(\(\dfrac{1}{3}\))².\(\dfrac{3}{7}\) + 1\(\dfrac{2}{3}\) : \(\dfrac{-7}{6}\)

= \(\dfrac{1}{9}\).\(\dfrac{3}{7}\)    + \(\dfrac{5}{3}\).\(\dfrac{6}{-7}\)

=    \(\dfrac{1}{21}\)    - \(\dfrac{10}{7}\)

= \(\dfrac{-29}{21}\) 

2,5+7,5x=-5

=>7,5x=-5-2,5=-7,5

=>x=-1

2,5 + 7,5\(x\) = -5

         7,5\(x\) = -5 - 2,5

         7,5\(x\) = -7,5

              \(x\) = -7,5 : 7,5

              \(x\) = -1

       Vậy \(x=-1\)

\(\dfrac{1}{6}x-\dfrac{3}{8}=\dfrac{1}{4}\)

=>\(\dfrac{1}{6}x=\dfrac{1}{4}+\dfrac{3}{8}=\dfrac{5}{8}\)

=>\(x=\dfrac{5}{8}:\dfrac{1}{6}=\dfrac{5}{8}\cdot6=\dfrac{30}{8}=\dfrac{15}{4}\)

\(121212:\left(x-2020\right)=12\)

=>x-2020=121212:12=10101

=>x=10101+2020=12121

\(125\%\cdot\left(-\dfrac{1}{2}\right)^2\cdot\left(\dfrac{15}{16}+1.5\right)+2016^0\)

\(=\dfrac{5}{4}\cdot\dfrac{1}{4}\cdot\left(\dfrac{15}{16}+\dfrac{24}{16}\right)+1\)

\(=\dfrac{5}{16}\cdot\dfrac{39}{16}+1=\dfrac{195}{256}+1=\dfrac{451}{256}\)

ĐKXĐ: x<>-1

\(\dfrac{x+1}{2}=\dfrac{8}{x+1}\)

=>\(\left(x+1\right)^2=2\cdot8=16\)

=>\(\left[{}\begin{matrix}x+1=4\\x+1=-4\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=3\left(nhận\right)\\x=-5\left(nhận\right)\end{matrix}\right.\)

a; \(\dfrac{x+1}{2}\) = \(\dfrac{8}{x+1}\) (\(x\ne\) -1)

    (\(x\) + 1)² = 8.2

   (\(x\) + 1)² = 16

    \(\left[{}\begin{matrix}x+1=-4\\x+1=4\end{matrix}\right.\)

     \(\left[{}\begin{matrix}x=-5\\x=3\end{matrix}\right.\)

     Vậy \(x\in\) {-5; 3}

\(M=1+2+2^2+2^3+...+2^{2022}+2^{2023}\)

\(=\left(1+2\right)+\left(2^2+2^3\right)+...+\left(2^{2022}+2^{2023}\right)\)

\(=\left(1+2\right)+2^2\left(1+2\right)+...+2^{2022}\left(1+2\right)\)

\(=3\left(1+2^2+...+2^{2022}\right)⋮3\)

-5/13 + 2/5 + -8/13 + 3/5

= (-5/13 + -8/13 ) + (2/5 + 3/5)

= - 1 + 1

= 0

0,5 . 7/13 + 0,5 . 9/13 - 0,5 . 3/13

= 0,5 . ( 7/13 + 9/13 - 3/13)

= 0,5 . 1

= 0,5

\(\dfrac{2x+1}{2}=\dfrac{8+x}{x}\)

x (2x +1) = 2 (8 +x)

2x² + x = 16 + 2x

2x² -x - 16 = 0

=> \(x=\dfrac{\sqrt{129}+1}{4}\)

\(x=\dfrac{-\sqrt{129}+1}{4}\)