Rút gọn biểu thức: a) ((t^4+4t^2+8)/(2t^#+2)).(t/(12t^2+1))

* Hướng dẫn giải

a) Ta có $\frac{{\mathrm{t}}^4+4{\mathrm{t}}^2+8}{2{\mathrm{t}}^3+2}.\frac{\mathrm{t}}{12{\mathrm{t}}^2+1}.\frac{3{\mathrm{t}}^3+3}{{\mathrm{t}}^4+4{\mathrm{t}}^2+8}$$=\frac{({\mathrm{t}}^4+4{\mathrm{t}}^2+8).\mathrm{t}.3({\mathrm{t}}^3+1)}{2({\mathrm{t}}^3+1).(12{\mathrm{t}}^2+1).({\mathrm{t}}^4+4{\mathrm{t}}^2+8)}$$=\frac{3\mathrm{t}}{2(12{\mathrm{t}}^2+1)}$ 

b) Ta có $\frac{\mathrm{y}-1}{2\mathrm{y}}.\left({\mathrm{y}}^2+\mathrm{y}+1+\frac{{\mathrm{y}}^3}{\mathrm{y}-1}\right)$$=\frac{\mathrm{y}-1}{2\mathrm{y}}.\left(\frac{{\mathrm{y}}^{3-1}}{\mathrm{y}-1}+\frac{{\mathrm{y}}^3}{\mathrm{y}-1}\right)=\frac{2{\mathrm{y}}^3-1}{2\mathrm{y}}$