Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
ĐKXĐ: \(x>0;a\ne9\)
\(A=\left(\dfrac{\sqrt{a}+3}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}+\dfrac{\sqrt{a}-3}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}\right)\left(\dfrac{\sqrt{a}-3}{\sqrt{a}}\right)\)
\(=\left(\dfrac{\sqrt{a}+3+\sqrt{a}-3}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}\right)\left(\dfrac{\sqrt{a}-3}{\sqrt{a}}\right)\)
\(=\dfrac{2\sqrt{a}\left(\sqrt{a}-3\right)}{\sqrt{a}\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}=\dfrac{2}{\sqrt{a}+3}\)
b.
\(A>\dfrac{1}{2}\Rightarrow\dfrac{2}{\sqrt{a}+3}>\dfrac{1}{2}\Rightarrow\sqrt{a}+3< 4\)
\(\Rightarrow\sqrt{a}< 1\Rightarrow a< 1\)
Kết hợp ĐKXĐ \(\Rightarrow0< a< 1\)
a.(ĐKXĐ: \(a\ge0,a\ne\dfrac{1}{9}\))
=> \(A=\left(\dfrac{\sqrt{a}-1}{3\sqrt{a}-1}-\dfrac{1}{1+3\sqrt{a}}+\dfrac{8\sqrt{a}}{\left(3\sqrt{a}-1\right)\left(3\sqrt{a}+1\right)}\right):\dfrac{3\sqrt{a}+1-3\sqrt{a}+2}{3\sqrt{a}+1}=\dfrac{\left(\sqrt{a}-1\right)\left(1+3\sqrt{a}\right)-3\sqrt{a}+1+8\sqrt{a}}{\left(3\sqrt{a}-1\right)\left(3\sqrt{a}+1\right)}.\dfrac{3\sqrt{a}+1}{3}\)
\(=\dfrac{\sqrt{a}+3a-1-3\sqrt{a}-3\sqrt{a}+1+8\sqrt{a}}{3\left(3\sqrt{a}-1\right)}=\dfrac{3\sqrt{a}+3a}{3\left(3\sqrt{a}-1\right)}=\dfrac{3\left(\sqrt{a}+a\right)}{3\left(3\sqrt{a}-1\right)}=\dfrac{\sqrt{a}+a}{3\sqrt{a}-1}\)
b. Để A \(=\dfrac{6}{5}\Leftrightarrow\dfrac{\sqrt{a}+a}{3\sqrt{a}-1}=\dfrac{6}{5}\)
\(\Leftrightarrow5\left(\sqrt{a}+a\right)=6\left(3\sqrt{a}-1\right)\)
\(\Leftrightarrow5\sqrt{a}+5a-18\sqrt{a}+6=0\)
\(\Leftrightarrow5a-13\sqrt{a}+6=0\)
\(\Leftrightarrow5a-10\sqrt{a}-3\sqrt{a}+6=0\)
\(\Leftrightarrow5\sqrt{a}\left(\sqrt{a}-2\right)-3\left(\sqrt{a}-2\right)=0\)
\(\Leftrightarrow\left(\sqrt{a}-2\right)\left(5\sqrt{a}-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{a}-2=0\\5\sqrt{a}-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{a}=2\\\sqrt{a}=\dfrac{3}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=4\\a=\dfrac{9}{25}\end{matrix}\right.\)(nhận)
Vậy ...
a) Ta có: \(A=\left(\dfrac{1}{\sqrt{x}-3}+\dfrac{1}{\sqrt{x}+3}\right)\left(1-\dfrac{3}{\sqrt{x}}\right)\)
\(=\dfrac{\sqrt{x}+3+\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}}\)
\(=\dfrac{2}{\sqrt{x}+3}\)
b) Để \(A>\dfrac{1}{2}\) thì \(A-\dfrac{1}{2}>0\)
\(\Leftrightarrow\dfrac{4-\sqrt{x}-3}{2\left(\sqrt{x}+3\right)}>0\)
\(\Leftrightarrow1-\sqrt{x}>0\)
\(\Leftrightarrow\sqrt{x}< 1\)
hay x<1
Kết hợp ĐKXĐ, ta được: 0<x<1
a) \(A=\left(\dfrac{\sqrt{a}}{2}-\dfrac{1}{2\sqrt{a}}\right).\left(\dfrac{a-\sqrt{a}}{\sqrt{a}+1}-\dfrac{a+\sqrt{a}}{\sqrt{a}-1}\right)\left(đk:a>0,x\ne1\right)\)
\(=\dfrac{a-1}{2\sqrt{a}}.\dfrac{\left(a-\sqrt{a}\right)\left(\sqrt{a}-1\right)-\left(a+\sqrt{a}\right)\left(\sqrt{a}+1\right)}{a-1}\)
\(=\dfrac{a\sqrt{a}-2a+\sqrt{a}-a\sqrt{a}-2a-\sqrt{a}}{2\sqrt{a}}\)
\(=\dfrac{-4a}{2\sqrt{a}}=-2\sqrt{a}\)
b) \(A=-2\sqrt{a}>-6\)
\(\Leftrightarrow\sqrt{a}< 3\Leftrightarrow0\le a< 9\) và \(a\ne1\)
c) \(a^2-3=0\Leftrightarrow a^2=3\Leftrightarrow\sqrt{a}=\sqrt[4]{3}\)
\(\Rightarrow A=-2\sqrt{a}=-2\sqrt[4]{3}\)
`a)đk:a>0,a ne 9`
`A=((sqrta+3+sqrta-3)/(a-9)).((sqrta-3)/sqrta)`
`=((2sqrtx)/(a-9)).((sqrta-3)/sqrta)`
`=2/(sqrta+3)`
`b)A>1/2`
`<=>2/(sqrta+3)>1/2`
`<=>sqrta+3<4`
`<=>sqrta<1`
`<=>a<1`
KẾt hợp đkxđ:`0<x<1`
ĐKXĐ: \(\left\{{}\begin{matrix}a>0\\a\ne9\end{matrix}\right.\)
a) Ta có: \(A=\left(\dfrac{1}{\sqrt{a}-3}+\dfrac{1}{\sqrt{a}+3}\right)\left(1-\dfrac{3}{\sqrt{a}}\right)\)
\(=\dfrac{\sqrt{a}+3+\sqrt{a}-3}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}\cdot\dfrac{\sqrt{a}-3}{\sqrt{a}}\)
\(=\dfrac{2\sqrt{a}}{\sqrt{a}+3}\cdot\dfrac{1}{\sqrt{a}}\)
\(=\dfrac{2}{\sqrt{a}+3}\)
b) Để \(A>\dfrac{1}{2}\) thì \(A-\dfrac{1}{2}>0\)
\(\Leftrightarrow\dfrac{2}{\sqrt{a}+3}-\dfrac{1}{2}>0\)
\(\Leftrightarrow\dfrac{4-\left(\sqrt{a}+3\right)}{2\left(\sqrt{a}+3\right)}>0\)
mà \(2\left(\sqrt{a}+3\right)>0\forall a\)
nên \(1-\sqrt{a}>0\)
\(\Leftrightarrow\sqrt{a}< 1\)
hay a<1
Kết hợp ĐKXĐ, ta được: 0<a<1
\(A=\dfrac{2-\sqrt{a}-\sqrt{a}-3}{2\sqrt{a}+1}=-1\)
\(B=\dfrac{1}{1-\sqrt{2+\sqrt{3}}}-\dfrac{1}{1-\sqrt{2-\sqrt{3}}}\)
\(=\dfrac{\sqrt{2}}{\sqrt{2}-\sqrt{3}-1}-\dfrac{\sqrt{2}}{\sqrt{2}-\sqrt{3}+1}\)
\(=\dfrac{2-\sqrt{6}+\sqrt{2}-2+\sqrt{6}+\sqrt{2}}{5-2\sqrt{6}-1}\)
\(=\dfrac{2\sqrt{2}}{4-2\sqrt{6}}=\dfrac{1}{\sqrt{2}-\sqrt{3}}=-\sqrt{2}-\sqrt{3}\)
a) \(M=3\sqrt{3}-\sqrt{12}-\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(M=3\sqrt{3}-2\sqrt{3}-\left|\sqrt{3}-1\right|\)
\(M=\sqrt{3}-\sqrt{3}+1\)
\(M=1\)
b) Ta có:
\(N=\left(\dfrac{1}{a-\sqrt{a}}+\dfrac{1}{\sqrt{a}-1}\right):\dfrac{\sqrt{a}+1}{a-2\sqrt{a}+1}\)
\(N=\left(\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}+\dfrac{\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\dfrac{\sqrt{a}+1}{\left(\sqrt{a}-1\right)^2}\)
\(N=\left(\dfrac{1+\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\right)\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}\)
\(N=\dfrac{\left(\sqrt{a}+1\right)\cdot\left(\sqrt{a}-1\right)^2}{\sqrt{a}\left(\sqrt{a}-1\right)\cdot\left(\sqrt{a}+1\right)}\)
\(N=\dfrac{\sqrt{a}-1}{\sqrt{a}}\)
Theo đề ta có: \(M=2N\)
Khi: \(1=2\cdot\left(\dfrac{\sqrt{a}-1}{\sqrt{a}}\right)\)
\(\Leftrightarrow1=\dfrac{2\sqrt{a}-2}{\sqrt{a}}\)
\(\Leftrightarrow\sqrt{a}=2\sqrt{a}-2\)
\(\Leftrightarrow2\sqrt{a}-\sqrt{a}=2\)
\(\Leftrightarrow\sqrt{a}=2\)
\(\Leftrightarrow a=4\left(tm\right)\)
a)A=\(\dfrac{x-4\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}=\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}\left(\sqrt{x}-2\right)}\)=\(\dfrac{\sqrt{x}-2}{\sqrt{x}}\)
b) Thay x=3+2\(\sqrt{2}\)
A=\(\dfrac{\sqrt{3+2\sqrt{2}}-2}{\sqrt{3+2\sqrt{2}}}\)=\(\dfrac{\sqrt{\left(\sqrt{2}+1\right)^2-2}}{\sqrt{\left(\sqrt{2}+1\right)^2}}\)=\(\dfrac{\sqrt{2}+1-2}{\sqrt{2}+1}\)
A=\(\dfrac{\sqrt{2}-1}{\sqrt{2}+1}\)
c)Ta có \(\dfrac{\sqrt{x}-2}{\sqrt{x}}=1-\dfrac{2}{\sqrt{x}}\)>0
\(\Rightarrow\dfrac{2}{\sqrt{x}}\)<1\(\Rightarrow\sqrt{x}\)>2\(\Rightarrow x>4\)
a) \(P=\dfrac{\sqrt{a}\left[\left(\sqrt{a}\right)^3+1\right]}{a-\sqrt{a}+1}-\dfrac{\sqrt{a}\left(2\sqrt{a}+1\right)}{\sqrt{a}}+1\)
\(P=\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{a-\sqrt{a}+1}-\left(2\sqrt{a}+1\right)+1\)
\(P=\sqrt{a}\left(\sqrt{a}+1\right)-\left(2\sqrt{a}+1\right)+1\)
\(P=a+\sqrt{a}-2\sqrt{a}-1+1\)
\(P=a-\sqrt{a}\)
b) Với a > 1 thì \(a>\sqrt{a}\) , do đó \(P=a-\sqrt{a}>0\), suy ra \(\left|P\right|=P\)
c) \(A=a-\sqrt{a}=\left(\sqrt{a}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)
Vậy A nhỏ nhất bằng \(-\dfrac{1}{4}\) khi cà chỉ khi \(\sqrt{a}=\dfrac{1}{2}\) hay \(a=\dfrac{1}{4}\)
a: \(P=\sqrt{a}\left(\sqrt{a}+1\right)-2\sqrt{a}-1+1=a-\sqrt{a}\)
b: a>1 nên P>0
\(\Leftrightarrow P=\left|P\right|\)