Bài 3:
a) Ta có: \(2x\left(x+2\right)^2-8x^2=2\left(x-2\right)\left(x^2+2x+4\right)\)
⇔\(2x\left(x^2+4x+4\right)-8x^2=2\left(x^3-8\right)\)
\(\Leftrightarrow2x^3+8x^2+8-8x^2=2x^3-16\)
\(\Leftrightarrow2x^3+8-2x^3+16=0\)
hay 24=0(vô lý)
Vậy: x∈∅
b) Ta có: \(\left(x-2\right)^3+\left(3x-1\right)\left(3x+1\right)=\left(x+1\right)^3\)
\(\Leftrightarrow x^3-6x^2+12x-8+9x^2-1=x^3+3x^2+3x+1\)
\(\Leftrightarrow x^3+3x^2+12x-9-x^3-3x^2-3x-1=0\)
\(\Leftrightarrow9x-10=0\)
\(\Leftrightarrow9x=10\)
hay \(x=\frac{10}{9}\)
Vậy: \(x=\frac{10}{9}\)
c) Ta có: \(\left(x-1\right)^3-x\left(x+1\right)^2=5x\left(2-x\right)-11\left(x+2\right)\)
\(\Leftrightarrow x^3-3x^2+3x-1-x\left(x^2+2x+1\right)=10x-5x^2-11x-22\)
\(\Leftrightarrow x^3-3x^2+3x-1-x^3-2x^2-x=-5x^2-x-22\)
\(\Leftrightarrow-5x^2+2x-1+5x^2+x+22=0\)
\(\Leftrightarrow3x+21=0\)
\(\Leftrightarrow3x=-21\)
hay x=-7
Vậy: x=-7
d) Ta có: (x-1)-(2x-1)=9-x
⇔x-1-2x+1-9+x=0
⇔-9=0(vô lý)
Vậy: x∈∅
e) Ta có: \(x\left(x+3\right)^2-3x=\left(x+2\right)^3+1\)
\(\Leftrightarrow x\left(x^2+6x+9\right)-3x=x^3+6x^2+12x+8+1\)
\(\Leftrightarrow x^3+6x^2+9x-3x-x^3-6x^2-12x-9=0\)
\(\Leftrightarrow-6x-9=0\)
\(\Leftrightarrow-6x=9\)
hay \(x=-\frac{3}{2}\)
Vậy: \(x=-\frac{3}{2}\)
