Ta có

x 4   +   64   =   ( x 2 ) 2   +   16 x 2   +   64   –   16 x 2 =   ( x 2 ) 2   +   2 . 8 . x   +   8 2   –   ( 4 x ) 2   =   ( x 2   +   8 ) 2   –   ( 4 x ) 2

Đáp án cần chọn là: C

D

a, \(x^5+x^4+1\)

\(\Leftrightarrow x^5+x^4-x^2+\frac{1}{4}-\frac{1}{4}+x^2\)

\(\Leftrightarrow x^5+\left(x^2-\frac{1}{2}\right)^2-\frac{1}{4}+x^2\)

\(\Leftrightarrow x^2\left(x^3+1\right)+\left(x^2-\frac{1}{2}\right)^2-\frac{1}{4}\)

\(\Leftrightarrow x^2\left(x+1\right)\left(x^2-x+1\right)+\left(x^2-\frac{1}{2}+\frac{1}{2}\right)\left(x^2-\frac{1}{2}-\frac{1}{2}\right)\)

ta có :x^5 +x^4 +1=x^5-x^2 +x^4 -x +x^2 +x +1=x^2(x^3-1) +x(x^3 -1)+x^2 +x +1=x^2(x-1)(x^2+x+1)+x(x-1)(x^2 +x+1) +x^2 +x +1=(x^2 +x +1)(x^3 -x^2 +x^2 -x +1)=(x^2 +x+1)(x^3-x+1)

ta có x^5 +x +1=x^5-x^2 +x^2+x+1=x^2(x^3-1) +x^2 +x+1=x^2(x-1)(x^2+x+1)+x^2+x+1=(x^2+x+1)(x^3-x^2+1)

\(x^2-6x+8\)

\(C1\)   \(=x^2-4x-2x+8\)

\(=\left(x^2-4x\right)-\left(2x-8\right)\)

\(=x\left(x-4\right)-2\left(x-4\right)\)

\(=\left(x-2\right)\left(x-4\right)\)

\(C2\):  \(x^2-6x+8\)

  \(=x^2-6x+9-1\)

\(=\left(x^2-6x+9\right)-1\)

\(=\left(x-3\right)^2-1\)

\(=\left(x-3-1\right)\left(x-3+1\right)\)

\(=\left(x-4\right)\left(x-2\right)\)

\(C3\) \(x^2-6x+8\)

 \(=x^2-2x-4x+8\)

\(=\left(x^2-2x\right)-\left(4x-8\right)\)

\(=x\left(x-2\right)-4\left(x-2\right)\)

\(=\left(x-2\right)\left(x-4\right)\)

\(x^4+4\)

\(=x^4+4x^2+4-4x^2\)

\(=\left(x^2+2\right)^2-4x^2\)

\(=\left(x^2-2x+2\right)\left(x^2+2x+2\right)\)

x⁴+4

=x⁴+4x²+4-4x²

=(x²+2)²-4x²

=(x²-2x+2)(x²+2x+2)

\(-8x^3+1=1^3-\left(2x\right)^3=\left(1-2x\right)\left(1+2x+4x^2\right)\)