{50}\)

Hình như đề sai thì phải

ko biết nữa

mik chỉ biết là cô cho vậy thôi

kb lqmb vs mk ko mk là P.A.D8a1

Làm hết thì đã lên " THÁNH "

a)(1,5.x+3/7):1,5-1,5=1                                                                     b)(2+x+4/7+3/7):1,4-5/7=-2

1,5.x:1,5+3/7:3/2=1+1,5                                                                       (2+x+1):1,4-5/7=-2

x+ 3/7x 2/3=2,5                                                                                   (3+x):1,4=-2+5

x+2/7=2,5                                                                                               (3+x):1,4=3

x+2/7=5/2                                                                                              3+x=3x1,4

x=5/2-2/7                                                                                               3+x=4,2

x=31/14                                                                                                 x=4,2-3=1,2

bai nay cung ko biet lam

mơn bạn