\(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x^2-9\right)=27\\ x.x^2-x.3x+x.9-x.x^2+x.9=27\\ x^3-3x^2+9x-x^3+9x=27\\ 3x^2+18x=27\\ 21x^2=27\\ x^2=\dfrac{9}{7}\\ \Rightarrow x=\sqrt{\dfrac{9}{7}}\)

\(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x^2-9\right)=27\\ x.x^2-x.3x+x.9+3.x^2-3.3x+3.9-x.x^2+x.9=27\\ x^3-3x^2+9x+3x^2-9x+27-x^3+9x=27\\ 9x+27=27\\ 9x=0\\ x=0\)

   (x + 3)(x² - 3x + 9) - x(x - 2)² = 27

\(\Leftrightarrow\) x³ + 27 - x( x² - 4x + 4) = 27

\(\Leftrightarrow\) x³ + 27 - x³ + 4x² - 4x - 27 = 0

\(\Leftrightarrow\) 4x² - 4x = 0

\(\Leftrightarrow\) 4x ( x - 1) = 0

khi 4x = 0 hoặc x - 1 = 0

\(\Leftrightarrow\)  x = 0        \(\Leftrightarrow\) x = 1

 Chúc bạn học tốt

\(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x-2\right)^2=27\\ x.x^2-x.3x+x.9+3.x^2-3.3x+3.9-x.x^2+x.2^2=27\\ x^3-3x^2+9x+3x^2-9x+27-x^3+4x=27\\ 4x+27=27\\ 4x=0\\ x=0\)

x³+27+(x+3).(x-9)=0 nha mk ghi nhầm 

\(x^3+27+\left(x+3\right)\left(x-9\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-3\end{matrix}\right.\)

a: Ta có: \(4x^2\left(x-2\right)-x+2=0\)

\(\Leftrightarrow\left(x-2\right)\left(4x^2-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)

b: Ta có: \(x^3+27+\left(x+3\right)\left(x-9\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9+x-9\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=0\\x=2\end{matrix}\right.\)

ĐKXĐ: \(x=\pm3\)

Nếu \(x=3\), phương trình tương đương 

\(x^3+\sqrt{x^2-9}-\sqrt{9-x^2}-27=0\)

\(\Leftrightarrow0=0\)

\(\Rightarrow x=3\) là nghiệm của phương trình 

Nếu \(x=-3\), phương trình tương đương

\(x^3+\sqrt{x^2-9}-\sqrt{9-x^2}-27=0\)

\(\Leftrightarrow-54=0\)

\(\Rightarrow x=-3\) không phải là nghiệm của phương trình

Vậy ...

\(a,\Leftrightarrow\left(x-2\right)^3-3x\left(x-2\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x-2-3x\right)=0\\ \Leftrightarrow\left(x-2\right)\left(-2x-2\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\\ b,\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\\ \Leftrightarrow\left(x+3\right)\left(x^2-2x\right)=0\\ \Leftrightarrow x\left(x-2\right)\left(x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-3\end{matrix}\right.\)

j mà hjhj thế nguyenthuy

j mà j mà hjhj thế nguyenthuy thế trieu dang

\(\frac{3^x}{27}=27\Rightarrow\frac{3^x}{27}=\frac{729}{27}=\frac{3^6}{27}\)

\(\Rightarrow3^x=3^6\Rightarrow x=6\)

\(\Leftrightarrow x\cdot\dfrac{1}{3}+2=-\dfrac{2}{3}\)

=>x*1/3=-8/3

hay x=-8