\(x^4+2x^3-16x^2-2x+15\\ =x^4+x^3+x^3+x^2-17x^2-17x+15x+15\\ =x^3\left(x+1\right)+x^2\left(x+1\right)-17x\left(x+1\right)+15\left(x+1\right)\\ =\left(x+1\right)\left(x^3+x^2-17x+15\right)\\ =\left(x+1\right)\left(x^3+5x^2-4x^2-20x+3x+15\right)\\ =\left(x+1\right)\left(x+5\right)\left(x^2-4x+3\right)\\ =\left(x+1\right)\left(x+5\right)\left(x-3\right)\left(x-1\right)\\ =\left[\left(x-1\right)\left(x+5\right)\right]\left[\left(x+1\right)\left(x+3\right)\right]\\ =\left(x^2+4x-5\right)\left(x^2+4x+3\right)\\ =\left(x^2+4x-1\right)^2-16\)
Đề sai hay sao ý, đâu có y với z đâu!
Đề thiếu dữ kiện, x là số tự nhiên lẻ.
Làm tiếp:
Với x = 2k+1, ta có:
\(\left(x+1\right)\left(x-1\right)\left(x+3\right)\left(x+5\right)\\ \Leftrightarrow\left(2k+1+1\right)\left(2k+1-1\right)\left(2k+1+3\right)\left(2k+1+5\right)\\ \Leftrightarrow\left(2k+2\right)\cdot2k\cdot\left(2k+4\right)\left(2k+6\right)\\ \Leftrightarrow16\left(k+1\right)\cdot k\cdot\left(k+2\right)\left(k+3\right)\\ \RightarrowĐpcm\)
