a)
\(x\left(x-2\right)-x+2=0\\ \Leftrightarrow-x\left(2-x\right)+\left(2-x\right)=0\\ \Leftrightarrow\left(2-x\right)\left(1-x\right)=0\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}2-x=0\\1-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=1\end{matrix}\right.\)
Vậy...
b)
\(x^2\left(x^2+1\right)-x^2-1=0\\ \Leftrightarrow x^2\left(x^2+1\right)-\left(x^2+1\right)=0\\ \Leftrightarrow\left(x^2+1\right)\left(x^2-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+1=0\\x^2-1=0\end{matrix}\right.\)
Vì \(x^2+1\) luôn lớn hơn 0 với mọi x
\(\Rightarrow x^2-1=0\\ \Leftrightarrow x=1\)
Vậy...
c)
\(5x(x-3)^2-5(x-1)^3+15(x+2)(x-2)=5\)
\(\Leftrightarrow x(x-3)^2-(x-1)^3+3(x+2)(x-2)=1\)
\(\Leftrightarrow x(x^2-6x+9)-(x^3-3x^2+3x-1)+3(x^2-4)=1\)
\(\Leftrightarrow 6x-12=0\Rightarrow x=2\)
c)
\(5x\left(x-3\right)^2-5\left(x-1\right)^3+15\left(x+2\right)\left(x-2\right)=5\\ \Leftrightarrow5x\left(x^2-6x+9\right)-5\left(x^3-3x^2+3x-1\right)+15\left(x^2-4\right)=5\\ \Leftrightarrow5x^3-30x^2+45x-5x^3+15x^2-15x+5+15x^2-60=5\\ \Leftrightarrow\left(5x^3-5x^3\right)+\left(-30x^2+15x^2+15x^2\right)+\left(45x-15x\right)=60\\\Leftrightarrow 30x=60\\ \Leftrightarrow x=2\)Vậy...
