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

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

=x³+3³+(x+3)(x-9)

=(x+3)(x²-3x+9)+(x+3)(x-9)

=(x+3)(x²-3x+9+x-9)

=(x+3)(x²-2x)

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

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

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

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

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

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

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

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

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

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

<=> (x+3)(x²+3x+9)+(x+3)(x - 9)

<=> (x+3)(x²-3x+9+x - 9)=0

<=> (x+3)(x²-2x)=0

<=> (x+3)x(x-2)=0

\(\Rightarrow\left[\begin{array}{nghiempt}x=0\\x=-3\\x=2\end{array}\right.\)

\(x^3+27+\left(x+3\right)\left(x-9\right)\)=0
\(\Leftrightarrow x^3+3^3+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9+x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-2x\right)=0\)
\(\Leftrightarrow x\left(x+3\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x+3=0\\x-2=0\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=-3\\x=2\end{array}\right.\)
Vậy x=0 hoặc x=-3 hoặc x=2

Đáp án của bạn Nguyễn Tiến Hải còn thiếu trường hợp:

(x + 1/2)² = 25/4

TH1: x + 1/2 = 5/2 và giải như bạn Hải

TH2: x + 1/2 = -5/2

         x = -3

=> (x+3)(x²-3x+9) + (x+3)(x-9) =0
=> (x+3)(x²-2x)=0 => (x+3)(x-2)x=0
=> x=-3 hoặc x=2 hoặc x=0

9^x : 3^x = 27⁵

(9:3)^x = (3³)⁵

3^x = 3¹⁵

Vậy x = 15              

27^5.3^x=9^10

3^x=9^10:27^5

3^x=3^5

=> x=5

27⁵.3^x=9¹⁰

=>3¹⁵.3^x=3³⁰

=>3^x=3³⁰:3¹⁵

=>3^x=3¹⁵

=>x=15

27⁵.3^x=9¹⁰

(3³)⁵.3^x = (3²)¹⁰

3¹⁵.3^x = 3²⁰

3^x = 3²⁰:3¹⁵ (3^20-15)

3^x = 3⁵

<=> x = 5

a) \(4x^3-36x=0\)

\(\Leftrightarrow4x\left(x^2-9\right)=0\)

\(\Leftrightarrow4x\left(x+3\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}4x=0\\x+3=0\\x-3=0\end{matrix}\right.\)

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

b) \(\left(x-2\right)^2-4x+8=0\)

\(\Leftrightarrow\left(x-2\right)^2-\left(4x-8\right)=0\)

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

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

\(\Leftrightarrow\left(x-2\right)\left(x-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-6=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=6\end{matrix}\right.\)

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

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

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

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

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

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

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

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