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\(\Rightarrow3x\left(x^2-25\right)=0\\ \Rightarrow3x\left(x-5\right)\left(x+5\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\)
\(\left(x^2-9\right)^2-\left(x-3\right)^2=0\)
\(\Leftrightarrow\left(x-3\right)^2\left(x+3\right)^2-\left(x-3\right)^2=0\)
\(\Leftrightarrow\left(x-3\right)^2\left[\left(x+3\right)^2-1\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-3\right)^2=0\\\left(x+3\right)^2=1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+3=1\\x+3=-1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\\x=-4\end{matrix}\right.\)
\(\Leftrightarrow4x^2+x-8x-2-4x^2-27x=1\)
=>-34x=3
hay x=-3/34
Ta có: \(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+9\left(x+1\right)^2=15\)
\(\Leftrightarrow x^3-9x^2+27x-27-x^3+27+9\left(x^2+2x+1\right)=15\)
\(\Leftrightarrow-9x^2+27x+9x^2+18x+9=15\)
\(\Leftrightarrow45x=6\)
hay \(x=\dfrac{2}{15}\)
\(3x\left(x-y\right)+x-y\)
\(=3x\left(x-y\right)+1\left(x-y\right)\)
\(=\left(x-y\right)\left(3x+1\right)\)
\(2x-1-x^2\\ =x+x-1-x^2\\ =\left(x-x^2\right)+\left(x-1\right)\\ =-x\left(x-1\right)+\left(x-1\right)\\ =\left(x-1\right)\left(1-x\right)\)
(x3-3x2)+(x-3)
x2(x-3)+(x-3)
(x-3)(x2+1)
x3-3x2+x-3=0
=(x3+x)-(3x2+3)
=x(x2+1)-3(x2+1)
=(x2+1)(x-3)
=>x2+1=0 hoặc x-3=0
=>x2=-1(vô nghiệm) hoặc x=3
Vậy x=3