Question

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= \(\sqrt{4x^2-4x+1}+\sqrt{4x^2-12x+9}\)

Answer 1

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 =)))