a) \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x-2\right)^2=27.\)

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

\(\Leftrightarrow x^3-x^3+4x^2-4x=0.\)

\(\Leftrightarrow4x\left(x-1\right)=0.\)

\(\Leftrightarrow\left[{}\begin{matrix}4x=0.\\x-1=0.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0.\\x=1.\end{matrix}\right.\)

Vậy \(S=\left\{0;1\right\}.\)

a) Ta biến đổi về x = 2 7 : 2 3

Thực hiện phép chia thu được x =  3 7

b ) x = 2 3 . c ) x = 8 7 . d ) x = 6 7 .

a) \(8x^3+27=\left(2x+3\right)\left(4x^2-6x+9\right)\)

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

c) \(x^4-2x^3+x^2-2x=x^3\left(x-2\right)+x\left(x-2\right)=x\left(x-2\right)\left(x^2-1\right)=x\left(x-2\right)\left(x-1\right)\left(x+1\right)\)

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

a) \(x+\dfrac{4}{9}=\dfrac{5}{27}\)  

    \(x=\dfrac{5}{27}-\dfrac{4}{9}\)

   \(x=-\dfrac{7}{27}\)

b) \(x-\dfrac{4}{11}=\dfrac{7}{33}\)

   \(x=\dfrac{7}{33}+\dfrac{4}{11}\)

   \(x=\dfrac{19}{33}\)

c) \(\dfrac{8}{5}-x=\dfrac{1}{3}\times\dfrac{2}{5}\)

  \(\dfrac{8}{5}-x=\dfrac{2}{15}\)

          \(x=\dfrac{8}{5}-\dfrac{2}{15}\)

          \(x=\dfrac{22}{15}\)

d) \(x-\dfrac{3}{4}=\dfrac{1}{2}+\dfrac{2}{6}\)

   \(x-\dfrac{3}{4}=\dfrac{5}{6}\)

   \(x=\dfrac{5}{6}+\dfrac{3}{4}\)

   \(z=\dfrac{19}{12}\)

Ví dụ : 5/2 = 5 phần 2

c: Ta có: \(\left(x+3\right)^3-x\left(3x+1\right)^2+\left(2x+1\right)\left(4x^2-2x+1\right)=28\)

\(\Leftrightarrow x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3+1=28\)

\(\Leftrightarrow3x^2+26x=0\)

\(\Leftrightarrow x\left(3x+26\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{26}{3}\end{matrix}\right.\)

\(a,\Leftrightarrow x^2+8x+16-x^3-12x^2=16\\ \Leftrightarrow x^3+11x^2-8x=0\\ \Leftrightarrow x\left(x^2+11x-8\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x^2+11x-8=0\left(1\right)\end{matrix}\right.\\ \Delta\left(1\right)=121+32=153\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-11-3\sqrt{17}}{2}\\x=\dfrac{-11+3\sqrt{17}}{2}\end{matrix}\right.\\ S=\left\{0;\dfrac{-11-3\sqrt{17}}{2};\dfrac{-11+3\sqrt{17}}{2}\right\}\)

\(c,\Leftrightarrow x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3+1=28\\ \Leftrightarrow3x^2+26x=0\\ \Leftrightarrow x\left(3x+26\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{26}{3}\end{matrix}\right.\\ d,\Leftrightarrow x^3-6x^2+12x-8-x^3-125-6x^2=11\\ \Leftrightarrow-12x^2+12x-144=0\\ \Leftrightarrow x^2-x+12=0\Leftrightarrow\left[{}\begin{matrix}x=4\\x=3\end{matrix}\right.\)

a: \(\left(x-1\right)^3+27\)

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

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

b: \(\left(x-2\right)^3-8\)

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

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

Lời giải:

a. 

$(25-2x)^3:5-3^2=4^2$

$(25-2x)^3:5=4^2+3^2=25$

$(25-2x)^3=25.5=5^3$

$\Rightarrow 25-2x=5$

$\Rightarrow 2x=20$

$\Rightarrow x=10$

b.

$2.3^x=10.3^{12}+8.27^4=10.3^{12}+8.3^{12}=18.3^{12}=2.3^{14}$

$\Rightarrow 3^x=3^{14}$

$\Rightarrow x=14$