Câu a:

(4\(\frac{5}{37}\) - 3\(\frac45\) + 8\(\frac{15}{29}\)) - (3\(\frac{5}{37}\) - 6\(\frac{14}{29}\))

= 4\(\frac{5}{37}\) - 3\(\frac45\) + 8\(\frac{15}{29}\) - 3\(\frac{15}{37}\) + 6\(\frac{14}{29}\)

= (4 - 3) + (\(\frac{5}{37}-\frac{5}{37}\)) + (8 + 6) + (\(\frac{15}{29}\) + \(\frac{14}{29}\)) - 3 - \(\frac45\)

= 1 + 0 + 14 + 1 - 3 - \(\frac45\)

= 1 + 14 + 1 - 3 - \(\frac45\)

= 15 + 1 - 3 - \(\frac45\)

= 16 - 3 - \(\frac45\)

= 13 - \(\frac45\)

= \(\frac{61}{5}\)

\(\frac{x+1}{35}+\frac{x+3}{33}=\frac{x+5}{31}+\frac{x+7}{29}\)

<=>\(\frac{x+1}{35}+1+\frac{x+3}{33}+1=\frac{x+5}{31}+1+\frac{x+7}{29}+1\)

<=>\(\frac{x+1+35}{35}+\frac{x+3+33}{33}=\frac{x+5+31}{31}+\frac{x+7+29}{29}\)

<=>\(\frac{x+36}{35}+\frac{x+36}{33}=\frac{x+36}{31}+\frac{x+36}{29}\)

<=>\(\frac{x+36}{35}+\frac{x+36}{33}-\frac{x+36}{31}-\frac{x+36}{29}=0\)

<=>\(\left(x+36\right)\left(\frac{1}{35}+\frac{1}{33}-\frac{1}{31}-\frac{1}{29}\right)=0\)

Vì \(\frac{1}{35}+\frac{1}{33}-\frac{1}{31}-\frac{1}{29}\ne0\Rightarrow x+36=0\Rightarrow x=-36\)

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!

a) \(x=\frac{3}{4}-\frac{1}{3}=\frac{5}{12}\)

b) \(x=\frac{5}{7}+\frac{2}{5}=\frac{39}{35}\)

c) \(-x=-\frac{6}{7}+\frac{2}{3}=-\frac{4}{21}\Leftrightarrow x=\frac{4}{21}\)

d) \(x=\frac{4}{7}-\frac{1}{3}=\frac{5}{21}\)

a/ \(x+\frac{1}{3}=\frac{3}{4}\)

\(x=\frac{3}{4}-\frac{1}{3}=\frac{5}{12}\)

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

\(x=\frac{5}{7}+\frac{2}{5}=\frac{39}{35}\)

c/\(-x-\frac{2}{3}=-\frac{6}{7}\)

\(-x=-\frac{6}{7}+\frac{2}{3}=-\frac{4}{21}\)

\(\rightarrow x=\frac{4}{21}\)

d/ \(-\frac{4}{7}-x=\frac{1}{3}\)

\(x=\left(-\frac{4}{7}\right)-\frac{1}{3}=-\frac{19}{21}\)

a) x : \(\left(-\frac{1}{3}\right)^3=-\frac{1}{3}\)

\(x:\frac{-1}{27}=\frac{-1}{3}\)

\(x=\frac{-1}{3}.\frac{-1}{27}\)

\(x=\frac{1}{81}\)

Vậy \(x=\frac{1}{81}\)

a) \(x:\left(-\frac{1}{3}\right)^3=-\frac{1}{3}\)

\(\Leftrightarrow x=\left(-\frac{1}{3}\right)\cdot\left(-\frac{1}{3}\right)^3\)

\(\Leftrightarrow x=\left(-\frac{1}{3}\right)^4\)

\(\Leftrightarrow x=\frac{1}{81}\)

b)\(\left(\frac{4}{5}\right)^5\cdot x=\left(\frac{4}{5}\right)^7\)

\(\Leftrightarrow x=\left(\frac{4}{5}\right)^7:\left(\frac{4}{5}\right)^5=\left(\frac{4}{5}\right)^2=\frac{16}{25}\)

c)\(\left(x+\frac{1}{2}\right)^2=\frac{1}{16}\)

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

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

d)\(\left(3x+1\right)^3=-27\)

\(\Leftrightarrow3x+1=-3\)

\(\Leftrightarrow3x=-4\)

\(\Leftrightarrow x=-\frac{4}{3}\)

Câu a đề thiếu vế phải rồi bạn

b: \(\Leftrightarrow x\cdot0+1=0\)

=>0x+1=0(vô lý)