Question

Bài 5: Tìm giá trị lớn nhất của biểu thức:

\(A=\sqrt{-9x^2+6x+3}\)

B=\(\sqrt{3-2x^2}\)

\(C=5+\sqrt{-4x_{ }^2-4x}\)

Answer 1

Lời giải:

Ta có:
\(A=\sqrt{-9x^2+6x+3}=\sqrt{4-(9x^2-6x+1)}=\sqrt{4-(3x-1)^2}\)

Ta thấy \((3x-1)^2\geq 0\Rightarrow 4-(3x-1)^2\leq 4\Rightarrow A=\sqrt{4-(3x-1)^2}\leq 2\)

Vậy \(A_{\max}=2\Leftrightarrow (3x-1)^2=0\Leftrightarrow x=\frac{1}{3}\)

-------------

Ta thấy:

\(2x^2\geq 0\Rightarrow 3-2x^2\leq 3\Rightarrow B=\sqrt{3-2x^2}\leq \sqrt{3}\)

Vậy \(B_{\max}=\sqrt{3}\Leftrightarrow x^2=0\Leftrightarrow x=0\)

--------------

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

Ta thấy \((2x+1)^2\geq 0\Rightarrow 1-(2x+1)^2\leq 1\)

\(\Rightarrow C=5+\sqrt{1-(2x+1)^2}\leq 5+\sqrt{1}=6\)

Vậy \(C_{\max}=6\Leftrightarrow (2x+1)^2=0\Leftrightarrow x=\frac{-1}{2}\)

Answer 2

Lời giải:

Ta có:
\(A=\sqrt{-9x^2+6x+3}=\sqrt{4-(9x^2-6x+1)}=\sqrt{4-(3x-1)^2}\)

Ta thấy \((3x-1)^2\geq 0\Rightarrow 4-(3x-1)^2\leq 4\Rightarrow A=\sqrt{4-(3x-1)^2}\leq 2\)

Vậy \(A_{\max}=2\Leftrightarrow (3x-1)^2=0\Leftrightarrow x=\frac{1}{3}\)

-------------

Ta thấy:

\(2x^2\geq 0\Rightarrow 3-2x^2\leq 3\Rightarrow B=\sqrt{3-2x^2}\leq \sqrt{3}\)

Vậy \(B_{\max}=\sqrt{3}\Leftrightarrow x^2=0\Leftrightarrow x=0\)

--------------

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

Ta thấy \((2x+1)^2\geq 0\Rightarrow 1-(2x+1)^2\leq 1\)

\(\Rightarrow C=5+\sqrt{1-(2x+1)^2}\leq 5+\sqrt{1}=6\)

Vậy \(C_{\max}=6\Leftrightarrow (2x+1)^2=0\Leftrightarrow x=\frac{-1}{2}\)