Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(R=\left[\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3\left(\sqrt{x}+3\right)}{x-9}\right]:\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
a/ \(R=\left[\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt[]{x-3}\right)}\right]:\left(\frac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\right)\)
=> \(R=\left[\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3}{\sqrt[]{x-3}}\right]:\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
=> \(R=\left[\frac{2\sqrt{x}}{\sqrt{x}-3}+1\right]:\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
=> \(R=\left[\frac{2\sqrt{x}+\sqrt{x}-3}{\sqrt{x}-3}\right].\frac{\sqrt{x}-3}{\sqrt{x}+1}\)
=> \(R=\frac{3\sqrt{x}-3}{\sqrt{x}-3}.\frac{\sqrt{x}-3}{\sqrt{x}+1}=\frac{3\left(\sqrt{x}-1\right)}{\sqrt{x}+1}\)
b/ Để R<-1 => \(\frac{3\left(\sqrt{x}-1\right)}{\sqrt{x}+1}< -1\)
<=> \(3\sqrt{x}-3< -\sqrt{x}-1\)
<=> \(4\sqrt{x}< 2\)=> \(\sqrt{x}< \frac{1}{2}\) => \(-\frac{1}{4}< x< \frac{1}{4}\)
Chỗ => R = \(\left(\frac{2\sqrt{x}}{\sqrt{x}-3}+1\right):\frac{\sqrt{x}+1}{\sqrt{x}-3}\) là sao vậy ạ?
ĐKXĐ: 5x-9>=0
=>5x>=9
=>x>=1,8
Ta có: \(K=\sqrt{5x+6\sqrt{5x-9}}+\sqrt{5x-6\sqrt{5x-9}}\)
\(=\sqrt{5x-9+2\cdot\sqrt{5x-9}\cdot3+9}+\sqrt{5x-9-2\cdot\sqrt{5x-9}\cdot3+9}\)
\(=\sqrt{\left(\sqrt{5x-9}+3\right)^2}+\sqrt{\left(\sqrt{5x-9}-3\right)^2}=\left|\sqrt{5x-9}+3\right|+\left|\sqrt{5x-9}-3\right|\)
=>\(K\le\left|\sqrt{5x-9}+3+3-\sqrt{5x-9}\right|=6\)
Dấu '=' xảy ra khi \(\left(\sqrt{5x-9}+3\right)\left(\sqrt{5x-9}-3\right)\le0\)
=>\(\sqrt{5x-9}-3\le0\)
=>\(\sqrt{5x-9}\le3\)
=>5x-9<=9
=>5x<=18
=>x<=3,6
=>1,8<=x<=3,6