Bài 4B:
a: Ta có: \(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+9\left(x+1\right)^2=4\)
\(\Leftrightarrow x^3-9x^2+27x-27-x^3+27+9x^2+18x+9=4\)
\(\Leftrightarrow45x=-5\)
hay \(x=-\dfrac{1}{9}\)
b: Ta có: \(x\left(x-5\right)\left(x+5\right)-\left(x+2\right)\left(x^2-2x+4\right)=17\)
\(\Leftrightarrow x^3-25x-x^3-8=17\)
\(\Leftrightarrow-25x=25\)
hay x=-1
Bài 4A:
a: Ta có: \(\left(x-1\right)^3+\left(2-x\right)\left(4+2x+x^2\right)+3x\left(x+2\right)=16\)
\(\Leftrightarrow x^3-3x^2+3x-1+8-x^3+3x^2+6x=16\)
\(\Leftrightarrow9x=9\)
hay x=1
b: Ta có: \(8\left(x-\dfrac{1}{2}\right)\left(x^2+\dfrac{1}{2}x+\dfrac{1}{4}\right)-4x\left(1-x+2x^2\right)+2=0\)
\(\Leftrightarrow8\left(x^3-\dfrac{1}{8}\right)-4x\left(2x^2-x+1\right)+2=0\)
\(\Leftrightarrow8x^3-1-8x^3+4x^2-4x+2=0\)
\(\Leftrightarrow4x^2-4x+1=0\)
\(\Leftrightarrow2x-1=0\)
hay \(x=\dfrac{1}{2}\)
Bài 3B:
a: Ta có: \(A=\left(4-x\right)\left(x^2+4x+16\right)+\left(x-1\right)\left(x^2+x+1\right)\)
\(=64-x^3+x^3-1\)
=63
b: Ta có: \(B=3\left(x-\dfrac{1}{3}y\right)\left(9x^2+3xy+y^2\right)+\left(x+y\right)\left(x^2-xy+y^2\right)-27x^3\)
\(=\left(3x-y\right)\left(9x^2+3xy+y^2\right)+x^3+y^3-27x^3\)
\(=27x^3-y^3-26x^3+y^3\)
\(=x^3\)
