Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
\(P=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\dfrac{\sqrt{x}-3}{\sqrt{x}+3}\)
\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\dfrac{\sqrt{x}-3}{\sqrt{x}+3}\)
\(=\dfrac{-3\sqrt{x}-3}{\left(\sqrt{x}+3\right)^2}\)
\(P=-\dfrac{1}{3}\)
\(\Rightarrow\left(\sqrt{x}+3\right)^2=3\sqrt{x}+3\)
\(\Leftrightarrow x-\sqrt{x}+6=0\)
\(\Leftrightarrow\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)=0\)
\(\Leftrightarrow x=9\left(Vì\sqrt{x}+2>0\right)\)
\(P=-\left(\dfrac{3\sqrt{x}+3}{\left(\sqrt{x}+3\right)^2}\right)=-\left(\dfrac{3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)^2}\right)< -3< -1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(A=\dfrac{2x+2+x+\sqrt{x}+1-x+\sqrt{x}-1}{\sqrt{x}}\)
\(=\dfrac{2x+2\sqrt{x}+2}{\sqrt{x}}\)
b: \(A-5=\dfrac{2x-4\sqrt{x}+2}{\sqrt{x}}=\dfrac{2\left(\sqrt{x}-1\right)^2}{\sqrt{x}}>=0\)
=>A>=5
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,B=\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2}{\sqrt{x}+3}-\dfrac{9\sqrt{x}-3}{x+\sqrt{x}-6}\left(x>0;x\ne6\right)\\ =\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2}{\sqrt{x}+3}-\dfrac{9\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}+\dfrac{2\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}-\dfrac{9\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{x+3\sqrt{x}+\sqrt{x}+3+2\sqrt{x}-4-9\sqrt{x}+3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{x-3\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\\\)
\(=\dfrac{x-\sqrt{x}-2\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)-2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{\sqrt{x}-1}{\sqrt{x}+3}\)
`b,` Tớ tính mãi ko ra, xl cậu nha=')
b) Xét hiệu:
\(\dfrac{\sqrt{x}-1}{\sqrt{x}+3}-3\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+3}-\dfrac{3\left(\sqrt{x}+3\right)}{\sqrt{x}+3}\)
\(=\dfrac{\sqrt{x}-1-3\sqrt{x}-9}{\sqrt{x}+3}\)
\(=\dfrac{-2\sqrt{x}-10}{\sqrt{x}+3}\)
\(=\dfrac{-2\left(\sqrt{x}+5\right)}{\sqrt{x}+3}\)
Mà: \(x>0\Rightarrow\left\{{}\begin{matrix}\sqrt{x}+5\ge5>0\\\sqrt{x}+3\ge3>0\end{matrix}\right.\)
\(\Rightarrow\dfrac{\sqrt{x}+5}{\sqrt{x}+3}>0\)
\(\Rightarrow\dfrac{-2\left(\sqrt{x}+5\right)}{\sqrt{x}+3}< 0\)
Vậy: \(\dfrac{\sqrt{x}-1}{\sqrt{x}+3}< 3\forall x>0\)
(giúp cậu nó nha)
![](https://rs.olm.vn/images/avt/0.png?1311)
câu 2 rút gọn A và tìm các giá trị nguyên của x để A nhận giá trị âm
1) So sánh:
N = \(\dfrac{5+\sqrt{5}}{\sqrt{5}+1}-\sqrt{6-2\sqrt{5}}\)
\(=\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{\sqrt{5}+1}-\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=\sqrt{5}-\left(\sqrt{5}-1\right)=1\)
M = \(\sqrt{18}-\sqrt{8}\)
\(=3\sqrt{2}-2\sqrt{2}\)
\(=\sqrt{2}\)
Ta có: \(1=\sqrt{1}\)
Mà 1 < 2
\(\Rightarrow\sqrt{1}< \sqrt{2}\)
Hay 1 \(< \sqrt{2}\)
Vậy N < M
![](https://rs.olm.vn/images/avt/0.png?1311)
a) ĐKXĐ: \(x>0\)
\(A=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}-\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+1\)
\(=x+\sqrt{x}-2\sqrt{x}-1+1=x-\sqrt{x}\)
\(A=x-\sqrt{x}=2\)
\(\Leftrightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)=0\)
\(\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\left(tm\right)\)(do \(\sqrt{x}+1\ge1>0\))
b) \(A=x-\sqrt{x}=\sqrt{x}\left(\sqrt{x}-1\right)>0\)(do \(x>1\))
\(\Leftrightarrow A=x-\sqrt{x}=\left|A\right|\)
c) \(A=x-\sqrt{x}=\left(x-\sqrt{x}+\dfrac{1}{4}\right)-\dfrac{1}{4}\)
\(=\left(\sqrt{x}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)
\(minA=-\dfrac{1}{4}\Leftrightarrow\sqrt[]{x}=\dfrac{1}{2}\Leftrightarrow x=\dfrac{1}{4}\left(tm\right)\)
\(a,A=\dfrac{x\left(x\sqrt{x}+1\right)}{x-\sqrt{x}+1}-\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+1\left(x>0\right)\\ A=\dfrac{x\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}-2\sqrt{x}-1+1\\ A=x+\sqrt{x}-2\sqrt{x}=x-\sqrt{x}\\ A=2\Leftrightarrow x-\sqrt{x}-2=0\\ \Leftrightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)=0\\ \Leftrightarrow\sqrt{x}=2\left(\sqrt{x}>0\right)\\ \Leftrightarrow x=4\left(tm\right)\)
\(b,x>1\Leftrightarrow\sqrt{x}-1>0\\ \Leftrightarrow\left|A\right|=\left|x-\sqrt{x}\right|=\left|\sqrt{x}\left(\sqrt{x}-1\right)\right|=\sqrt{x}\left(\sqrt{x}-1\right)=A\left(\sqrt{x}>0\right)\)
\(c,A=x-\sqrt{x}+\dfrac{1}{4}-\dfrac{1}{4}=\left(\sqrt{x}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\\ A_{min}=-\dfrac{1}{4}\Leftrightarrow\sqrt{x}=\dfrac{1}{2}\Leftrightarrow x=\dfrac{1}{4}\left(tm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) A= \(\dfrac{\sqrt{x}}{\sqrt{x-2}}-\dfrac{4}{x-2\sqrt{x}}=\dfrac{\sqrt{x}\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}-\dfrac{4}{\sqrt{x}\left(\sqrt{x}-2\right)}=\dfrac{x-4}{\sqrt{x}\left(\sqrt{x}-2\right)}=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}+2}{\sqrt{x}}=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\sqrt{x}\sqrt{x}}=\dfrac{x+2\sqrt{x}}{x}\)
b) Ta có x >0 nên \(\sqrt{x}\) >0
<=> \(2\sqrt{x}\) > 0
<=> \(x+2\sqrt{x}\) > x
<=> \(\dfrac{x+2\sqrt{x}}{x}\) > \(\dfrac{x}{x}\)
hay A > 1
c)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(P=1:\left(\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\right)\)
\(=1:\dfrac{x+2+x-1-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}=\dfrac{x+\sqrt{x}+1}{\sqrt{x}}\)
b: \(P-3=\dfrac{x-2\sqrt{x}+1}{\sqrt{x}}=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}>0\)
=>P>3
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(P=\dfrac{x+2+x-1-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\dfrac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\)
b: Thay x=1/9 vào P, ta được:
\(P=\dfrac{1}{3}:\left(\dfrac{1}{9}+\dfrac{1}{3}+1\right)=\dfrac{1}{3}:\dfrac{1+3+9}{9}=\dfrac{1}{3}\cdot\dfrac{9}{13}=\dfrac{3}{13}\)
a) P < 1; P < 2
b) P < 1 --> P(1 - P) > 0
--> P > P^2