Bài 1: Phân tích đa thức thành nhân tử
a)Sửa đề: \(\left(x+1\right)\left(2x-1\right)\left(3x+2\right)\left(6x-5\right)-4\)
Ta có: \(\left(x+1\right)\left(2x-1\right)\left(3x+2\right)\left(6x-5\right)-4\)
\(=\left(x+1\right)\left(6x-5\right)\left(2x-1\right)\left(3x+2\right)-4\)
\(=\left(6x^2-5x+6x-5\right)\left(6x^2+4x-3x-2\right)-4\)
\(=\left(6x^2+x-5\right)\left(6x^2+x-2\right)-4\)
\(=\left(6x^2+x\right)^2-7\left(6x^2+x\right)+10-4\)
\(=\left(6x^2+x\right)^2-7\left(6x^2+x\right)+6\)
\(=\left(6x^2+x\right)^2-\left(6x^2+x\right)-6\left(6x^2+x\right)+6\)
\(=\left(6x^2+x\right)\left(6x^2+x-1\right)-6\left(6x^2+x-1\right)\)
\(=\left(6x^2+x-1\right)\left(6x^2+x-6\right)\)
\(=\left(6x^2+3x-2x-1\right)\left(6x^2+x-6\right)\)
\(=\left[3x\left(2x+1\right)-\left(2x+1\right)\right]\left(6x^2+x-6\right)\)
\(=\left(2x+1\right)\left(3x-1\right)\left(6x^2+x-6\right)\)
b) Ta có: \(x^3+y^3-z^3+3xyz\)
\(=\left(x+y\right)^3-3xy\left(x+y\right)-z^3+3xyz\)
\(=\left(x+y-z\right)\left[\left(x+y\right)^2+z\left(x+y\right)+z^2\right]-3xy\left(x+y-z\right)\)
\(=\left(x+y-z\right)\left(x^2+2xy+y^2+xz+yz+z^2-3xy\right)\)
\(=\left(x+y-z\right)\left(x^2+y^2+z^2-xy+xz+yz\right)\)
c) Ta có: \(\left(x+2\right)\left(x+3\right)\left(x+8\right)\left(x+12\right)-4x^2\)
\(=\left(x^2+14x+24\right)\left(x^2+11x+24\right)-4x^2\)
\(=\left(x^2+24\right)^2+25x\left(x^2+24\right)+154x^2-4x^2\)
\(=\left(x^2+24\right)^2+25x\left(x^2+24\right)+150x^2\)
\(=\left(x^2+24\right)^2+10x\left(x^2+24\right)+15x\left(x^2+24\right)+150x^2\)
\(=\left(x^2+24\right)\left(x^2+24+10x\right)+15x\left(x^2+24+10x\right)\)
\(=\left(x^2+15x+24\right)\left(x^2+10x+24\right)\)
\(=\left(x+6\right)\left(x+4\right)\left(x^2+15x+24\right)\)
