\(-32:\left(-2\right)^n=4\\ =>\left(-2\right)^n=\left(-32\right):4=-8=\left(-2\right)^3\\ =>n=3\)

\(\left|x\right|+x=\dfrac{1}{3}\left(1\right)\)

TH1 : \(x\ge0\)

\(\left(1\right)=>x+x=\dfrac{1}{3}\\ =>2x=\dfrac{1}{3}\\ =>x=\dfrac{1}{3}:2=\dfrac{1}{6}\left(TMDK\right)\)

\(TH2:x< 0\)

\(\left(1\right)=>-x+x=\dfrac{1}{3}\\ =>0=\dfrac{1}{3}\)( Vô lí )

Vậy `x=1/6`

\(9.27\le3n\le243\\ =>9.27:3\le3n:3\le243:3\\=>81\le n\le81\\ =>n=81\)

\(\dfrac{1}{2}.2^{n+4}.2^n=2^5\\ =>2^{n+4+n}=2^5:\dfrac{1}{2}\\ =>2^{2n+4}=2^5.2\\ =>2^{2n+4}=2^6\\ =>2n+4=6\\ =>2n=2=>n=1\)

Sửa đề : \(2^x+2^{x+3}=144\\ =>2^x.\left(1+2^3\right)=144\\ =>2^x=\dfrac{144}{9}=16=2^4\\ =>x=4\)

\(\dfrac{x-\dfrac{x}{12}+\dfrac{x}{18}}{7-\dfrac{7}{12}+\dfrac{7}{18}}=-\dfrac{4}{7}\\ \dfrac{x\left(1-\dfrac{1}{12}+\dfrac{1}{18}\right)}{7\left(1-\dfrac{1}{12}+\dfrac{1}{18}\right)}=-\dfrac{4}{7}\\ \dfrac{x}{7}=-\dfrac{4}{7}\\ x=-4\)

\(\dfrac{\dfrac{2}{3}-\dfrac{2}{5}+\dfrac{2}{7}}{\dfrac{8}{3}-\dfrac{8}{5}+\dfrac{8}{7}}=\dfrac{2\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{7}\right)}{8\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{7}\right)}\\ =\dfrac{2}{8}=\dfrac{1}{4}\)

Sửa đề : \(\left(2x-y\right)\left(4x^2+2xy+y^2\right)-8x^3+y^3+2023\)

\(=\left(2x\right)^3-y^3-8x^3+y^3+2023\\ =8x^3-y^3-8x^3+y^3+2023=2023\)

Do `2023` không phụ thuộc vào biến

Vậy nên bt B không phụ thuộc vào biến (DPCM)

\(\dfrac{11}{-13}=-\dfrac{11}{13}=-\dfrac{13}{13}+\dfrac{2}{13}=-1+\dfrac{2}{13}\\ -\dfrac{14}{15}=-\dfrac{15}{15}+\dfrac{1}{15}=-1+\dfrac{1}{15}\)

Ta thấy : \(\dfrac{1}{15}< \dfrac{1}{13}< \dfrac{2}{13}=>-1+\dfrac{1}{15}< -1+\dfrac{2}{13}\)

hay \(\dfrac{11}{-13}>-\dfrac{14}{15}\)

\(\dfrac{15}{19}\times\dfrac{38}{5}< x< \dfrac{67}{15}-\dfrac{56}{16}\\ \dfrac{5\times3\times19\times2}{19\times5}< x< \dfrac{67\times16-56\times15}{15\times16}\\ 3\times2< x< \dfrac{232}{240}\\ 6< x< \dfrac{232}{240}\) (*)

Mà : \(\dfrac{232}{240}< \dfrac{240}{240}=1\\ \)

Nên (*) là vô lí ( Do `6>232/240` )

Vậy \(x\in\varnothing\)