a) Ta có: \(\left|2.5-x\right|=1.3\)

\(\Leftrightarrow\left|x-2.5\right|=1.3\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2.5=1.3\\x-2.5=-1.3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3.8\\x=-1.3+2.5=1.2\end{matrix}\right.\)

Vậy: \(x\in\left\{3.8;1.2\right\}\)

b) Ta có: \(1.6-\left|x-0.2\right|=0\)

\(\Leftrightarrow\left|x-0.2\right|=1.6\)

\(\Leftrightarrow\left[{}\begin{matrix}x-0.2=1.6\\x-0.2=-1.6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1.8\\x=-1.4\end{matrix}\right.\)

Vậy: \(x\in\left\{1.8;-1.4\right\}\)

c) Ta có: \(13^x=169\)

\(\Leftrightarrow13^x=13^2\)

\(\Leftrightarrow x=2\)

Vậy: x=2

d) Ta có: \(\dfrac{-2}{x}=\dfrac{-x}{\dfrac{8}{25}}\)

\(\Leftrightarrow\dfrac{2}{x}=\dfrac{x}{\dfrac{8}{25}}\)

\(\Leftrightarrow x^2=\dfrac{16}{25}\)

hay \(x\in\left\{\dfrac{4}{5};-\dfrac{4}{5}\right\}\)

Vậy: \(x\in\left\{\dfrac{4}{5};-\dfrac{4}{5}\right\}\)

\(|2,5-x|=1,3\)

\(\Rightarrow2,5-x=1,3\)    hoặc      -(2,5-x)=1,3

=>x=2,5-1,3                          x-2,5=1,3

=>x=1,2                                x=1,3+2,5=3,8

Vậy \(x\in\left\{1,2;3,8\right\}\)

\(1,6-|x-0,2|=0\Rightarrow|x-0,2|=1,6\)

=> x-0,2=1,6       hoặc       -(x-0,2)=1,6

=>x=1,6+0,2                      0,2-x=1,6

=>x=3,8                              x=0,2-1,6=-1,4

Vậy \(x\in\left\{3,8;-1,4\right\}\)

\(13^x=169\Rightarrow13^x=13^2\Rightarrow x=2\)

\(\dfrac{-2}{x}=\dfrac{-x}{\dfrac{8}{25}}\Rightarrow-2\times\dfrac{8}{25}=-x\times x\Rightarrow\dfrac{-16}{25}=-x^2\Rightarrow\dfrac{16}{25}=x^2\Rightarrow x^2=\left(\dfrac{4}{5}\right)^2=\left(-\dfrac{4}{5}\right)^2\)

\(\Rightarrow x=\dfrac{4}{5}\) hoặc \(x=-\dfrac{4}{5}\)

\(x=\dfrac{35.16}{-28}\)

\(x=-20\)

\(\dfrac{-28}{35}=\dfrac{16}{x}\\ \Rightarrow-28\cdot x=35\cdot16=560\\ x=560:-28\\ x=-20\)

\(x=\dfrac{35.16}{-28}=20\)

\(x=\dfrac{3}{5}-\dfrac{7}{8}=-\dfrac{11}{40}\)

x = 3/5 - 7/8

x =  -11/40

\(\dfrac{7}{8}+x=\dfrac{3}{5}\)

       \(x=\dfrac{3}{5}-\dfrac{7}{8}\)

      \(x=-\dfrac{11}{40}\)

x = 3/4

\(x=\dfrac{5}{8}+\dfrac{1}{8}\)

\(x=\dfrac{6}{8}=\dfrac{3}{4}\)

x = 3/4

\(\dfrac{3}{x-2}=\dfrac{-2}{x-4}\left(dk:x\ne2;x\ne4\right)\)

\(\Rightarrow3\cdot\left(x-4\right)=-2\cdot\left(x-2\right)\)

\(\Rightarrow3x-12=-2x+4\)

\(\Rightarrow3x+2x=4+12\)

\(\Rightarrow5x=16\)

\(\Rightarrow x=\dfrac{16}{5}\left(tm\right)\)

\(ĐK:x\ne2;x\ne4\\ Có:\dfrac{3}{x-2}=\dfrac{-2}{x-4}\\ \Leftrightarrow3\left(x-4\right)=-2\left(x-2\right)\\ \Leftrightarrow3x-12=-2x+4\\ \Leftrightarrow3x+2x=4+12\\ \Leftrightarrow5x=16\\ \Leftrightarrow x=\dfrac{16}{5}\left(TM\right)\\ Vậy:x=\dfrac{16}{5}\)

\(x=\dfrac{4}{27}-\dfrac{2}{3}\)

\(x=-\dfrac{14}{27}\)

\(x=\dfrac{4}{27}-\dfrac{2}{3}\)

\(x=-\dfrac{14}{27}\)

x=−1427

=> 1/x = 5/6 - y/3

1/x = 5-2y/6

=> x(5-2y) = 1.6 = 6

Do x ∈ N => x >= 0

Mà 6>0 => 5-2y > 0

Vì y ∈ N => 5-2y ∈ N*

Ta có bảng:

Do x,y ∈ N => (x,y) = (6,2) (thử lại thỏa mãn)

Vậy x=6; y = 2

=>6/x=32/35

=>x=6:32/35=6*35/32=210/32=105/16