\(6x^2-\left(2x-3\right)\left(3x-2\right)=1\)

\(\Rightarrow6x^2-\left(6x^2-4x-9x+4\right)=1\)

\(\Rightarrow6x^2-6x^2-13x+6=1\)

\(\Rightarrow-13x+6=1\)

\(\Rightarrow-13x=-7\)

\(\Rightarrow x=\frac{7}{13}\)

\(\left(x+1\right)^3-\left(x-1\right)\left(x^2+x+1\right)-2=0\)

\(\Rightarrow x^3+3x^2+3x+1-x^3+1-2=0\)

\(\Rightarrow3x^2+3x=0\)

\(\Rightarrow3x\left(x+1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}3x=0\\x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)

Sửa a)

\(6x^2-\left(2x-3\right)\left(3x-2\right)=1\)

\(\Rightarrow6x^2-6x^2+13x-6=1\)

\(\Rightarrow13x-6=1\)

\(\Rightarrow13x=7\)

\(\Rightarrow x=\frac{7}{13}\)

\(5x\left(x-3\right)-x+3=0\)

\(\Rightarrow5x\left(x-3\right)-\left(x-3\right)=0\)

\(\Rightarrow\left(x-3\right)\left(5x-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-3=0\\5x-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=\frac{1}{5}\end{cases}}\)

`5x  (x-3)-x+3=0`

`->5x (x-3)-(x-3)=0`

`-> (x-3)(5x-1)=0`

TH1 : `x-3=0 ->x=3`

TH2 : `5x-1=0 ->x=1/5`

Vậy `x=3,x=1/5`

`(2x^2 +1/3)^3`

`= (2x^2)^3  + 3 . (2x^2)^2 . 1/3 + 3 . 2x^2 . (1/3)^2 + (1/3)^3`

`= 8x^6 + 4x^4 + 6/9 x^2 + 1/27`

b) \(\left(3x+y\right)\left(9x^2-3xy+y^2\right)-\left(3x-y\right)\left(9x^2+3xy+y\right)\)

\(=\left(27x^3+y^3\right)-\left(27x^3-y^3\right)=27x^3+y^3-27x^3+y^3=2y^3\)

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

\(=\left(8x^3+12x^2+6x+1\right)+\left(8x^3-12x^2+6x-1\right)-2\)

\(=8x^3+12x^2+6x+1+8x^3-12x^2+6x-1-2=16x^3+12x-2\)

a) (Sửa đề) \(\left(x-3\right)\left(x^2-6x+9\right)\left(x+3\right)\left(x^2+6x+9\right)\)

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

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

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

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