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tranducduy75215

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howare

câu trả lời

ta có:\frac{x^2+x+1}{x^2+2x+1}=\frac{(x^2+2x+1)-(x+1)+1}{x^2+2x+1}
=1+\frac{1}{x+1}-\frac{1}{(x+1)^2}
=(\frac{1}{(x+1)^2}-\frac{1}{x+1}+\frac{1}{4})+\frac{3}{4}
=(\frac{1}{x+1}-\frac{1}{2})^2+\frac{3}{4}\geq\frac{3}{4}
dấu "=" xảy ra khi \frac{1}{x+1}-\frac{1}{2}=0
\Rightarrow x=2
\RightarrowminBT=\frac{3}{4}\Leftrightarrowx=2
 
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