Question

tìm cả GTNN và GTLN của bt:

1, A=\(\sqrt{1-x}+\sqrt{1+x}\)

2, B=\(\sqrt{x-2}+\sqrt{6-x}\) (bđt bunhiacovxki)

3, A=2x+3y biết \(2x^2+3y^2\le5\)

Answer 1

\(A=\sqrt{1-x}+\sqrt{x+1}\)

\(A^2=\left(\sqrt{1-x}\cdot1+\sqrt{x+1}\cdot1\right)^2\)

Áp dụng BĐT Bunhiacospki ta có:
\(A^2\le\left(1^2+1^2\right)\left(1-x+1+x\right)\)

\(A^2\le4\)

\(A\le2\)

\(A_{max}=2\Leftrightarrow x=0\)

E ms tìm dc MAX thôi ah

Answer 2

ĐKXĐ: ....

a/ \(A\le\sqrt{2\left(1-x+1+x\right)}=2\Rightarrow A_{max}=2\) khi \(x=0\)

\(A\ge\sqrt{1-x+1+x}=\sqrt{2}\Rightarrow A_{min}=\sqrt{2}\) khi \(\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

b/ \(B\le\sqrt{2\left(x-2+6-x\right)}=2\sqrt{2}\Rightarrow B_{max}=2\sqrt{2}\) khi \(x=4\)

\(B\ge\sqrt{x-2+6-x}=2\Rightarrow B_{min}=2\) khi \(\left[{}\begin{matrix}x=2\\x=6\end{matrix}\right.\)

c/ \(A^2=\left(2x+3y\right)^2=\left(\sqrt{2}.\sqrt{2}x+\sqrt{3}.\sqrt{3}y\right)^2\)

\(\Rightarrow A^2\le\left(2+3\right)\left(2x^2+3y^2\right)\le5.5=25\)

\(\Rightarrow-5\le A\le5\)

\(A_{max}=5\) khi \(x=y=1\)

\(A_{min}=-5\) khi \(x=y=-1\)