Với \(x\ge-\frac{1}{2}\)

2f(x) = \(2\sqrt{\left(2x+1\right)\left(x+2\right)}+4\sqrt{x+3}-4x\)

\(=-\left(2x+1\right)+2\sqrt{\left(2x+1\right)\left(x+2\right)}-\left(x+2\right)-\left(x+3\right)+4\sqrt{x+3}-4+10\)

\(=-\left(\sqrt{2x+1}-\sqrt{x+2}\right)^2-\left(\sqrt{x+3}-2\right)^2+10\le10\)

Dấu "=" xảy ra <=> \(\hept{\begin{cases}2x+1=x+2\\x+3=4\end{cases}}\Leftrightarrow x=1\)

=> min 2f(x) = 10 tại x = 1

=> min f(x) = 5 tại x = 1

Lời giải:
Áp dụng BĐT Bunhiacopxky, với $x\geq \frac{-1}{2}$ ta có:

\((\sqrt{2x^2+5x+2}+2\sqrt{x+3})^2=(\sqrt{(2x+1)(x+2)}+2\sqrt{x+3})^2\)

\(\leq [(2x+1)+2^2][(x+2)+(x+3)]=(2x+5)^2\)

\(\Rightarrow \sqrt{2x^2+5x+2}+2\sqrt{x+3}\leq 2x+5\)

\(\Rightarrow A\leq 5\)

Vậy $A_{\max}=5$. Giá trị này đạt tại $x=1$

Bài 1: 

Ta có: \(D=\sqrt{16x^4}-2x^2+1\)

\(=4x^2-2x^2+1\)

\(=2x^2+1\)

1:

ĐKXĐ: x≠4

Ta có: \(x=\sqrt{7-4\sqrt{3}}+\sqrt{7+4\sqrt{3}}\)

\(=\sqrt{3-2\cdot\sqrt{3}\cdot2+4}+\sqrt{3+2\cdot\sqrt{3}\cdot2+4}\)

\(=\sqrt{\left(\sqrt{3}-2\right)^2}+\sqrt{\left(\sqrt{3}+2\right)^2}\)

\(=\left|\sqrt{3}-2\right|+\left|\sqrt{3}+2\right|\)

\(=2-\sqrt{3}+\sqrt{3}+2\)

\(=4\)(ktm ĐKXĐ)

Vậy: Khi x=4 thì A không có giá trị

2: Ta có: P=A+B

\(\Leftrightarrow P=\frac{2}{\sqrt{x}-2}+\frac{\sqrt{x}}{x+1}-\frac{4\sqrt{x}+2}{x\sqrt{x}-2x+\sqrt{x}-2}\)

\(\Leftrightarrow P=\frac{2\left(x+1\right)}{\left(\sqrt{x}-2\right)\left(x+1\right)}+\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{\left(x+1\right)\left(\sqrt{x}-2\right)}-\frac{4\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(x+1\right)}\)

\(\Leftrightarrow P=\frac{2x+2+x-2\sqrt{x}-4\sqrt{x}-2}{\left(x+1\right)\left(\sqrt{x}-2\right)}\)

\(\Leftrightarrow P=\frac{3x-6\sqrt{x}}{\left(x+1\right)\left(\sqrt{x}-2\right)}=\frac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(x+1\right)\left(\sqrt{x}-2\right)}\)

\(\Leftrightarrow P=\frac{3\sqrt{x}}{x+1}\)