Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
ĐK \(\hept{\begin{cases}x\ge0\\x\ne9\end{cases}}\)
a, \(R=\frac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\frac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)
\(=\frac{3x-6\sqrt{x}-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}+1}=\frac{3\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\frac{3\left(\sqrt{x}-3\right)}{\sqrt{x}+3}\)
b. \(R< -1\Rightarrow R+1< 0\Rightarrow\frac{3\sqrt{x}-9+\sqrt{x}+3}{\sqrt{x}+3}< 0\Rightarrow\frac{4\sqrt{x}-6}{\sqrt{x}+3}< 0\)
\(\Rightarrow0\le x< \frac{9}{4}\)
c. \(R=\frac{3\left(\sqrt{x}-3\right)}{\sqrt{x}+3}=3+\frac{-18}{\sqrt{x}+3}\)
Ta thấy \(\sqrt{x}+3\ge3\Rightarrow\frac{-18}{\sqrt{x}+3}\ge-6\Rightarrow3+\frac{-18}{\sqrt{x}+3}\ge-3\Rightarrow R\ge-3\)
Vậy \(MinR=-3\Leftrightarrow x=0\)
a: \(A=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}\)
\(=\sqrt{a}-\sqrt{b}-\sqrt{a}-\sqrt{b}=-2\sqrt{b}\)
b: \(B=\dfrac{2\sqrt{x}-x-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{x+\sqrt{x}+1}{x-1}\)
\(=\dfrac{-2x+\sqrt{x}-1}{\sqrt{x}-1}\cdot\dfrac{1}{x-1}\)
c: \(C=\dfrac{x-9-x+3\sqrt{x}}{x-9}:\left(\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}-2}{\sqrt{x}+3}+\dfrac{x-9}{x+\sqrt{x}-6}\right)\)
\(=\dfrac{3\left(\sqrt{x}-3\right)}{x-9}:\dfrac{9-x+x-4\sqrt{x}+4+x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3}{\sqrt{x}+3}\cdot\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{x-4\sqrt{x}+4}\)
\(=\dfrac{3}{\sqrt{x}-2}\)
a) \(ĐKXĐ:\hept{\begin{cases}x>0\\x\ne4\\x\ne9\end{cases}}\)
\(P=\left(\frac{\sqrt{x}-3}{2-\sqrt{x}}+\frac{\sqrt{x}+2}{3+\sqrt{x}}-\frac{9-x}{x+\sqrt{x}-6}\right):\left(1-\frac{3\sqrt{x}-9}{x-9}\right)\)
\(=\left[\frac{-\left(\sqrt{x}-3\right)}{\sqrt{x}-2}+\frac{\sqrt{x}+2}{\sqrt{x}+3}+\frac{x-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right]:\left[1-\frac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right]\)
\(=\left[\frac{-\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}+\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}+\frac{x-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right]:\left(1-\frac{3}{\sqrt{x}+3}\right)\)
\(=\left[\frac{-x+9+x-4+x-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right]:\left(\frac{\sqrt{x}+3-3}{\sqrt{x}+3}\right)\)
\(=\frac{x-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}:\frac{\sqrt{x}}{\sqrt{x}+3}\)
\(=\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}.\frac{\sqrt{x}+3}{\sqrt{x}}=\frac{\sqrt{x}+2}{\sqrt{x}}\)
b) Ta có: \(P=\frac{\sqrt{x}+2}{\sqrt{x}}=1+\frac{2}{\sqrt{x}}\)
Vì \(x\inℤ\)\(\Rightarrow\)Để P nguyên thì \(\frac{2}{\sqrt{x}}\inℤ\)
\(\Rightarrow2⋮\sqrt{x}\)\(\Rightarrow\sqrt{x}\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
Vì \(\sqrt{x}>0\)\(\Rightarrow\sqrt{x}\in\left\{1;2\right\}\)
\(\Rightarrow x\in\left\{1;4\right\}\)
So sánh với ĐKXĐ ta thấy \(x=1\)thỏa mãn
\(\Rightarrow P=\frac{\sqrt{1}+2}{\sqrt{1}}=\frac{1+2}{1}=3\)
Vậy \(x=1\)khi đó \(P=3\)
\(P=\left(\frac{\sqrt{x}-3}{2-\sqrt{x}}+\frac{\sqrt{x}+2}{3+\sqrt{x}}-\frac{9-x}{x+\sqrt{x}-6}\right)\div\left(1-\frac{3\sqrt{x}-9}{x-9}\right)\)
a) ĐK : \(\hept{\begin{cases}x\ge0\\x\ne4\\x\ne9\end{cases}}\)
\(=\left(\frac{3-\sqrt{x}}{\sqrt{x}-2}+\frac{\sqrt{x}+2}{\sqrt{x}+3}-\frac{9-x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right)\div\left(1-\frac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right)\)
\(=\left(\frac{\left(3-\sqrt{x}\right)\left(x+\sqrt{3}\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}+\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}-\frac{9-x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right)\div\left(1-\frac{3}{\sqrt{x}+3}\right)\)
\(=\left(\frac{9-x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}+\frac{x-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}-\frac{9-x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right)\div\left(\frac{\sqrt{x}+3}{\sqrt{x}+3}-\frac{3}{\sqrt{x}+3}\right)\)
\(=\left(\frac{9-x+x-4-9+x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right)\div\left(\frac{\sqrt{x}}{\sqrt{x}+3}\right)\)
\(=\frac{x-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\times\frac{\sqrt{x}+3}{\sqrt{x}}\)
\(=\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}=\frac{\sqrt{x}+2}{\sqrt{x}}\)
b) Ta có : \(\frac{\sqrt{x}+2}{\sqrt{x}}=1+\frac{2}{\sqrt{x}}\)
Để P nguyên => \(\frac{2}{\sqrt{x}}\)nguyên
=> \(2⋮\sqrt{x}\)
=> \(\sqrt{x}\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
=> \(\sqrt{x}\in\left\{1;2\right\}\)( vì x ≥ 0 )
=> \(x\in\left\{1;4\right\}\Rightarrow x=1\)( vì x ≠ 4 )
Vậy với x = 1 thì P có giá trị nguyên
a/ \(P=\left[1-\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right]:\left[\frac{3-\sqrt{x}}{\sqrt{x}-2}+\frac{\sqrt{x}-2}{\sqrt{x}+3}-\frac{9x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right]\)
\(=\left(1-\frac{\sqrt{x}}{\sqrt{x}+3}\right):\left[\frac{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)+\left(\sqrt{x}-2\right)^2-9x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right]\)
\(=\left(\frac{\sqrt{x}+3-\sqrt{x}}{\sqrt{x}+3}\right):\left[\frac{9-x+x-4\sqrt{x}+4-9x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right]\)
\(=\frac{3}{\sqrt{x}+3}.\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{13-4\sqrt{x}-9x}\)
\(=\frac{3\sqrt{x}-6}{13-4\sqrt{x}-9x}\)
b/ \(P=1\Rightarrow\frac{3\sqrt{x}-6}{13-4\sqrt{x}-9x}=1\Rightarrow3\sqrt{x}-6=13-4\sqrt{x}-9x\)
\(\Rightarrow9x+7\sqrt{x}-19=0\)
Mình k biết mình sai chỗ nào nữa, bạn xem giúp mình với
