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\(1.a.A=\left(1-\dfrac{\sqrt{x}}{1+\sqrt{x}}\right):\left(\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{\sqrt{x}+2}{3-\sqrt{x}}+\dfrac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right)=\dfrac{1}{\sqrt{x}+1}:\dfrac{x-9-x+4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{1}{\sqrt{x}+1}.\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{\sqrt{x}-3}=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\left(x\ge0;x\ne4;x\ne9\right)\)
\(b.A< 0\Leftrightarrow\dfrac{\sqrt{x}-2}{\sqrt{x}+1}< 0\)
\(\Leftrightarrow\sqrt{x}-2< 0\)
\(\Leftrightarrow x< 4\)
Kết hợp với ĐKXĐ , ta có : \(0\le x< 4\)
KL............
\(2.\) Tương tự bài 1.
\(3a.A=\dfrac{1}{x-\sqrt{x}+1}=\dfrac{1}{x-2.\dfrac{1}{2}\sqrt{x}+\dfrac{1}{4}+\dfrac{3}{4}}=\dfrac{1}{\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\le\dfrac{4}{3}\)
\(\Rightarrow A_{Max}=\dfrac{4}{3}."="\Leftrightarrow x=\dfrac{1}{4}\)
có phải/....
1) \(A=\dfrac{x+3}{\sqrt{x}-2}\)
\(B=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}+\dfrac{5\sqrt{x}-2}{x-4}\) hay \(B=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}+\dfrac{5\left(\sqrt{x}-2\right)}{x-4}\)
2) \(A=\dfrac{\sqrt{x}+2}{\sqrt{x}+3}\)
Bài 2:
a: \(A=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}-\dfrac{\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}-\dfrac{\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{15\sqrt{x}-11-3x-9\sqrt{x}+2\sqrt{x}+6-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-\left(5\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\dfrac{-5\sqrt{x}+2}{\sqrt{x}+3}\)
b: Thay \(x=5-2\sqrt{6}\) vào A, ta được:
\(A=\dfrac{-5\left(\sqrt{3}-\sqrt{2}\right)+2}{\sqrt{3}-\sqrt{2}+3}=\dfrac{-5\sqrt{3}+5\sqrt{2}+2}{\sqrt{3}-\sqrt{2}+3}\simeq0,124\)
d: Để A=1/2 thì \(\sqrt{x}+3=-10\sqrt{x}+4\)
\(\Leftrightarrow11\sqrt{x}=1\)
hay x=1/121
ĐKXĐ \(x\ge0,x\ne4\)
a) \(B=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)-\left(\sqrt{x}+3\right)\cdot\left(\sqrt{x}+1\right)-\left(x+5\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x-2\sqrt{x}-\sqrt{x}+2-\left(x+\sqrt{x}+3\sqrt{x}+3\right)-x-5}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{-x-7\sqrt{x}-6}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}=\dfrac{-\left(\sqrt{x}+1\right)\cdot\left(\sqrt{x}+6\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\sqrt{x}+6}{2-\sqrt{x}}\)
b) B > -1 <=> B + 1 > 0.
\(\Leftrightarrow\dfrac{\sqrt{x}+6}{2-\sqrt{x}}+1>0\Leftrightarrow\dfrac{8}{2-\sqrt{x}}>0\)
=> \(2-\sqrt{x}>0\Leftrightarrow\sqrt{x}< 2\Rightarrow x< 4\)
Vậy \(0\le x< 4\) thì B > -1.
c) \(B=\dfrac{\sqrt{x}+6}{2-\sqrt{x}}=-1-\dfrac{8}{2-\sqrt{x}}\in Z\)
\(\Rightarrow2-\sqrt{x}\inƯ_{\left(8\right)}=\left\{\pm1;\pm2;\pm4;\pm8\right\}\)
\(\Rightarrow\sqrt{x}\in\left\{1;3;0;4;-2;6;-6;10\right\}\)
\(\Rightarrow x\in\left\{1;9;0;16;36;100\right\}\)thì \(B\in Z\)
a) đk : \(x\ne4;x\ge0\)
B = \(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{x+5}{x-\sqrt{x}-2}\)
B = \(\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)-\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)-\left(x+5\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)
B = \(\dfrac{x-2\sqrt{x}-\sqrt{x}+2-\left(x+\sqrt{x}+3\sqrt{x}+3\right)-x-5}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)
B = \(\dfrac{x-2\sqrt{x}-\sqrt{x}+2-x-\sqrt{x}-3\sqrt{x}-3-x-5}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)
B = \(\dfrac{-x-7\sqrt{x}-6}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\) = \(\dfrac{\left(-\sqrt{x}-6\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)
B = \(\dfrac{-\sqrt{x}-6}{\sqrt{x}-2}\)
a) ĐKXĐ: \(x\ge0;x\ne9\) . Rút gọn: \(A=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}+\dfrac{\sqrt{x}+3}{\sqrt{x}-3}-\dfrac{x-4\sqrt{x}+7}{x-2\sqrt{x}-3}=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}+\dfrac{\sqrt{x}+3}{\sqrt{x}-3}-\dfrac{x-4\sqrt{x}+7}{x+\sqrt{x}-3\sqrt{x}-3}=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}+\dfrac{\sqrt{x}+3}{\sqrt{x}-3}-\dfrac{x-4\sqrt{x}+7}{\sqrt{x}\left(\sqrt{x}+1\right)-3\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}+\dfrac{\sqrt{x}+3}{\sqrt{x}-3}-\dfrac{x-4\sqrt{x}+7}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)+\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)-x+4\sqrt{x}-7}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\dfrac{x-3\sqrt{x}-2\sqrt{x}+6+x+\sqrt{x}+3\sqrt{x}+3-x+4\sqrt{x}-7}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\dfrac{x+3\sqrt{x}+2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\dfrac{x+\sqrt{x}+2\sqrt{x}+2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)+2\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\dfrac{\sqrt{x}+2}{\sqrt{x}-3}\)
A>-1\(\Leftrightarrow\dfrac{\sqrt{x}+2}{\sqrt{x}-3}\)>-1\(\Leftrightarrow\dfrac{\sqrt{x}+2}{\sqrt{x}-3}+1>0\Leftrightarrow\dfrac{\sqrt{x}+2+\sqrt{x}-3}{\sqrt{x}-3}>0\Leftrightarrow\dfrac{2\sqrt{x}-1}{\sqrt{x}-3}>0\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2\sqrt{x}-1>0\\\sqrt{x}-3>0\end{matrix}\right.\\\left\{{}\begin{matrix}2\sqrt{x}-1< 0\\\sqrt{x}-3< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\sqrt{x}>0,5\\\sqrt{x}>3\end{matrix}\right.\\\left\{{}\begin{matrix}\sqrt{x}< 0,5\\\sqrt{x}< 3\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>0,25\\x>9\end{matrix}\right.\\\left\{{}\begin{matrix}x< 0,25\\x< 9\end{matrix}\right.\end{matrix}\right.\Leftrightarrow}}\left[{}\begin{matrix}x>9\\0\le x< 0,25\end{matrix}\right.\)
Bài 1:
a: \(A=\left(\dfrac{x-1}{2\sqrt{x}}\right)^2\cdot\dfrac{x-2\sqrt{x}+1-x-2\sqrt{x}-1}{x-1}\)
\(=\dfrac{\left(x-1\right)^2}{4x}\cdot\dfrac{-4\sqrt{x}}{x-1}=\dfrac{-\left(x-1\right)}{\sqrt{x}}\)
b: Để B<0 thì -x+1<0
=>-x<-1
hay x>1
c: Để B=2 thì \(-\left(x-1\right)=2\sqrt{x}\)
\(\Leftrightarrow-x+1-2\sqrt{x}=0\)
\(\Leftrightarrow x+\sqrt{x}-1=0\)
\(\Leftrightarrow\sqrt{x}=\dfrac{\sqrt{5}-1}{2}\)
hay \(x=\dfrac{6-2\sqrt{5}}{4}\)
a) ĐKXĐ: \(x\ge0;x\ne9\), kết quả B sau khi rút: \(B=\dfrac{-x+\sqrt{x}-4}{x-\sqrt{x}-6}\) (mình đã làm rồi nhưng rất tiếc lại ko đc gửi đi, bạn cứ làm sẽ ra). b) ta có: \(B=\dfrac{-x+\sqrt{x}-4}{x-\sqrt{x}-6}=-1-\dfrac{10}{x-\sqrt{x}-6}\) , do đó B nguyên khi \(\dfrac{10}{x-\sqrt{x}-6}\) là số nguyên, tức \(x-\sqrt{x}-6\inƯ\left(10\right)=\left\{\pm1;\pm2;\pm5;\pm10\right\}\) (Chắc sai r bạn)
T làm ra: \(B=\dfrac{x-\sqrt{x}-8}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\) ,b có thể giải chi tiết ko?