\(-\dfrac{17}{14}:\left(-\dfrac{34}{7}\right)+\dfrac{10}{3}\left(\dfrac{1}{5}-\dfrac{3}{4}\right)-\dfrac{6}{5}\left(\dfrac{1}{3}-\dfrac{5}{6}\right)\)

\(=\dfrac{17}{14}.\dfrac{7}{34}+\dfrac{10}{3}.\left(-\dfrac{11}{20}\right)-\dfrac{6}{5}.\left(-\dfrac{9}{18}\right)\)

\(=\dfrac{1}{4}-\dfrac{11}{6}+\dfrac{9}{15}\)

\(=\dfrac{15-110+36}{60}\)

\(=-\dfrac{59}{60}\)

Ta thấy chỉ có \(n=3\) thỏa mãn đẳng thức \(\left(n+5\right)^2=64\left(n-2\right)^3\)vì

- \(\left(n+5\right)^2\) là 1 số chính phương

- \(64\) là 1 số chính phương

- \(\left(n-2\right)^3\) không phải số chính phương

- \(\)\(\left(n+5\right)^2< 64\left(n-2\right)^3,\forall n>3\)

\(\sqrt{\dfrac{4}{81}}\div\sqrt{\dfrac{25}{81}}-1\dfrac{3}{5}=\dfrac{2}{9}\div\dfrac{5}{9}-\dfrac{8}{5}\)

                                 \(=\dfrac{2}{5}-\dfrac{8}{5}=\dfrac{-6}{5}\)

\(\dfrac{1}{27}-x^8=\dfrac{1}{3}\\ x^8=\dfrac{1}{27}-\dfrac{1}{3}\\ \Leftrightarrow x^8=-\dfrac{8}{27}\)

Mà \(x^8\ge0\Rightarrow\) phương trình vô nghiệm

F(x)=6²+5x+8+3x-3x²+3x³

      =(36+8)+(5x+3x)-3x²+3x³

      =3x³-3x²+8x+44

G(x)=12x²-6-9x²+3x³

       =3x³+(12x²-9x²)-6

       =3x³+3x²-6

F(x)+G(x)=3x³-3x²+8x+44+3x³+3x²-6

                =(3x³+3x³)+(-3x²+3x²)+8x+(44-6)

                =6x³+8x+38

\(F\left(x\right)=G\left(x\right)\\ \Rightarrow6^2-5x+8+3x-3x^2+3x^3=12x^2-6-9x^2+3x^3\\ \Leftrightarrow-3x^2-2x+44=3x^2-6\\ \Leftrightarrow6x^2+2x-50=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1+\sqrt{301}}{6}\\x=\dfrac{-1-\sqrt{301}}{6}\end{matrix}\right.\)

\(P=\left(x+5\right)\left(ax^2+bx+25\right)\) (Sửa =25 thành +25)

\(Q=x^3+125=x^3+5^3=\left(x+5\right)\left(x^2-5x+25\right)\) (Sửa =25 thành +25)

Để \(P=Q\Rightarrow\left(ax^2+bx+25\right)=\left(x^2-5x+25\right)\)

\(\Rightarrow\left\{{}\begin{matrix}a=1\\b=-5\end{matrix}\right.\)