Câu 4:
ĐKXĐ: \(\left\{{}\begin{matrix}2x-3>=0\\23-2x>=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=\dfrac{3}{2}\\x< =\dfrac{23}{2}\end{matrix}\right.\)
=>3/2<=x<=23/2
\(A=\sqrt{2x-3}+\sqrt{23-2x}\)
=>\(A^2=1\cdot\sqrt{2x-3}+1\cdot\sqrt{23-2x}< =\left(1^2+1^2\right)\cdot\left(2x-3+23-2x\right)\)
=>\(A^2< =2\cdot20=40\)
=>\(-2\sqrt{10}< =A< =2\sqrt{10}\)
Vậy: \(A_{max}=2\sqrt{10}\) khi 2x-3=23-2x
=>4x=26
=>x=6,5
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