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hoangkhuongpro
![](https://blog.hocmai.vn/wp-content/uploads/2017/07/hot.gif)
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[TEX]ln(x+1)-ln(x+2)+1/(x+2)=0[/TEX]
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[TEX]pt\Leftrightarrow{f(x)=ln\frac{x+1}{x+2}+\frac{1}{x+2}=0\ \ \ \ (x>-1)[/TEX]
[TEX]f^'(x)=\frac{x+2}{x+1}.\frac{1}{(x+2)^2}-\frac{1}{(x+2)^2}=\frac{1}{(x+1)(x+2)^2}>0\ \ \ \ \ \forall{x>{-1(1)[/TEX]
[TEX]\left{\lim_{x \to {-1}}f(x)=-\infty\\ \lim_{x \to {+\infty}}f(x)=0\ \ (2)[/TEX]
[TEX](1)(2)\Rightarrow{f(x)<0\forall{x>-1\Rightarrow{pt:\ VN[/TEX]
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