toán hay GTNN đây

H

hoang_duythanh

A=$\frac{(x+6\sqrt[]{x}+9)-6(\sqrt[]{x}+3)+25}{\sqrt[]{x}+3}$
=$([\sqrt[]{x}+3)+\frac{25}{\sqrt[]{x}+3}]-6$\geq$2\sqrt[]{25}-6=4$
=>Min A=4
Đạt khi$ \sqrt[]{x}+3=\frac{25}{\sqrt[]{x}+3}$ \Leftrightarrow $(\sqrt[]{x}+3)^2=25$
\Leftrightarrow $\sqrt[]{x}+3$=5\Leftrightarrowx=4
Xong!!!!!
 
You must log in or register to reply here.
Top Bottom