定义域：X²+2X-3≥0(X+3)(X-1)≥0解得X≤-3或X≥1y=√u在定义域内单调递增u=X²+2X-3=(X+1)²-4 （X≤-3或X≥1）对称轴为X=-1u在（-∞,-3]单调递减,在[1,+∞）单调递增所以y=根号下X²+2X-3的单调...