1: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\notin\left\{4;9\right\}\end{matrix}\right.\)

Ta có: \(A=\dfrac{2\sqrt{x}-9-x+9+2x-4\sqrt{x}+\sqrt{x}-2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)

\(1,A=\dfrac{2\sqrt{x}-9-x+9+2x-3\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\\ A=\dfrac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\\ A=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\left(x\ge0;x\ne4;x\ne9\right)\\ 2,A< 1\Leftrightarrow\dfrac{\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}-3}< 0\\ \Leftrightarrow\dfrac{4}{\sqrt{x}-3}< 0\Leftrightarrow\sqrt{x}-3< 0\Leftrightarrow0\le x< 9\)

\(A\ge0\)
Dấu "=" xảy ra <=> \(\sqrt{x^2-4x+5}=\sqrt{x^2+6x+13}\) 
\(\Leftrightarrow x^2-4x+5=x^2+6x+13\)
\(\Leftrightarrow10x=-8\)
\(\Leftrightarrow x=-0.8\)
 

GTNN và GTLN của cả A và B hay của A + B vậy bạn...

\(x=\sqrt{5+\sqrt{13+\sqrt{5+\sqrt{13+\sqrt{5+....}}}}}\)

\(\Rightarrow x^2=5+\sqrt{13+\sqrt{5+\sqrt{13+\sqrt{5+...}}}}\)

\(\Rightarrow x^4=25+10\sqrt{13+\sqrt{5+\sqrt{13+\sqrt{5+....}}}}+13+\sqrt{5+\sqrt{13+\sqrt{5+...}}}\)

\(\Leftrightarrow x^4=38+10x^2+x\)

\(\Leftrightarrow x^4-10x^2-x-38=0\)

giải ra tìm x xong

a) tương tự : https://hoc24.vn/hoi-dap/question/650070.html

b) ta có : \(A=\dfrac{x\sqrt{x}-6x+9\sqrt{x}}{4\left(\sqrt{x}-1\right)}.\left(\dfrac{3\sqrt{x}-5}{\sqrt{x}-3}-1\right)\)

\(=\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)^2}{4\left(\sqrt{x}-1\right)}.\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}\right)=\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{2}=\dfrac{x-3\sqrt{x}}{2}\)

\(\Rightarrow x-3\sqrt{x}-2A=0\)

vì phương trình này luôn có nghiệm \(\Rightarrow\Delta\ge0\)

\(\Leftrightarrow3^2-4\left(-2A\right)=9+8A\ge0\Leftrightarrow A\ge\dfrac{-9}{8}\)

\(\Rightarrow\) GTNN của \(A=\dfrac{-9}{8}\) khi \(\sqrt{x}=\dfrac{-b}{2a}=\dfrac{3}{2}\)\(\Leftrightarrow x=\dfrac{9}{4}\)