y=2x/(2x^2+1). (2x^2+1)y=2x. 2yx^2+y-2x=0. 2yx^2-2x+y=0.(1). 判别式△=(-2)^2-4*2y*y. =4-8y^2. ∵方程(1)有实数解,∴判别式△≥0. ∴4-8y^2≥0,y^2≤1/2. |y|≤√2/2. ∴-√2/2≤y√2/2.----即为所求函数y的值域.
