Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=\left(\frac{2}{\sqrt{x}-2}+\frac{3}{2\sqrt{x}+1}-\frac{5\sqrt{x}-7}{2x-3\sqrt{x}-2}\right):\)\(\frac{2\sqrt{x}+3}{5x-10\sqrt{x}}\)
\(=\left(\frac{2}{\sqrt{x}-2}+\frac{3}{2\sqrt{x}+1}-\frac{5\sqrt{x}-7}{\left(\sqrt{x}-2\right)\left(2\sqrt{x}+1\right)}\right)\)\(:\frac{2\sqrt{x}+3}{5x-10\sqrt{x}}\)
\(=\frac{2\left(2\sqrt{x}+1\right)+3\left(\sqrt{x}-2\right)-5\sqrt{x}+7}{\left(\sqrt{x}-2\right)\left(2\sqrt{x}+1\right)}\)\(:\frac{2\sqrt{x}+3}{5\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\frac{4\sqrt{x}+2+3\sqrt{x}-6-5\sqrt{x}+7}{\left(\sqrt{x}-2\right)\left(2\sqrt{x}+1\right)}\)\(.\frac{5\sqrt{x}\left(\sqrt{x}-2\right)}{2\sqrt{x}+3}\)
\(=\frac{2\sqrt{x}+3}{2\sqrt{x}+1}.\frac{5\sqrt{x}}{2\sqrt{x}+3}=\frac{5\sqrt{x}}{2\sqrt{x}+1}\)
\(A\in Z\Leftrightarrow\frac{5\sqrt{x}}{2\sqrt{x}+1}\in Z\Leftrightarrow\frac{10\sqrt{x}}{2\sqrt{x}+1}\in Z\)
\(\Rightarrow\frac{10\sqrt{x}+5-5}{2\sqrt{x}+1}\in Z\Leftrightarrow5-\frac{5}{2\sqrt{x}+1}\in Z\)
\(\Rightarrow\frac{5}{2\sqrt{x}+1}\in Z\Rightarrow2\sqrt{x}+1\inƯ_5\)
Mà \(Ư_5=\left\{\pm1;\pm5\right\}\)
Nhưng \(2\sqrt{x}+1\ge1\)
\(\Rightarrow\orbr{\begin{cases}2\sqrt{x}+1=1\\2\sqrt{x}+1=5\end{cases}\Rightarrow\orbr{\begin{cases}2\sqrt{x}=0\\2\sqrt{x}=4\end{cases}}}\)
\(\Rightarrow\orbr{\begin{cases}\sqrt{x}=0\\\sqrt{x}=2\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=4\end{cases}}}\)
Vậy \(x\in\left\{0;4\right\}\)
+) \(P=\frac{\sqrt{x}+\sqrt{x^2-2x+1}+1}{\sqrt{x^2-2x+1}}=\frac{\sqrt{x}+\left|x-1\right|+1}{\left|x-1\right|}\)
+) \(x=a+1-\sqrt{1+a^2+\frac{a^2}{\left(a+1\right)^2}}\)
\(=a+1-\sqrt{\left(a+1\right)^2-2a+\frac{a^2}{\left(a+1\right)^2}}\)
\(=a+1-\sqrt{\left(a+1-\frac{a}{a+1}\right)^2}\) vì a > 0 => \(a+1-\frac{a}{a+1}=\frac{a^2+a+1}{a+1}>0\)
\(=a+1-\left(a+1-\frac{a}{a+1}\right)=\frac{a}{a+1}\)
=> \(\left|x-1\right|=\left|\frac{a}{a+1}-1\right|=\left|-\frac{1}{a+1}\right|=\frac{1}{a+1}\)
=> \(P=\frac{\sqrt{\frac{a}{a+1}}+\frac{1}{a+1}+1}{\frac{1}{a+1}}=\sqrt{a\left(a+1\right)}+a+2\)
\(P=\frac{B}{A}=\frac{3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)}.\frac{\sqrt{x}}{\left(\sqrt{x}+1\right)}=\frac{3\sqrt{x}}{\sqrt{x}-2}\)
Để \(\left|P\right|>P\Rightarrow P< 0\)
\(\Rightarrow\frac{3\sqrt{x}}{\sqrt{x}-2}< 0\Rightarrow\sqrt{x}-2< 0\Rightarrow\sqrt{x}< 2\Rightarrow x< 4\)
Mà \(x\) nguyên \(\Rightarrow x=\left\{2;3\right\}\)