Bài này mik giải đc rồi nha !

\(\left(2x-12\right)+\left(8x-19\right)-\left(9x+16\right)=4\)

\(\Leftrightarrow2x-12+8x-19-9x-16=4\)

\(\Leftrightarrow x=51\)Vậy x = 51 

Ta có: \(\left(2x-12\right)+\left(8x-19\right)-\left(9x+16\right)=4\)

     \(\Leftrightarrow2x-12+8x-19-9x-16=4\)

     \(\Leftrightarrow2x+8x-9x=4+12+16+19\)

     \(\Leftrightarrow x=51\)

Vậy............

\(P=\frac{2n-1}{n-1}=\frac{2\left(n-1\right)+1}{n-1}=\frac{1}{n-1}\)

hay \(n-1\inƯ\left(1\right)=\left\{\pm1\right\}\)

Ta có: \(P=\frac{2n-1}{n-1}\)

    \(\Leftrightarrow P=\frac{2n-2+1}{n-1}\)

    \(\Leftrightarrow P=2+\frac{1}{n-1}\)

Để \(P\inℤ\)\(\Rightarrow\)\(2+\frac{1}{n-1}\inℤ\)mà \(2\)nguyên 

\(\Rightarrow\)\(1⋮n-1\)\(\Rightarrow\)\(n-1\inƯ\left(1\right)\in\left\{\pm1\right\}\)

+ \(n-1=1\)\(\Leftrightarrow\)\(n=2\)

+ \(n-1=-1\)\(\Leftrightarrow\)\(n=0\)

Vậy ......

Ta có: \(B=\frac{3^{2021}-3^{2019}}{3^{2021}+3^{2020}}\)

    \(\Leftrightarrow B=\frac{3^{2019}.\left(3^2-1\right)}{3^{2020}.\left(3+1\right)}\)

    \(\Leftrightarrow B=\frac{8}{3.4}\)

    \(\Leftrightarrow B=\frac{2}{3}\)

Vậy \(B=\frac{2}{3}\)

Ta có : \(B=\frac{3^{2021}-3^{2019}}{3^{2021}+3^{2020}}\)

\(=\frac{3^{2019}\left(3^2-1\right)}{3^{2020}\left(3+1\right)}\)

\(=\frac{3^{2019}.8}{3^{2020}.4}\)

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