Bài 1 $y=\frac{2x}{2+x}=>y'=\frac{4}{(x+2)^2}$$(C)$ có tâm đối xứng $I(-2;2)$$M(x_0;2-\frac4{x+2})\in (C)$Tiếp tuyến qua $M d: y=\frac4{(x_0+2)^2}(x-x_0)+2-\frac4{x_0+2}$$d(I;d)=\frac{|\frac{8}{(x_0+2)}|}{\sqrt{1+\frac{16}{(x_0+2)^4}}}=\frac{8}{\sqrt{(x_0+2)^2+\frac{16}{(x_0+2)^2}}}\leq \sqrt8$$d(I;d) max\Leftrightarrow (x_0+2)^4=16\Leftrightarrow x_0=0 \veebar x_0=-4 $
Bài 1 $y=\frac{2x}{2+x}=>y'=\frac{4}{(x+2)^2}$$(C)$ có tâm đối xứng $I(-2;2)$$M(x_0;2-\frac4{x+2})\in (C)$Tiếp tuyến qua $M d: y=\frac4{(x_0+2)^2}(x-x_0)+2-\frac4{x_0+2}$$d(I;d)=\frac{|\frac{8}{(x_0+2)}|}{\sqrt{1+\frac{16}{(x_0+2)^4}}}=\frac{8}{\sqrt{(x_0+2)^2+\frac{16}{(x_0+2)^2}}}\leq \sqrt8$$d(I;d) min\Leftrightarrow (x_0+2)^4=16\Leftrightarrow x_0=0 \veebar x_0=-4 $
Bài 1 $y=\frac{2x}{2+x}=>y'=\frac{4}{(x+2)^2}$$(C)$ có tâm đối xứng $I(-2;2)$$M(x_0;2-\frac4{x+2})\in (C)$Tiếp tuyến qua $M d: y=\frac4{(x_0+2)^2}(x-x_0)+2-\frac4{x_0+2}$$d(I;d)=\frac{|\frac{8}{(x_0+2)}|}{\sqrt{1+\frac{16}{(x_0+2)^4}}}=\frac{8}{\sqrt{(x_0+2)^2+\frac{16}{(x_0+2)^2}}}\leq \sqrt8$$d(I;d) m
ax\Leftrightarrow (x_0+2)^4=16\Leftrightarrow x_0=0 \veebar x_0=-4 $