Question

Tìm GTLN của:

Answer 1

\(A^2=\left(\sqrt{6-x}+\sqrt{x+2}\right)^2\)

Áp dụng Bu-nhi-a-cốp-xki:

\(A^2=\left(\sqrt{6-x}+\sqrt{x+2}\right)^2\le\left(1+1\right)8=16\)

=> A ≤ 4

Dấu = xảy ra khi: \(\dfrac{1}{\sqrt{6-x}}=\dfrac{1}{\sqrt{x+2}}\)

\(\Rightarrow6-x=x+2\)

=> x = 2

Answer 2

Áp dụng BĐT: \(a+b\le\sqrt{2\left(a^2+b^2\right)}\)

\(A\le\sqrt{2\left(6-x+x+2\right)}=4\)

\(A_{max}=4\) khi \(6-x=x+2\Rightarrow x=2\)