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

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

minh@@22@22222@@@@? 

Thiếu đề rồi bạn

\(\dfrac{x}{x^2-9}+\dfrac{2}{x^2+6x+9}=\dfrac{x\left(x+3\right)+2\left(x-3\right)}{\left(x-3\right)\left(x+3\right)^2}\\ =\dfrac{x^2+5x-6}{\left(x-3\right)\left(x+3\right)^2}=\dfrac{\left(x-1\right)\left(x+6\right)}{\left(x-3\right)\left(x+3\right)^2}\)

\(1.x-\dfrac{2}{3}\times\left(x+9\right)=1\)

\(x-\dfrac{2}{3}\times x-6=1\)

\(x\times\left(1-\dfrac{2}{3}\right)=7\)

\(x\times\dfrac{1}{3}=7\)

\(x=21\)

\(2.x-\dfrac{11}{15}=\dfrac{3+x}{5}\)

\(\dfrac{15x}{15}-\dfrac{11}{15}=\dfrac{9+3x}{15}\)

\(15x-11=9+3x\)

\(12x=20\)

\(x=\dfrac{5}{3}\)

giải giúp mk vs ạ

bn ơi mk rất mún giải hộ pạn nhưng mk k bít để giải xin lỗi pạn nhìu

k mk nhé

ahihu tk k ai pít giải à 

   (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

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

\(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x^2-2\right)=15\)

\(\Leftrightarrow x^3+27-x^3+2x=15\)

\(\Leftrightarrow2x=-12\)

\(\Leftrightarrow x=-6\)

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

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

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

Ta có:(x-3)(x²+3x+9)-x(x²-4)=1

     => x³-27-x³+4x=1

     =>4x=28=>x=7

= 0