a) Ta có: \(x\left(x-2\right)-x^2+4=0\)
\(\Leftrightarrow x\left(x-2\right)-\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow-2\left(x-2\right)=0\)
\(\Leftrightarrow x-2=0\)
hay x=2
b) Ta có: \(5x^2-5x=x-1\)
\(\Leftrightarrow5x\left(x-1\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(5x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\)
c) Ta có: \(x+x^2-x^3-x^4=0\)
\(\Leftrightarrow x\left(1+x\right)-x^3\left(1+x\right)=0\)
\(\Leftrightarrow x\left(x+1\right)\left(1-x^2\right)=0\)
\(\Leftrightarrow x\left(x+1\right)^2\cdot\left(1-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=1\end{matrix}\right.\)
d) Ta có: \(2x^3+3x^2+2x+3=0\)
\(\Leftrightarrow x^2\left(2x+3\right)+\left(2x+3\right)=0\)
\(\Leftrightarrow2x+3=0\)
hay \(x=-\dfrac{3}{2}\)
e: Ta có: \(4x^2-25-\left(2x-5\right)\left(3x+7\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(2x+5\right)-\left(2x-5\right)\left(3x+7\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(2x+5-3x-7\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-2\end{matrix}\right.\)
f) Ta có: \(x^3+27+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x-9\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9+x-9\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-3\end{matrix}\right.\)
g) Ta có: \(3x^3-14-7x^2+6x=0\)
\(\Leftrightarrow3x\left(x^2+2\right)-7\left(x^2+2\right)=0\)
\(\Leftrightarrow3x-7=0\)
hay \(x=\dfrac{7}{3}\)
