d) x2(x−3)+27−9x=0
x2 (x - 3) + 27 - 9x = 0
x2 ( x - 3 ) +( 27 - 9x) = 0
x2 ( x - 3) + 9 ( 3 - x ) = 0
x2 ( x - 3) + 9 \(\left[-\left(x-3\right)\right]\) = 0
x2 ( x - 3) - 9 ( x - 3) = 0
(x - 3 )( x2 - 9) = 0
(x - 3) ( x-3) ( x + 3) = 0
(x - 3)2 ( x + 3) = 0
\(\Rightarrow\) x - 3 = 0 hoặc x+ 3 =0
+) x - 3 = 0 \(\Rightarrow\) x = 0+ 3=3
+) x+3= 0 \(\Rightarrow\) x = 0-3 = -3
Vậy: x= 3 hoặc x= -3
\(x^2\left(x-3\right)+27-9x=0\)
\(\Rightarrow\)\(x^2\left(x-3\right)+\left(x-3\right).9\)\(=0\)
\(\Rightarrow\)(\(x-3\))\(\left(x^2+9\right)=0\)
\(\Rightarrow\)x-3=0
\(\Rightarrow\)x=3
=>x^2(x-3)-9(x-3)=0
=>(x-3)^2(x+3)=0
=>x=3 hoặc x=-3
