Ta có :\(x^{20}=9.\left(x^2\right)^9\)

\(\Rightarrow x^{20}=9.x^{18}\)

Ta lại có: \(x^{20}=x^{18}.x^2=9.x^{18}\)

\(\Rightarrow x^2=9\)(x^18:x^`8 là =1 ko cần ghi)

\(\Rightarrow x=\sqrt{9}\)

\(\Rightarrow x=\)3 hoặc \(x=\)-3

Vậy: \(x=3;-3\)

\(\frac{x}{3\frac{1}{2}.2\frac{2}{3}}=\frac{9}{56}\Rightarrow\frac{x}{\frac{7}{2}.\frac{8}{3}}=\frac{9}{56}\Rightarrow x=\frac{9}{56}.\frac{28}{3}=\frac{3}{2}\)

\(x:\left(3\frac{1}{2}.2\frac{2}{3}\right)=\frac{9}{56}\)

\(x:\left(\frac{7}{2}.\frac{8}{3}\right)=\frac{9}{56}\)

\(x:\left(\frac{7.4}{3}\right)=\frac{9}{56}\)

\(x.\frac{3}{28}=\frac{9}{56}\)

\(x=\frac{9}{56}.\frac{28}{3}=\frac{3}{2}\)

Vậy \(x=\frac{3}{2}\)

\(2x-1\left(x+2\right)-3\left(x+2\right)=0\)

\(\Rightarrow\left(2x-4\right)\left(x+2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x-4=0\\x+2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2\\x=-2\end{cases}}}\)

(2x-1)(x+2)-3(x+2)=0

<=>2x²+3x-2-3x-6=0

<=>2x²-8=0

<=>2(x²-4)=0

<=>x²-4=0

<=>(x+2)(x-2)=0

=>\(\orbr{\begin{cases}x+2=0\\x-2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-2\\x=2\end{cases}}}\)

Vậy...

ta có \(x+3⋮x^2+1\Rightarrow\left(x+3\right)\left(x-3\right)⋮x^2+1\Rightarrow x^2-9⋮x^2+1\)

=> \(x^2+1-10⋮x^2+1\Rightarrow10⋮x^2+1\)

Em lập bảng tìm x và nhở thử lại đấy

Nhầm rồi phần e 

e, \(\frac{4}{9}-\frac{7}{8}.x=\frac{-2}{3}\) 

              \(\frac{7}{8}.x=\frac{4}{9}-\frac{-2}{3}\) 

               \(\frac{7}{8}.x=\frac{-10}{9}\)

                      \(x=\frac{-10}{9}:\frac{7}{8}\)

                      \(x=\frac{-80}{63}\)