判断函数y=根号x在区间【0,正无穷大)上的单调性,并证明你的结论解;任取0≤x1<x2则f(x1)-f(x2)=√x1-√x2=(√x1-√x2)(√x1+√x2)/(√x1+√x2)=(x1-x2)/(√x1+√x2)因为x1-x2<0.√x1+√x2﹥0所以f(x1)-f(x2)﹤0
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![判断函数y=根号x在区间【0,正无穷大)上的单调性,并证明你的结论解;任取0≤x1<x2则f(x1)-f(x2)=√x1-√x2=(√x1-√x2)(√x1+√x2)/(√x1+√x2)=(x1-x2)/(√x1+√x2)因为x1-x2<0.√x1+√x2﹥0所以f(x1)-f(x2)﹤0](/uploads/image/z/3857173-61-3.jpg?t=%E5%88%A4%E6%96%AD%E5%87%BD%E6%95%B0y%3D%E6%A0%B9%E5%8F%B7x%E5%9C%A8%E5%8C%BA%E9%97%B4%E3%80%900%2C%E6%AD%A3%E6%97%A0%E7%A9%B7%E5%A4%A7%EF%BC%89%E4%B8%8A%E7%9A%84%E5%8D%95%E8%B0%83%E6%80%A7%2C%E5%B9%B6%E8%AF%81%E6%98%8E%E4%BD%A0%E7%9A%84%E7%BB%93%E8%AE%BA%E8%A7%A3%3B%E4%BB%BB%E5%8F%960%E2%89%A4x1%EF%BC%9Cx2%E5%88%99f%28x1%29-f%28x2%29%3D%E2%88%9Ax1-%E2%88%9Ax2%3D%28%E2%88%9Ax1-%E2%88%9Ax2%29%28%E2%88%9Ax1%2B%E2%88%9Ax2%29%2F%28%E2%88%9Ax1%2B%E2%88%9Ax2%29%3D%28x1-x2%29%2F%28%E2%88%9Ax1%2B%E2%88%9Ax2%29%E5%9B%A0%E4%B8%BAx1-x2%EF%BC%9C0.%E2%88%9Ax1%2B%E2%88%9Ax2%EF%B9%A50%E6%89%80%E4%BB%A5f%28x1%29-f%28x2%29%EF%B9%A40)
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