Question

Tìm GTLN của Q= căn x+3 + căn 10-x

Answer 1

\(Q=\sqrt{x+3}+\sqrt{10-x}\)

\(\Leftrightarrow Q^2=\left(\sqrt{x+3}+\sqrt{10-x}\right)^2\le\left(1^2+1^2\right)\left[\left(\sqrt{x+3}\right)^2+\left(\sqrt{10-x}\right)^2\right]\)

\(\Leftrightarrow Q^2\le2\left(x+3+10-x\right)=2.13=26\)

\(\Leftrightarrow Q\le\sqrt{26}\)

\(maxQ=\sqrt{26}\Leftrightarrow x+3=10-x\Leftrightarrow x=\dfrac{7}{2}\)

Answer 2

Áp dụng BĐT Bunhiacopski:

\(Q=\sqrt{x+3}+\sqrt{10-x}\\ \Leftrightarrow Q^2=\left(\sqrt{x+3}+\sqrt{10-x}\right)^2\le\left(1^2+1^2\right)\left(x+3+10-x\right)=2\cdot13=26\\ \Leftrightarrow Q\le\sqrt{26}\\ Q_{max}=\sqrt{26}\Leftrightarrow x+3=10-x\Leftrightarrow x=\dfrac{7}{2}\)