Tìm GTLN:
a) A= \(\sqrt{3-2x^2}\)
b) B= \(\sqrt{-9x^2+6x+3}\)
c) B= \(5+\sqrt{-4x^2-4x}\)
d) C= \(\sqrt{-x^2+x+\frac{3}{4}}\)
e) D= \(\sqrt{x^2+2x+1}+\sqrt{x^2-2x+1}\)
g) G= \(\sqrt{25x^2-20x+4}+\sqrt{25x^2}\)
f) F= \(\sqrt{4x^2-4x+1}+\sqrt{4x^2-12x+9}\)
c)\(C=5+\sqrt{-4x^2-4x}\)
\(C=5+\sqrt{1-\left(4x^2+4x+1\right)}\)
\(C=5+\sqrt{1-\left(2x+1\right)^2}\)
Ta có: \(-\left(2x+1\right)^2\le0\)
\(\sqrt{1-\left(2x+1\right)^2}\le1\)
\(\sqrt{1-\left(2x+1\right)^2}+5\le6\Leftrightarrow C\le6\)
Vậy \(C_{max}=6\) khi \(2x+1=0\Leftrightarrow x=-\frac{1}{2}\)
f) \(F=\sqrt{4x^2-4x+1}+\sqrt{4x^2-12x+9}\)
\(F=\sqrt{\left(2x-1\right)^2}+\sqrt{\left(2x-3\right)^2}\)
\(F=\left|2x-1\right|+\left|3-2x\right|\ge\left|2x+1+3-2x\right|=4\)
\(F_{min}=4\) khi \(\left(2x-1\right)\left(3-2x\right)\ge0\Leftrightarrow\frac{1}{2}\le x\le\frac{3}{2}\)
Mấy còn lại tương tự =)))