Giải bất phương trình : (left( {dfrac{{x + 1}}{{sqrt x + 1}} + dfrac{1}{{x + sqrt x }} - dfrac{1}{{sqrt x }}} right):dfrac{x}{{x + 2sqrt x + 1}} ge 2017 + sqrt {2017} )
Giải chi tiết:
Điều kiện: (x > 0).
(begin{array}{l}left( {dfrac{{x + 1}}{{sqrt x + 1}} + dfrac{1}{{sqrt x + x}} - dfrac{1}{{sqrt x }}} right):dfrac{x}{{x + 2sqrt x + 1}} ge 2017 + sqrt {2017} Leftrightarrow dfrac{{sqrt x left( {x + 1} right) + 1 - left( {sqrt x + 1} right)}}{{sqrt x left( {sqrt x + 1} right)}}.dfrac{{{{left( {sqrt x + 1} right)}^2}}}{x} ge 2017 + sqrt {2017} Leftrightarrow dfrac{{xsqrt x .}}{{sqrt x left( {sqrt x + 1} right)}}.dfrac{{{{left( {sqrt x + 1} right)}^2}}}{x} ge 2017 + sqrt {2017} Leftrightarrow sqrt x + 1 ge 2017 + sqrt {2017} Leftrightarrow sqrt x ge 2016 + sqrt {2017} Leftrightarrow x ge {left( {2016 + sqrt {2017} } right)^2}end{array})
Vậy (x ge {left( {2016 + sqrt {2017} } right)^2})