\(-5\left(x+\frac{1}{5}\right)-\frac{1}{2}\left(x-\frac{2}{3}\right)=\frac{3}{2}x-\frac{5}{6}\)

\(-5x+\left(-1\right)-\frac{1}{2}x+\frac{1}{3}=\frac{3}{2}x-\frac{5}{6}\)

\(-5x-\frac{1}{2}x-\frac{3}{2}x=1-\frac{1}{3}-\frac{5}{6}\)

\(x.\left(-5-\frac{1}{2}-\frac{3}{2}\right)=\frac{-1}{6}\)

\(x.\left(-7\right)=\frac{-1}{6}\)

x=\(\frac{1}{42}\)

\(\frac{x_1-1}{5}=\frac{x_2-2}{4}=\frac{x_3-3}{3}=\frac{x_4-4}{2}=\frac{x_5-5}{1}=\frac{x_1+x_2+x_3+x_4+x_5-\left(1+2+3+4+5\right)}{5+4+3+2+1}\)

\(=\frac{30-\left(1+2+3+4+5\right)}{15}=1\)

Vậy \(\frac{x_{1-1}}{5}=1\)

 phần sau nữa bạn tự làm nhé

a)Ta có: \(\frac{-2}{5}+\frac{6}{5}.\left(y-\frac{2}{3}\right)=\frac{-4}{15}\)

\(\Rightarrow\frac{6}{5}.\left(y-\frac{2}{3}\right)=\frac{-4}{15}-\frac{-2}{15}\)

\(\Rightarrow\frac{6}{5}.\left(y-\frac{2}{3}=\right)\frac{-2}{5}\)

\(\Rightarrow y-\frac{2}{3}=\frac{-2}{5}:\frac{6}{5}=\frac{-1}{3}\)

\(\Rightarrow y=\frac{-1}{3}+\frac{2}{3}=\frac{1}{3}\)

Vậy x = \(\frac{1}{3}\)

b) Ta có: \(\frac{-2}{5}+\frac{2}{3}x+\frac{1}{6}x=\frac{-4}{15}\)

        \(\Rightarrow\frac{-2}{5}+x.\left(\frac{2}{3}+\frac{1}{6}\right)=\frac{-4}{15}\)

        \(\Rightarrow x.\frac{5}{6}=\frac{-4}{15}-\frac{-2}{15}\)

         \(x.\frac{5}{6}=\frac{-2}{15}\)

\(\Rightarrow x=\frac{-2}{15}:\frac{5}{6}=\frac{-4}{25}\)

Vậy x = \(\frac{-4}{25}\)

c) Ta có: \(\frac{3}{2}x+\frac{-2}{5}-\frac{2}{3}.x=\frac{-4}{15}\)

\(\Rightarrow\frac{3}{2}x-\frac{2}{3}x+\frac{-2}{5}=\frac{-4}{15}\)

\(\Rightarrow x.\left(\frac{3}{2}-\frac{2}{4}\right)=\frac{-4}{15}-\frac{-2}{15}\)

\(\Rightarrow x.\frac{5}{6}=\frac{-2}{15}\)

\(\Rightarrow x=\frac{-2}{15}:\frac{5}{6}=\frac{-4}{25}\)

Vậy x = \(\frac{-4}{25}\)

Ủng hộ tớ nha m.n

Ơ !!??!?!!?!! chả nhìn thấy x ở đâu cả

\(\frac{10}{3}:\frac{5}{2}=\frac{10}{3}.\frac{2}{5}=\frac{20}{15}=\frac{3}{4}\)

=3/4.ung ho mk nghe

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 \(\frac{1}{2004}+\frac{1}{2005}+\frac{1}{2006}+\frac{1}{2007}\ne0\))

\(\frac{1}{3}x+\frac{2}{5}\left(x-1\right)=0\)

\(\frac{1}{3}x+\frac{2}{5}x-\frac{2}{5}=0\)

\(x\left(\frac{1}{3}+\frac{2}{5}\right)=\frac{2}{5}\)

\(x.\frac{11}{15}=\frac{2}{5}\)

x=6/11