Bài 3:
1: ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x\ne9\end{matrix}\right.\)
\(M=\left(\dfrac{x+\sqrt{x}+10}{x-9}-\dfrac{1}{\sqrt{x}-3}\right):\dfrac{1}{\sqrt{x}-3}\)
\(=\left(\dfrac{x+\sqrt{x}+10}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}-\dfrac{1}{\sqrt{x}-3}\right)\cdot\dfrac{\sqrt{x}-3}{1}\)
\(=\dfrac{x+\sqrt{x}+10-\sqrt{x}-3}{\left(\sqrt{x}+3\right)\cdot\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}-3}{1}=\dfrac{x+7}{\sqrt{x}+3}\)
2: Thay x=25 vào M, ta được:
\(M=\dfrac{25+7}{\sqrt{25}+3}=\dfrac{32}{5+3}=\dfrac{32}{8}=4\)
3: \(M< \sqrt{x}+1\)
=>\(\dfrac{x+7}{\sqrt{x}+3}< \sqrt{x}+1\)
=>\(\dfrac{x+7-\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}{\sqrt{x}+3}< 0\)
=>\(x+7-x-4\sqrt{x}-3< 0\)
=>\(-4\sqrt{x}+4< 0\)
=>\(-4\sqrt{x}< -4\)
=>\(\sqrt{x}>1\)
=>x>1
kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}x>1\\x\ne9\end{matrix}\right.\)
bài 5:
1: Thay x=49 vào B, ta được:
\(B=\dfrac{1}{\sqrt{49}-1}=\dfrac{1}{7-1}=\dfrac{1}{6}\)
2:
ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x\ne1\end{matrix}\right.\)
S=A-B
\(=\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\)
\(=\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\)
\(=\dfrac{x+2+\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{-\sqrt{x}+1+x-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\dfrac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\)
3: \(S-\dfrac{1}{3}=\dfrac{\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{1}{3}\)
\(=\dfrac{3\sqrt{x}-x-\sqrt{x}-1}{3\left(x+\sqrt{x}+1\right)}=\dfrac{-x+2\sqrt{x}-1}{3\left(x+\sqrt{x}+1\right)}\)
\(=-\dfrac{\left(x-2\sqrt{x}+1\right)}{3\left(x+\sqrt{x}+1\right)}=\dfrac{-\left(\sqrt{x}-1\right)^2}{3\left(x+\sqrt{x}+1\right)}< 0\)
=>\(S< \dfrac{1}{3}\)
