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)
a) \(P=\dfrac{1-2\sqrt{a}+a}{1-\sqrt{a}}\cdot\dfrac{1+2\sqrt{a}+a}{1+\sqrt{a}}\) \(=\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)\) \(=1-a\)
b) ĐKXĐ: \(\left\{{}\begin{matrix}a\ge0\\a\ne1\end{matrix}\right.\)
Để P>0 \(\Leftrightarrow1-a>0\) \(\Leftrightarrow a< 1\)
Vậy \(0\le a< 1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) ĐKXĐ: \(a\ge0;a\ne1\)
\(P=\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\dfrac{1+a\sqrt{a}}{1+\sqrt{a}}-\sqrt{a}\right)\)
\(=\left[\dfrac{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}{1-\sqrt{a}}+\sqrt{a}\right]\left[\dfrac{\left(1+\sqrt{a}\right)\left(1-\sqrt{a}+a\right)}{1+\sqrt{a}}\right]\)
\(=\left(1+2\sqrt{a}+a\right)\left(1-2\sqrt{a}+a\right)\)
\(=\left(1-a\right)^2\)
b) Để \(P< 7-4\sqrt{3}\)
\(\Rightarrow\left(1-a\right)^2< 7-4\sqrt{3}\)
\(\Leftrightarrow\left|1-a\right|< \left(2-\sqrt{3}\right)^2\)
\(\Leftrightarrow\sqrt{3}-2< a-1< 2-\sqrt{3}\)
\(\Leftrightarrow\sqrt{3}-1< a< 3-\sqrt{3}\)
Vậy \(\sqrt{3}-1< a< 3-\sqrt{3}\) thì \(P< 7-4\sqrt{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: ĐKXĐ: \(\left\{{}\begin{matrix}a>0\\a< >1\end{matrix}\right.\)
\(P=\dfrac{a\sqrt{a}-1}{a-\sqrt{a}}-\dfrac{a\sqrt{a}+1}{a+\sqrt{a}}+\left(\sqrt{a}-\dfrac{1}{\sqrt{a}}\right)\cdot\left(\dfrac{3\sqrt{a}}{\sqrt{a}-1}-\dfrac{\sqrt{a}+2}{\sqrt{a}+1}\right)\)
\(=\dfrac{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}-\dfrac{\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}+1\right)}+\dfrac{a-1}{\sqrt{a}}\cdot\dfrac{3\sqrt{a}\left(\sqrt{a}+1\right)-\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}{a-1}\)
\(=\dfrac{a+\sqrt{a}+1-\left(a-\sqrt{a}+1\right)}{\sqrt{a}}+\dfrac{3a+3\sqrt{a}-a-\sqrt{a}+2}{\sqrt{a}}\)
\(=\dfrac{2\sqrt{a}+2a+2\sqrt{a}+2}{\sqrt{a}}=\dfrac{2\left(\sqrt{a}+1\right)^2}{\sqrt{a}}\)
b: \(P=\sqrt{a}+7\)
=>\(2\left(a+2\sqrt{a}+1\right)=a+7\sqrt{a}\)
=>\(2a+4\sqrt{a}+2-a-7\sqrt{a}=0\)
=>\(a-3\sqrt{a}+2=0\)
=>\(\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)=0\)
=>\(\left[{}\begin{matrix}a=1\left(loại\right)\\a=4\left(nhận\right)\end{matrix}\right.\)
c: \(P-6=\dfrac{2\left(\sqrt{a}+1\right)^2-6\sqrt{a}}{\sqrt{a}}\)
\(=\dfrac{2a+4\sqrt{a}+2-6\sqrt{a}}{\sqrt{a}}=\dfrac{2a-2\sqrt{a}+2}{\sqrt{a}}\)
\(=\dfrac{2\left(a-\sqrt{a}+\dfrac{1}{4}+\dfrac{3}{4}\right)}{\sqrt{a}}=\dfrac{2\left[\left(\sqrt{a}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\right]}{\sqrt{a}}>0\)
=>P>6
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(P=\dfrac{a-\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\sqrt{a}-1}{1}=\sqrt{a}-1\)
b: Để P<0 thì căn a-1<0
=>căn a<1
=>0<a<1
![](https://rs.olm.vn/images/avt/0.png?1311)
\(P=\left(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{\sqrt{a}}{a-\sqrt{a}}\right):\dfrac{\sqrt{a}+1}{a-1}\left(a>0;a\ne1\right)\)
\(a,P=\left(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{\sqrt{a}}{a-\sqrt{a}}\right):\dfrac{\sqrt{a}+1}{a-1}\)
\(=\left[\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\right]:\dfrac{\sqrt{a}+1}{a-1}\)
\(=\left(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}-1}\right):\dfrac{\sqrt{a}+1}{a-1}\)
\(=1:\dfrac{\sqrt{a}+1}{a-1}\)
\(=\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\sqrt{a}+1}\)
\(=\sqrt{a}-1\)
\(b,P< 0\Rightarrow\sqrt{a}-1< 0\Leftrightarrow\sqrt{a}< 1\Leftrightarrow a< 1\)
Kết hợp điều kiện \(a>0;a\ne1\)
\(\Rightarrow0< a< 1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Sửa đề: \(Q=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}-1}{\sqrt{a}+2}\right)\)
a) ĐKXĐ: \(\left\{{}\begin{matrix}a\ge0\\a\notin\left\{1;4\right\}\end{matrix}\right.\)
Ta có: \(Q=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}-1}{\sqrt{a}+2}\right)\)
\(=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a+3\sqrt{a}+2-a+3\sqrt{a}-2}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\)
\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}{6\sqrt{a}}\)
\(=\dfrac{a-4}{6a\left(\sqrt{a}-1\right)}\)
c) Thay \(a=9-4\sqrt{5}\) vào Q, ta được:
\(Q=\dfrac{5-4\sqrt{5}}{6\left(9-4\sqrt{5}\right)\left(\sqrt{5}-3\right)}\)
\(=\dfrac{5-4\sqrt{5}}{6\left(9\sqrt{5}-27-20+12\sqrt{5}\right)}\)
\(=\dfrac{5-4\sqrt{5}}{6\left(21\sqrt{5}-47\right)}\)
\(=\dfrac{\left(5-4\sqrt{5}\right)\left(21\sqrt{5}+47\right)}{-24}\)
\(=\dfrac{105\sqrt{5}+235-420-188\sqrt{5}}{-24}\)
\(=\dfrac{-83\sqrt{5}-185}{-24}=\dfrac{83\sqrt{5}+185}{24}\)
`a)P=((1-asqrta)/(1-sqrta)+sqrta).((1+asqrta)/(1+sqrta)-sqrta)`
`=(((1-sqrta)(a+sqrta+1))/(1-sqrta)+sqrta).(((1+sqrta)(a-sqrta+1))/(1+sqrta)-sqrta)`
`=(a+sqrta+1+sqrta)(a-sqrta+1-sqrta)`
`=(a+2sqrta+1)(a-2sqrta+1)`
`=(sqrta+1)^2(sqrta-1)^2`
`=(a-1)^2`
`b)a<7-4sqrt3`
`<=>(a-1)^2<(2-sqrt3)^2`
`<=>sqrt3-2<a-1<2-sqrt3`
`<=>sqrt3-1<a<3-sqrt3`