已知(fx)=alnx/x+1+b/x,曲线y=f(x)在点(1,f(1))处的切线方程为x+2y-3=0.求a,b的值;当x>0已知(fx)=alnx/(x+1)+b/x,曲线y=f(x)在点(1,f(1))处的切线方程为x+2y-3=0.求a,b的值;当x>0,且x≠1时,f
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![已知(fx)=alnx/x+1+b/x,曲线y=f(x)在点(1,f(1))处的切线方程为x+2y-3=0.求a,b的值;当x>0已知(fx)=alnx/(x+1)+b/x,曲线y=f(x)在点(1,f(1))处的切线方程为x+2y-3=0.求a,b的值;当x>0,且x≠1时,f](/uploads/image/z/5439294-54-4.jpg?t=%E5%B7%B2%E7%9F%A5%EF%BC%88fx%EF%BC%89%3Dalnx%2Fx%2B1%2Bb%2Fx%2C%E6%9B%B2%E7%BA%BFy%3Df%28x%29%E5%9C%A8%E7%82%B9%EF%BC%881%2Cf%EF%BC%881%EF%BC%89%EF%BC%89%E5%A4%84%E7%9A%84%E5%88%87%E7%BA%BF%E6%96%B9%E7%A8%8B%E4%B8%BAx%2B2y-3%3D0.%E6%B1%82a%2Cb%E7%9A%84%E5%80%BC%EF%BC%9B%E5%BD%93x%EF%BC%9E0%E5%B7%B2%E7%9F%A5%EF%BC%88fx%EF%BC%89%3Dalnx%2F%EF%BC%88x%2B1%EF%BC%89%2Bb%2Fx%2C%E6%9B%B2%E7%BA%BFy%3Df%28x%29%E5%9C%A8%E7%82%B9%EF%BC%881%2Cf%EF%BC%881%EF%BC%89%EF%BC%89%E5%A4%84%E7%9A%84%E5%88%87%E7%BA%BF%E6%96%B9%E7%A8%8B%E4%B8%BAx%2B2y-3%3D0.%E6%B1%82a%2Cb%E7%9A%84%E5%80%BC%EF%BC%9B%E5%BD%93x%EF%BC%9E0%2C%E4%B8%94x%E2%89%A01%E6%97%B6%2Cf)
已知(fx)=alnx/x+1+b/x,曲线y=f(x)在点(1,f(1))处的切线方程为x+2y-3=0.求a,b的值;当x>0已知(fx)=alnx/(x+1)+b/x,曲线y=f(x)在点(1,f(1))处的切线方程为x+2y-3=0.求a,b的值;当x>0,且x≠1时,f
已知(fx)=alnx/x+1+b/x,曲线y=f(x)在点(1,f(1))处的切线方程为x+2y-3=0.求a,b的值;当x>0
已知(fx)=alnx/(x+1)+b/x,曲线y=f(x)在点(1,f(1))处的切线方程为x+2y-3=0.求a,b的值;当x>0,且x≠1时,f(x)>lnx/(x-1)+k/x,求k的取值范围
已知(fx)=alnx/x+1+b/x,曲线y=f(x)在点(1,f(1))处的切线方程为x+2y-3=0.求a,b的值;当x>0已知(fx)=alnx/(x+1)+b/x,曲线y=f(x)在点(1,f(1))处的切线方程为x+2y-3=0.求a,b的值;当x>0,且x≠1时,f
(1) 切线方程变形为 y=(-1/2)(x-1)+1
可见斜率k=-1/2,f(1)=1
f'(x)=[a(x+1)/x-alnx]/(x+1)^2-b/x^2
已知k=f'(1)=(2a)/4-b=-1/2 即a-2b=-1 (1)
f(1)=b=1
代入(1)得 a=1
(2) 当x>0时 f(x)=lnx/(x+1)+1/x
f(x)-lnx/(x+1)-k/x=(1-k)/x>0
只要1-k>0即可
所以k
我只求出b=1,其它的就要问各位高手了^^
a=1;b=1
同学,你两个问题不一样啊,x+1到底有括号没有啊?算出来不一样