a: \(\dfrac{\left(x^2+2x+1\right)-x-1}{3}=\dfrac{6\left(x+1\right)^2-5x-5}{6}\)
=>\(\dfrac{x^2+x}{3}=\dfrac{6\left(x^2+2x+1\right)-5x-5}{6}\)
=>\(\dfrac{2x^2+2x}{6}=\dfrac{6x^2+12x+6-5x-5}{6}\)
=>\(\dfrac{6x^2+7x+1}{6}=\dfrac{2x^2+2x}{6}\)
=>\(6x^2+7x+1=2x^2+2x\)
=>\(4x^2+5x+1=0\)
=>(x+1)(4x+1)=0
=>\(\left[{}\begin{matrix}x=-1\\x=-\dfrac{1}{4}\end{matrix}\right.\)
b: Đề thiếu vế phải rồi bạn
c: \(\left(x-2\right)^2=\left(2x-3\right)^2-\left(x+1\right)^2\)
=>\(\left(x-2\right)^2=\left(2x-3-x-1\right)\left(2x-3+x+1\right)\)
=>\(\left(x-2\right)^2=\left(x-4\right)\left(3x-2\right)\)
=>\(3x^2-2x-12x+8=x^2-4x+4\)
=>\(3x^2-14x+8-x^2+4x-4=0\)
=>\(2x^2-10x+4=0\)
=>\(x^2-5x+2=0\)
=>\(x=\dfrac{5\pm\sqrt{17}}{2}\)
d: Sửa đề: \(\left(x-2\right)\left(x^2-3x+5\right)=x^3-2x^2\)
=>\(\left(x-2\right)\left(x^2-3x+5\right)-x^2\left(x-2\right)=0\)
=>\(\left(x-2\right)\left(x^2-3x+5-x^2\right)=0\)
=>\(\left(x-2\right)\left(-3x+5\right)=0\)
=>(x-2)(3x-5)=0
=>\(\left[{}\begin{matrix}x=2\\x=\dfrac{5}{3}\end{matrix}\right.\)