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tt预测站
f(x1)-f(x2)=x1/(x²1+2x1+3)-x2(x²2+x2+3) =[x1(x²2+x2+3)-x2(x²1+2x1+3)]/[(x²1+2x1+3)(x²2+x2+3)] =[x1(x²2+x2+3)-x2(x²1+2x1+3)]/[(x²1+2x1+3)(x²2+x2+3)] =[(x1x²2-x²1x2)+3(x1-x2)]/[(x²1+2x1+3)(x²2+x2+3)] =[x1x2(x2-x1)-3(x1-x2)]/[(x²1+2x1+3)(x²2+x2+3)] =[(x2-x1)(x1x2+3)]/[(x²1+2x1+3)(x²2+x2+3)]>0 f(x)是减函数 y=(3x+5)/√(x-3) 就怕不加括号 设√(x-3)=t>0 ∴x-3=t² x=t²+3 ∴y=[3(t²+3)+5]/t =(3t²+14)/t =3t+14/t 用均值定理: 3t+14/t≥2√42 ∴y≥2√42 值域[2√42,+∞) 若解析式理解不对,再追问