Question

Chứng minh đẳng thức :

\(2x^2y'=x^2y^2+1\) với \(y=\frac{1+\ln x}{x\left(1-\ln x\right)}\)

Answer 1

Ta có \(y'=\frac{\frac{1}{x}x\left(1-\ln x\right)-\left[1-\ln x+x\left(-\frac{1}{x}\right)\right]\left(1+\ln x\right)}{x^2\left(1-\ln x\right)^2}=\frac{1-\ln x+\ln x\left(1+\ln x\right)}{x^2\left(1-\ln x\right)^2}=\frac{1+\ln^2x}{x^2\left(1-\ln x\right)^2}\)

\(\Rightarrow\begin{cases}2x^2y'=2x^2\frac{1+\ln^2x}{x^2\left(1-\ln x\right)^2}=\frac{2\left(1+\ln^2x\right)}{\left(1-\ln x\right)^2}\\x^2y^2+1=x^2\frac{1+\ln^2x}{x^2\left(1-\ln x\right)^2}+1=\frac{\left(1+\ln^2x\right)}{\left(1-\ln x\right)^2}+1=\frac{2\left(1+\ln^2x\right)}{\left(1-\ln x\right)^2}\end{cases}\)

\(\Rightarrow2x^2y'=x^2y^2+1\Rightarrow\) Điều phải chứng minh