3x³ - 27 x = 0

=> 3x ( x² - 9 ) = 0

\(=>\orbr{\begin{cases}3x=0\\x^2-9=0\end{cases}}\) 

\(=>\orbr{\begin{cases}x=0\\x^2=9\end{cases}}\)

=> x = + 3

hoặc x = 0

\(3x^2-27x=0\)

\(\Leftrightarrow3x\left(x+3\right)\left(x-3\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x=\pm3\end{cases}}\)

      \(a\left(b^2-c^2\right)+b\left(c^2-a^2\right)+c\left(a^2-b^2\right)\)

\(=a\left(b^2-c^2\right)+b\left(c^2-b^2\right)+b\left(b^2-a^2\right)+c\left(a^2-b^2\right)\)

\(=\left(a-b\right)\left(b^2-c^2\right)+\left(c-b\right)\left(a^2-b^2\right)\)

\(=\left(a-b\right)\left(b-c\right)\left(b+c\right)+\left(c-b\right)\left(a-b\right)\left(a+b\right)\)

\(=\left(a-b\right)\left(b-c\right)\left(b+c-a-b\right)\)

\(=\left(a-b\right)\left(b-c\right)\left(c-a\right)\)

      \(2x^2+y^2+2xy-8x-6y+10=0\)

\(\Rightarrow2.\left(2x^2+y^2+2xy-8x-6y+10\right)=0\)

\(\Rightarrow4x^2+2y^2+4xy-16x-12y+20=0\)

\(\Rightarrow\left(4x^2+y^2+16+4xy-8y-16x\right)+\left(y^2-4y+4\right)=0\)

\(\Rightarrow\left(2x+y-4\right)^2+\left(y-2\right)^2=0\left(1\right)\)

Ta có: \(\hept{\begin{cases}\left(2x+y-4\right)^2\ge0\forall x;y\\\left(y-2\right)^2\ge0\forall y\end{cases}\Rightarrow\left(2x+y-4\right)^2+\left(y-2\right)^2\ge0\forall x;y\left(2\right)}\)

Từ (1) và (2) \(\Rightarrow\hept{\begin{cases}2x+y-4=0\\y-2=0\end{cases}\Rightarrow\hept{\begin{cases}2x+y=4\\y=2\end{cases}\Rightarrow}\hept{\begin{cases}2x+2=4\\y=2\end{cases}\Rightarrow}\hept{\begin{cases}x=1\\y=2\end{cases}}}\)

Chúc bạn học tốt.

\(x^2-14x+13=0\)

\(\Rightarrow x^2-2x.7+7^2-7^2+13=0\)

\(\Rightarrow\left(x^2-2x.7+7^2\right)-7^2+13=0\)

\(\Rightarrow\left(x-7\right)^2-49+13=0\)

\(\Rightarrow\left(x-7\right)^2-36=0\)

\(\Rightarrow\left(x-7\right)^2=36\)

\(\Rightarrow\left(x-7\right)^2=\pm6^2\)

\(\Rightarrow\orbr{\begin{cases}x-7=6\\x-7=-6\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=13\\x=1\end{cases}}\)

Vậy ...

\(x^2-14x+13=0\)

\(x^2-14x+49-36=0\)

\(\left(x^2-14x+19\right)-36=0\)

\(\left(x-7\right)^2-6^2=0\)

\(\left(x-7-6\right)\left(x-7+6\right)=0\)

\(\left(x-13\right)\left(x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-13=0\\x-1=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=13\\x=1\end{cases}}\)