Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(P=A:B=\dfrac{\sqrt{x}+2}{\sqrt{x}}:\dfrac{\sqrt{x}+2}{\sqrt{x}+1}=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
P>3/2
=>P-3/2>0
=>\(\dfrac{\sqrt{x}+1}{\sqrt{x}}-\dfrac{3}{2}>0\)
=>\(\dfrac{2\sqrt{x}+2-3\sqrt{x}}{2\sqrt{x}}>0\)
=>-căn x+2>0
=>-căn x>-2
=>0<x<4
Lời giải:
a. ĐKXĐ: $x\geq 0$
$P< \frac{1}{2}\Leftrightarrow \frac{\sqrt{x}}{\sqrt{x}+2}< \frac{1}{2}$
$\Leftrightarrow \frac{\sqrt{x}}{\sqrt{x}+2}-\frac{1}{2}<0$
$\Leftrightarrow \frac{\sqrt{x}-2}{2(\sqrt{x}+2)}<0$
$\Leftrightarrow \sqrt{x}-2<0$ (do mẫu dương rồi)
$\Leftrightarrow 0\leq x< 4$
Kết hợp đkxđ suy ra $0\leq x< 4$
b.
Với $x\geq 0$ thì $P\geq 0$
Lại có: $P<1$ (do tử nhỏ hơn mẫu)
$\Rightarrow P$ nguyên khi mà $P=0$
$\Leftrightarrow x=0$
Ta có: \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}+2}-\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{4\sqrt{x}-1}{x-4}\right):\dfrac{1}{\sqrt{x}+2}\)
\(=\dfrac{x-2\sqrt{x}-x-2\sqrt{x}+4\sqrt{x}-1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+2}{1}\)
\(=\dfrac{-1}{\sqrt{x}-2}\)
Để P nguyên thì \(\sqrt{x}-2\in\left\{-1;1\right\}\)
\(\Leftrightarrow\sqrt{x}\in\left\{1;3\right\}\)
hay \(x\in\left\{1;9\right\}\)
1) Ta có: \(P=\dfrac{1}{\sqrt{x}-1}-\dfrac{x\sqrt{x}-\sqrt{x}}{x+1}\left(\dfrac{1}{x-2\sqrt{x}+1}+\dfrac{1}{1-x}\right)\)
\(=\dfrac{1}{\sqrt{x}-1}-\dfrac{\sqrt{x}\left(x-1\right)}{x+1}\cdot\left(\dfrac{\sqrt{x}+1-\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)^2\cdot\left(\sqrt{x}+1\right)}\right)\)
\(=\dfrac{1}{\sqrt{x}-1}-\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{x+1}\cdot\dfrac{2}{\left(\sqrt{x}-1\right)^2\cdot\left(\sqrt{x}+1\right)}\)
\(=\dfrac{1}{\sqrt{x}-1}-\dfrac{2\sqrt{x}}{\left(x+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{x-2\sqrt{x}+1}{\left(x+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\sqrt{x}-1}{x+1}\)
Để \(P=-\dfrac{2}{5}\) thì \(\dfrac{\sqrt{x}-1}{x+1}=\dfrac{-2}{5}\)
\(\Leftrightarrow-2x-2=5\sqrt{x}-5\)
\(\Leftrightarrow-2x-2-5\sqrt{x}+5=0\)
\(\Leftrightarrow-2x-5\sqrt{x}+3=0\)
\(\Leftrightarrow-2x-6\sqrt{x}+\sqrt{x}+3=0\)
\(\Leftrightarrow-2\sqrt{x}\left(\sqrt{x}+3\right)+\left(\sqrt{x}+3\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}+3\right)\left(-2\sqrt{x}+1\right)=0\)
\(\Leftrightarrow-2\sqrt{x}+1=0\)
\(\Leftrightarrow-2\sqrt{x}=-1\)
\(\Leftrightarrow x=\dfrac{1}{4}\)(thỏa ĐK)
\(a,P=\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}+1}{2}=\dfrac{\sqrt{x}+1}{2\sqrt{x}}\\ b,P=1\Leftrightarrow\sqrt{x}+1=2\sqrt{x}\\ \Leftrightarrow\sqrt{x}=1\Leftrightarrow x=1\\ c,P=\dfrac{\sqrt{x}+1}{2\sqrt{x}}\in Z\\ \Leftrightarrow\sqrt{x}+1⋮2\sqrt{x}\\ \Leftrightarrow2\sqrt{x}+2⋮2\sqrt{x}\\ \Leftrightarrow2\sqrt{x}\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\\ \Leftrightarrow\sqrt{x}=1\left(\sqrt{x}>0\right)\\ \Leftrightarrow x=1\)
a: \(P=\left(\dfrac{2+\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\dfrac{\sqrt{x}+1-\sqrt{x}}{\sqrt{x}+1}\)
\(=\dfrac{1}{\sqrt{x}-1}\cdot\dfrac{\sqrt{x}+1}{1}=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
b: Để P nguyên thì \(\sqrt{x}+1⋮\sqrt{x}-1\)
\(\Leftrightarrow\sqrt{x}-1\in\left\{-1;1;2\right\}\)
hay \(x\in\left\{0;4;9\right\}\)
P<1/2
=>P-1/2<0
=>\(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{1}{2}< 0\)
=>\(\dfrac{2\sqrt{x}-\sqrt{x}+2}{2\left(\sqrt{x}-2\right)}< 0\)
=>căn x-2<0
=>0<=x<4
=>căn x-2<0
=>0<=x<4
dòng này là sao ạ em chx hiểu