\(\left|a+2\right|=a\)

\(\Rightarrow a+2=\hept{\begin{cases}a\\-a\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}a-a=2\\-a-a=2\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}0=2\left(loai\right)\\-2a=2\end{cases}}\)

\(\Rightarrow a=-1\)

\(\frac{a+4}{20}=\frac{5}{a+4}\)

\(\Rightarrow\left(a+4\right)^2=20\cdot5\)

\(\Rightarrow\left(a+4\right)^2=100\)

\(\Rightarrow\hept{\begin{cases}a+4=10\\a+4=-10\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}a=6\\a=-14\end{cases}}\)

\(\left|a+2\right|=a\)

\(\Rightarrow a\ge0\Rightarrow a=-1\left(loai\right)\)

a, \(-\frac{2}{5}+\frac{5}{3}\left(\frac{3}{2}-\frac{4}{15}x\right)=\frac{7}{6}\)

\(\frac{5}{3}\left(\frac{3}{2}-\frac{4}{15}x\right)=\frac{47}{30}\)

\(\frac{3}{2}-\frac{4}{15}x=\frac{47}{50}\)

\(\frac{4}{15}x=\frac{14}{25}\)

\(x=\frac{21}{10}\)

bạn ơi đề thiếu

\(5X\left(X-2020\right)+X=2020\)

\(\Leftrightarrow5X^2-10100X+X=2020\)

\(\Leftrightarrow5X^2-10099X=2020\)

\(\Leftrightarrow5X^2-10099X-2020=0\)

\(\Leftrightarrow5X^2-10100X+x-2020=0\)

\(\Leftrightarrow5X\left(X-2020\right)+X-2020=0\)

\(\Leftrightarrow\left(X-2020\right)\left(5X+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=2020\\x=-\frac{1}{5}\end{cases}}\)

\(4\left(x-5\right)^2-\left(2x+1\right)^2=0\)

\(\Leftrightarrow\left[2\left(x-5\right)\right]^2-\left(2x+1\right)^2=0\)

\(\Leftrightarrow\left[2\left(x-5\right)-2x-1\right]\left[2\left(x-5\right)+2x+1\right]=0\)

\(\Leftrightarrow\left(2x-10-2x-1\right)\left(2x-10+2x+1\right)=0\)

\(\Leftrightarrow-11\left(4x-9\right)=0\)

\(\Leftrightarrow x=\frac{9}{4}\)

\(a,5x\left(x-2020\right)+x=2020\)

\(< =>5x\left(x-2020\right)+x-2020=0\)

\(< =>\left(5x+1\right)\left(x-2020\right)=0\)

\(< =>\orbr{\begin{cases}5x+1=0\\x-2020=0\end{cases}}\)

\(< =>\orbr{\begin{cases}5x=-1\\x=2020\end{cases}< =>\orbr{\begin{cases}x=-\frac{1}{5}\\x=2020\end{cases}}}\)

\(b,4\left(x-5\right)^2-\left(2x+1\right)^2=0\)

\(< =>4\left(x^2-20x+25\right)-\left(4x^2+4x+1\right)=0\)

\(< =>4x^2-80x+100-4x^2-4x-1=0\)

\(< =>-84x+99=0< =>84x=99< =>x=\frac{99}{84}\)

a, (x-15):5+22=24

    ( x - 15 ) : 5 = 2

      x-15  = 10

       x = 25

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

\(\frac{3}{2}\times\frac{4}{5}\times\frac{2}{6}\times\frac{15}{4}\)

\(=\frac{6}{5}\times\frac{2}{6}\times\frac{15}{4}\)

\(=\frac{2}{5}\times\frac{15}{4}\)

\(=\frac{3}{2}\)

3/2 x 4/5 x 2/6 x 15/4 = 3/2

\(\frac{3x4x2x15}{2x5x6x4}\)

Loại 4 x 2 

\(\frac{3x15}{5x6}=\frac{3x3x5}{5x3x2}\)

Loại 3 x 5 

=>\(\frac{3}{2}\)