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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
\(a,B=\dfrac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{1+\sqrt{x}}-\sqrt{x}\\ B=x-\sqrt{x}+1-\sqrt{x}=\left(\sqrt{x}-1\right)^2\)
Mà \(x=4-2\sqrt{3}=\left(\sqrt{3}-1\right)^2\)
\(\Rightarrow B=\left(\sqrt{3}-1-1\right)^2=\left(\sqrt{3}-2\right)^2=7-4\sqrt{3}\)
\(b,P=AB=\dfrac{2x+1-x+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\left(\sqrt{x}-1\right)^2\\ P=\dfrac{\left(x+\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{x+\sqrt{x}+1}=\sqrt{x}-1\\ c,Q=\sqrt{x}+\dfrac{1}{P}=\sqrt{x}+\dfrac{1}{\sqrt{x}-1}\\ Q=\sqrt{x}-1+\dfrac{1}{\sqrt{x}-1}+1\ge2\sqrt{1}+1=3\\ Q_{min}=3\Leftrightarrow\left(\sqrt{x}-1\right)^2=1\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-1=1\\1-\sqrt{x}=1\end{matrix}\right.\Leftrightarrow\sqrt{x}=2\left(x>1\Leftrightarrow\right)x=4\left(tm\right)\)
a: \(B=\left(\sqrt{x}-1\right)^2=\left(\sqrt{3}-2\right)^2=7-4\sqrt{3}\)
b: \(A=\dfrac{2x+1-x+\sqrt{x}}{x\sqrt{x}-1}\cdot\left(\sqrt{x}-1\right)^2=\sqrt{x}-1\)
\(P=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\dfrac{2\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\)
\(=\sqrt{x}\left(\sqrt{x}-1\right)-\left(2\sqrt{x}+1\right)+2\left(\sqrt{x}+1\right)\)
\(=x-\sqrt{x}+1\)
\(=\left(\sqrt{x}-\dfrac{1}{2}\right)^3+\dfrac{3}{4}\ge\dfrac{3}{4}\)
\(\Rightarrow\left\{{}\begin{matrix}a=3\\b=4\end{matrix}\right.\) \(\Rightarrow a+b=7\)
a: Thay x=36 vào B, ta được:
\(B=\dfrac{6}{6-3}=\dfrac{6}{3}=2\)