Bài 3:
a: Ta có: \(A=\left(x+1\right)^3-4x\left(x^2+2x-5\right)\)
\(=x^3+3x^2+3x+1-4x^3-8x^2+20x\)
\(=-3x^3-5x^2+23x+1\)
b: Ta có: \(B=2x^2\left(x+3\right)-\left(x^2+x-5\right)+\left(x-2\right)^3\)
\(=2x^3+6x^2-x^2-x+5+x^3-6x^2+12x-8\)
\(=3x^3-x^2+11x-3\)
c: Ta có: \(C=\left(x-2\right)\left(x^2+2x+4\right)-\left(x^2+2\right)\left(x-1\right)\)
\(=x^3-8-x^3+x^2-2x+2\)
\(=x^2-2x-6\)
d: Ta có: \(\left(x-3\right)\left(x^2+3x+9\right)-\left(x+2\right)\left(x^2-2x+4\right)\)
\(=x^3-27-x^3+8\)
=-19
Bài 4:
a: Ta có: \(\left(x-2\right)\left(x^2+2x+4\right)-x\left(x^2-5\right)=-7\)
\(\Leftrightarrow x^3-8-x^3+5x=-7\)
\(\Leftrightarrow5x=1\)
hay \(x=\dfrac{1}{5}\)
b: Ta có: \(2x\left(4x^2+5\right)-\left(2x-3\right)\left(4x^2+6x+9\right)=3x-4\)
\(\Leftrightarrow8x^3+10x-8x^3+27-3x+4=0\)
\(\Leftrightarrow7x=-31\)
hay \(x=-\dfrac{31}{7}\)
c: Ta có: \(\left(2x+1\right)^3-8x\left(x^2+1\right)=0\)
\(\Leftrightarrow8x^3+12x^2+6x+1-8x^3-8x=0\)
\(\Leftrightarrow-2x=-1\)
hay \(x=\dfrac{1}{2}\)
