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a) \(P=\left[\frac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-\left(3x+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right]:\left[\frac{\left(2\sqrt{x}-2\right)-\left(\sqrt{x}-3\right)}{\sqrt{x}-3}\right]\left(ĐK:x\ge0;x\ne9\right)\)
\(=\frac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\frac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)
\(=\frac{-3\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\frac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\frac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\frac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\frac{-3}{\sqrt{x}+3}\)
1. \(x=\frac{1}{9}\) thỏa mãn đk: \(x\ge0;x\ne9\)
Thay \(x=\frac{1}{9}\) vào A ta có:
\(A=\frac{\sqrt{\frac{1}{9}}+1}{\sqrt{\frac{1}{9}}-3}=-\frac{1}{2}\)
2. \(B=...\)
\(B=\frac{3\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\frac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\frac{4x+6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(B=\frac{3x-9\sqrt{x}+x+3\sqrt{x}-4x-6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(B=\frac{-6\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
3. \(P=A:B=\frac{\sqrt{x}+1}{\sqrt{x}-3}:\frac{-6\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(P=\frac{\sqrt{x}+3}{-6}\)
Vì \(\sqrt{x}+3\ge3\forall x\)\(\Rightarrow\frac{\sqrt{x}+3}{-6}\le\frac{3}{-6}=-\frac{1}{2}\)
hay \(P\le-\frac{1}{2}\)
Dấu "=" xảy ra <=> x=0
\(x=9\Rightarrow\sqrt{x}=3\Rightarrow A=\frac{3+2}{3-5}=\frac{5}{-2}=-\frac{5}{2}\\ \)
\(B=\frac{3}{\sqrt{x}+5}+\frac{20-2\sqrt{x}}{x-25}=\frac{3.\left(\sqrt{x}-5\right)}{\left(\sqrt{x}+5\right).\left(\sqrt{x}-5\right)}+\frac{20-2\sqrt{x}}{\left(x+\sqrt{5}\right).\left(x-\sqrt{5}\right)}\)
\(=\frac{3\sqrt{x}-15+20-2\sqrt{x}}{\left(\sqrt{x}+5\right).\left(\sqrt{x}-5\right)}=\frac{\sqrt{x}+5}{\left(\sqrt{x}+5\right).\left(\sqrt{x}-5\right)}=\frac{1}{\sqrt{x}-5}\)
\(A=B.\left|x-4\right|\Leftrightarrow\left|x-4\right|=A:B=\frac{\sqrt{x}+2}{\sqrt{x}-5}:\frac{1}{\sqrt{x}-5}=\sqrt{x}+2\)
\(\Rightarrow\left(x-4\right)^2=\left(\sqrt{x}+2\right)^2\Leftrightarrow x^2-8x+16=x+4\sqrt{x}+4\)
\(\Leftrightarrow x^2-9x-4\sqrt{x}+12=0\Leftrightarrow x.\left(x-9\right)-4.\left(\sqrt{x}-3\right)=0\)
\(\Leftrightarrow x.\left(\sqrt{x}-3\right).\left(\sqrt{x}+3\right)-4.\left(\sqrt{x}-3\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-3\right).\left(x\sqrt{x}+3x-4\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-3\right).\left(\left(x\sqrt{x}-x\right)+\left(4x-4\right)\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-3\right).\left(x.\left(\sqrt{x}-1\right)+4.\left(\sqrt{x}-1\right).\left(\sqrt{x}+1\right)\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-3\right).\left(\sqrt{x}-1\right).\left(x+4\sqrt{x}+4\right)=0\Leftrightarrow\left(\sqrt{x}-3\right).\left(\sqrt{x}-1\right).\left(\sqrt{x}+2\right)^2=0\)
\(\Rightarrow\orbr{\begin{cases}\sqrt{x}-3=0\\\sqrt{x}-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=9\\x=1\end{cases}}}\)(Vì \(\sqrt{x}\ge0\Rightarrow\sqrt{x}+2\ge2\Rightarrow\left(\sqrt{x}+2\right)^2\ge4>0\))
Với x >= 0 ; x khác 9
\(B=\frac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{x-9}=\frac{-3\sqrt{x}-3}{x-9}=\frac{-3\left(\sqrt{x}+1\right)}{x-9}\)
\(\frac{B}{A}=\frac{-3\left(\sqrt{x}+1\right)}{x-9}:\frac{\sqrt{x}+1}{\sqrt{x}-3}=\frac{-3}{\sqrt{x}+3}+\frac{1}{2}< 0\)
\(\Leftrightarrow\frac{-6+\sqrt{x}+3}{2\left(\sqrt{x}+3\right)}< 0\Rightarrow\sqrt{x}-3< 0\Leftrightarrow x< 9\)
Kết hợp đk vậy 0 =< x < 9