Giải bài 2.22 trang 50

Đề bài

Tính nhanh: $\frac{t}{t^2+1}.\frac{x-2y+z}{x+y+z}+\frac{t}{t^2+1}.\frac{x+y-2z}{x+y+z}+\frac{t}{t^2+1}.\frac{y+z-2x}{x+y+z}$

Lời giải chi tiết

$\begin{array}{c}\frac{t}{t^2+1}.\frac{x-2y+z}{x+y+z}+\frac{t}{t^2+1}.\frac{x+y-2z}{x+y+z}+\frac{t}{t^2+1}.\frac{y+z-2x}{x+y+z}\\ =\frac{t}{t^2+1}.\left(\frac{x-2y+z}{x+y+z}+\frac{x+y-2z}{x+y+z}+\frac{y+z-2x}{x+y+z}\right)\\ =\frac{t}{t^2+1}.\left(\frac{x-2y+z+x+y-2z+y+z-2x}{x+y+z}\right)\\ =\frac{t}{t^2+1}.\left(\frac{\mathrm{0.0.0}}{x+y+z}\right)=0\end{array}$