1) \(\dfrac{1}{27}+a^3=\left(\dfrac{1}{3}+a\right)\left(\dfrac{1}{9}-\dfrac{a}{3}+a^2\right)\)

2) \(=\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)\)

3) \(=\left(\dfrac{1}{2}x+2y\right)\left(\dfrac{1}{4}x-xy+4y^2\right)\)

4) \(=\left(x^2+1\right)\left(x^4-x^2+1\right)\)

5) \(=\left(x^3+1\right)\left(x^6-x^3+1\right)\)

6) \(=\left(x-4\right)\left(x^2+4x+16\right)\)

7) \(=\left(x-5\right)\left(x^2+5x+25\right)\)

8) \(=\left(2x^2-3y\right)\left(4x^4+6x^2y+9y^2\right)\)

9) \(=\left(\dfrac{1}{4}x^2-5y\right)\left(\dfrac{1}{16}x^4+\dfrac{5}{4}x^2y+25y^2\right)\)

10) \(=\left(\dfrac{1}{2}x-2\right)\left(\dfrac{1}{4}x^2+x+4\right)\)

11) \(=\left(x+2\right)^3\)

12) \(=\left(x+3\right)^3\)

\(\left(x-\dfrac{1}{2}\right)^3=\dfrac{1}{27}\)

\(\Rightarrow\left(x-\dfrac{1}{2}\right)=\dfrac{1}{3}\)

\(\Rightarrow x=\dfrac{1}{3}+\dfrac{1}{2}\)

Vậy : \(x=\dfrac{5}{6}\)

\(x=\dfrac{4}{27}-\dfrac{2}{3}\)

\(x=-\dfrac{14}{27}\)

\(x=\dfrac{4}{27}-\dfrac{2}{3}\)

\(x=-\dfrac{14}{27}\)

x=−1427

Sửa đề: \(\dfrac{2x-1}{3}=\dfrac{27}{2x-1}\)

ĐKXĐ: x<>1/2

\(\dfrac{2x-1}{3}=\dfrac{27}{2x-1}\)

=>\(\left(2x-1\right)^2=3\cdot27=81\)

=>\(\left[{}\begin{matrix}2x-1=9\\2x-1=-9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=10\\2x=-8\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=5\left(nhận\right)\\x=-4\left(nhận\right)\end{matrix}\right.\)