\(a,A=\dfrac{\sqrt{x}-2}{\sqrt{x}+9}\left(ĐK:x>0;x\ne4\right)\)
Thay \(x=1\) vào \(A\), ta được:
\(A=\dfrac{\sqrt{1}-2}{\sqrt{1}+9}\)
\(=\dfrac{1-2}{1+9}\)
\(=-\dfrac{1}{10}\)
Vậy \(A=-\dfrac{1}{10}\) khi \(x=1\).
\(b,B=\dfrac{3\sqrt{x}-6}{x-2\sqrt{x}}+\dfrac{\sqrt{x}-3}{\sqrt{x}}-\dfrac{1}{2-\sqrt{x}}\left(ĐK:x>0;x\ne4\right)\)
\(=\dfrac{3\sqrt{x}-6}{\sqrt{x}\left(\sqrt{x}-2\right)}+\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}+\dfrac{1}{\sqrt{x}-2}\)
\(=\dfrac{3\sqrt{x}-6+x-5\sqrt{x}+6}{\sqrt{x}\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x-2\sqrt{x}+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}\)
Vậy \(B=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}\).
\(c,P=A\cdot B=\dfrac{\sqrt{x}-2}{\sqrt{x}+9}\cdot\dfrac{\sqrt{x}-1}{\sqrt{x}-2}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+9}\)
Khi đó: \(P< \dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}+9}< \dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}+9}-\dfrac{1}{2}< 0\)
\(\Leftrightarrow\dfrac{2\cdot\left(\sqrt{x}-1\right)}{2\cdot\left(\sqrt{x}+9\right)}-\dfrac{\sqrt{x}+9}{2\cdot\left(\sqrt{x}+9\right)}< 0\)
\(\Leftrightarrow\dfrac{2\sqrt{x}-2-\sqrt{x}-9}{2\sqrt{x}+18}< 0\)
\(\Leftrightarrow\dfrac{\sqrt{x}-11}{2\sqrt{x}+18}< 0\)
\(\Leftrightarrow\sqrt{x}-11< 0\)
\(\Leftrightarrow\sqrt{x}< 11\)
\(\Leftrightarrow x< 121\)
Kết hợp với điều kiện của \(x\), ta được: \(0< x< 121;x\ne4\)
mà \(x\) là số nguyên lớn nhất
nên \(x=120\)
Vậy \(P< \dfrac{1}{2}\) khi \(x=120\).
