a,( 2x-3)2-(x+5)2 = 0
=> (2x-3-x-5)(2x-3+x+5)=0
=> (x-8)(3x+2)=0
=> \(\left[{}\begin{matrix}x-8=0\\3x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=8\\3x=-2\Rightarrow x=\dfrac{-2}{3}\end{matrix}\right.\)
vậy x=8 hoăc x=\(\dfrac{-2}{3}\)
b, (x3-x2) - 4x2+8x-4 =0
=> x2(x-1)-(4x2-8x+4)=0
=> x2(x-1)-4(x2-2x+1)=0
=> x2(x-1)-4(x-1)2=0
=> (x-1)[x2-4(x-1)]=0
=> (x-1)(x2-4x+4)=0
=> (x-1)(x-2)2=0
=> \(\left[{}\begin{matrix}x-1=0\\x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
vậy x=1; x=2
