bài mới $(y-3)^{4}+(y-1)^{4}-16$
$\left (y-3\right )^4+\left (y-1\right )^4-16$
$=\left (3-y\right )^4+\left (y-1\right )^4-\left [\left (3-y\right )+\left (y-1\right ) \right ]^4$
$=\left (3-y\right )^4+\left (y-1\right )^4 -\left [\left (3-y\right )^4 + 4\left (3-y\right )^3\left (y-1\right ) + 6\left (3-y\right )^2\left (y-1\right )^2 + 4\left (3-y\right )\left (y-1\right )^3 + \left (y-1\right )^4 \right ]$
$=-\left [4\left (3-y\right )^3\left (y-1\right )+6\left (3-y\right )^2\left (y-1\right )^2+4\left (3-y\right )\left (y-1\right )^3 \right ]$
$=-\left (3-y\right )\left (y-1\right )\left [4\left (3-y\right )^2+6\left (3-y\right )\left (y-1\right )+4\left (y-1\right )^2 \right ]$
$=\left (y-3\right )\left (y-1\right )\left [4\left (y-3\right )^2-6\left (y-3\right )\left (y-1\right )+4\left (y-1\right )^2 \right ]$
$=\left (y-3\right )\left (y-1\right )\left (4y^2-24y+36-6y^2+24y-18+4y^2-8y+4 \right )$
$=\left (y-3\right )\left (y-1\right )\left (2y^2-8y+22 \right )$
$=2\left (y-3\right )\left (y-1\right )\left (y^2-4y+11\right )$
Bài viết đã được chỉnh sửa nội dung bởi huykinhcan99: 17-10-2013 - 15:08