a) Ta có: A = B
\(\Leftrightarrow\left(x-3\right)\left(x+4\right)-2\left(3x-2\right)=\left(x-4\right)^2\\ \Leftrightarrow x^2+4x-3x-12-6x+4=x^2-8x+16\\ \Leftrightarrow x^2-x^2+4x-3x-6x+8x=16+12-4\\ \Leftrightarrow3x=24\Leftrightarrow x=8\)
Vậy với x = 8 thì A = B
b) Ta có: A = B
\(\Leftrightarrow\left(x+2\right)\left(x-2\right)+3x^2=\left(2x+1\right)^2+2x\\ \Leftrightarrow x^2-4+3x^2=4x^2+4x+1+2x\\ \Leftrightarrow x^2+3x^2-4x^2-4x-2x=1+4\\ \Leftrightarrow-6x=5\Leftrightarrow x=-\frac{5}{6}\)
Vậy với \(x=-\frac{5}{6}\) thì A = B
c) Ta có: A = B
\(\Leftrightarrow\left(x-1\right)\left(x^2+x+1\right)-2x=x\left(x-1\right)\left(x+1\right)\\ \Leftrightarrow x^3-1-2x=x\left(x^2-1\right)\\ \Leftrightarrow x^3-1-2x=x^3-x\\ \Leftrightarrow x^3-x^3-2x+x=1\\ \Leftrightarrow-x=1\Leftrightarrow x=-1\)
Vậy với x = -1 thì A = B
d) Ta có: A = B
\(\Leftrightarrow\left(x+1\right)^3-\left(x-2\right)^3=\left(3x-1\right)\left(3x+1\right)\\ \Leftrightarrow x^3+3x^2+3x+1-x^3+6x^2-12x+8=9x^2-1\\ \Leftrightarrow x^3-x^3+3x^2+6x^2-9x^2+3x-12x=-1-1-8\\ \Leftrightarrow-9x=-10\Leftrightarrow x=\frac{10}{9}\)
Vậy với \(x=\frac{10}{9}\) thì A = B.
