![](https://rs.olm.vn/images/avt/0.png?1311)
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)
![](https://rs.olm.vn/images/avt/0.png?1311)
c/\(P=\frac{\frac{2\left(\sqrt{x}-1\right)}{x\sqrt{x}-1}}{1-\frac{x+2}{x+\sqrt{x}+1}}\)\(\Leftrightarrow P=\frac{2\left(\sqrt{x}-1\right)}{x\sqrt{x}-1}:\frac{\sqrt{x}-1}{x+\sqrt{x}+1}\)
\(\Leftrightarrow\frac{2\left(x+\sqrt{x}+1\right)}{x\sqrt{x}-1}\)
Xét P-1 ta có \(\frac{2x+2\sqrt[]{x}+2-x\sqrt{x}+1}{x\sqrt{x}-1}=\frac{2x+2\sqrt{x}-x\sqrt{x}+3}{x\sqrt{x}-1}\)
với x<1 thì tử dương, mẫu âm, với x>1 thì tử âm và mẫu dương
Từ đó ta luuon có P-1\(\le0\RightarrowĐPCM\)
a/\(\Leftrightarrow x=\frac{5-\sqrt{5}}{1-\sqrt{5}}+\frac{5+\sqrt{5}}{1+\sqrt{5}}-\frac{25-5}{1-5}-1\)
\(\Leftrightarrow x=0+5-1\Leftrightarrow x=4\)
Thay vào B đc \(B=\frac{4+2}{4+2+1}=\frac{6}{7}\)
b/
![](https://rs.olm.vn/images/avt/0.png?1311)
- Các ĐKXĐ tự tìm dùm mình hen :)
Ta có : \(D=\left(\frac{5}{x-\sqrt{x}-6}+\frac{1}{\sqrt{x}+2}\right):\frac{1}{\sqrt{x}-3}\)
=> \(D=\left(\frac{5}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}+\frac{1}{\sqrt{x}+2}\right)\left(\sqrt{x}-3\right)\)
=> \(D=\left(\frac{5+\sqrt{x}-3}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\right)\left(\sqrt{x}-3\right)\)
=> \(D=\left(\frac{\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\right)\left(\sqrt{x}-3\right)\)
=> \(D=\left(\frac{1}{\sqrt{x}-3}\right)\left(\sqrt{x}-3\right)=1\)
Ta có : \(E=\left(\frac{1}{a-\sqrt{a}}+\frac{1}{\sqrt{a}-1}\right):\frac{\sqrt{a+1}}{a-2\sqrt{a}+1}\)
=> \(E=\left(\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}+\frac{1}{\sqrt{a}-1}\right):\frac{\sqrt{a+1}}{\left(\sqrt{a}-1\right)^2}\)
=> \(E=\left(\frac{\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\frac{\sqrt{a}+1}{\left(\sqrt{a}-1\right)^2}\)
=> \(E=\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)^2}{\sqrt{a}\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}=\frac{\sqrt{a}-1}{\sqrt{a}}\)
( làm đến đây thôi câu còn lại bạn tự làm hen )
Ghét nhất mấy câu viết sai đề b, c sai rất nhiều bạn ới
đấy là mình đánh máy tính nên kéo dài hơi nhầm bạn ơi chứ không phải sai đề :))
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\left(\frac{2+\sqrt{x}}{x-5\sqrt{x}+6}-\frac{\sqrt{x}+3}{2-\sqrt{x}}-\frac{\sqrt{x}+2}{\sqrt{x}-3}\right)\) \(:\left(2-\frac{\sqrt{x}}{\sqrt{x}+1}\right)\)
\(A=\left[\frac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{\sqrt{x}+2}{\sqrt{x}-3}\right]\)
\(:\left[\frac{2\left(\sqrt{x}+1\right)-\sqrt{x}}{\sqrt{x}+1}\right]\)
\(A=\left[\frac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\frac{x-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{\sqrt{x}+2}{\sqrt{x}-3}\right]\)
\(:\left[\frac{2\sqrt{x}+2-\sqrt{x}}{\sqrt{x}+1}\right]\)
\(A=\left[\frac{\sqrt{x}+2+x-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\right]\) \(:\left[\frac{\sqrt{x}+2}{\sqrt{x}+1}\right]\)
\(A=\left[\frac{\sqrt{x}+x-7-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right]:\frac{\sqrt{x}+2}{\sqrt{x}+1}\)
\(A=\frac{\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}:\frac{\sqrt{x}+2}{\sqrt{x}+1}\)
\(A=\frac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Cái này là toán lớp 9 chứ.
a)
ĐKXĐ : \(x\ne\pm4\)
\(A=\left(\frac{x-\sqrt{x}+7}{x-4}+\frac{\sqrt{x}+2}{x-4}\right):\left(\frac{\left(\sqrt{x}+2\right)^2}{x-4}-\frac{\left(\sqrt{x}-2\right)^2}{x-4}-\frac{2\sqrt{x}}{x-4}\right)\)
\(=\left(\frac{x-\sqrt{x}+7+\sqrt{x}+2}{x-4}\right):\left(\frac{x+4\sqrt{x}+4-x+4\sqrt{x}-4-2\sqrt{x}}{x-4}\right)\)
\(=\frac{x+9}{x-4}\cdot\frac{x-4}{6\sqrt{x}}=\frac{x+9}{6\sqrt{x}}\)
b)
Ta có
\(x+9-6\sqrt{x}=\left(\sqrt{x}-3\right)^2\ge0\)
\(\Rightarrow x+9\ge6\sqrt{x}\)
\(\Rightarrow\frac{x+9}{6\sqrt{x}}\ge1\)
\(\Leftrightarrow A\ge1\)
\(\Leftrightarrow\frac{1}{A}\le1\)
\(\Rightarrow A\ge\frac{1}{A}\)
\(A=\left(\frac{1}{\sqrt{x}-2}+\frac{1}{\sqrt{x}+2}\right):\frac{5x}{x-5}\)
\(=\left(\frac{\sqrt{x}+2}{x-4}+\frac{\sqrt{x}-2}{x-4}\right):\frac{5x}{x-5}\)
\(=\frac{2\sqrt{x}}{x-4}.\frac{x-5}{5x}=\frac{2\sqrt{x}\left(x-5\right)}{5x^2-20x}\)