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a: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
b: Thay x=9 vào A, ta được:
\(A=\dfrac{3-1}{3+1}=\dfrac{1}{2}\)
c: Ta có: P=AB
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\left(\dfrac{\sqrt{x}+3}{\sqrt{x}+1}+\dfrac{4}{\sqrt{x}-1}+\dfrac{5-x}{x-1}\right)\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\left(\dfrac{x+2\sqrt{x}-3+4\sqrt{x}+4+5-x}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\cdot\dfrac{6\sqrt{x}+6}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{6}{\sqrt{x}+1}\)
a: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
Ta có: \(A=\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{3}{\sqrt{x}+1}-\dfrac{6\sqrt{x}-4}{x-1}\)
\(=\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
Thay \(x=6-2\sqrt{5}\) vào A, ta được:
\(A=\dfrac{\sqrt{5}-1-1}{\sqrt{5}-1+1}=\dfrac{\sqrt{5}-2}{\sqrt{5}}=\dfrac{5-2\sqrt{5}}{5}\)
b: Để \(A< \dfrac{1}{2}\) thì \(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{1}{2}< 0\)
\(\Leftrightarrow2\sqrt{x}-2-\sqrt{x}-1< 0\)
\(\Leftrightarrow x< 9\)
Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}0\le x< 9\\x\ne1\end{matrix}\right.\)
a: ĐKXĐ: x=0; x<>1
\(M=\left(2+\sqrt{x}\right)\left(1-2\sqrt{x}-x+1+\sqrt{x}+x\right)\)
\(=\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)=4-x\)
b: Sửa đề: P=1/M
P=1/4-x=-1/x-4
Để P nguyên thì x-4 thuộc {1;-1}
=>x thuộc {5;3}
a/ đkxđ: x > 0; x≠1
b/ \(A=\left(\dfrac{\sqrt{x}}{2}-\dfrac{1}{2\sqrt{x}}\right):\left(\dfrac{x-\sqrt{x}}{\sqrt{x}+1}-\dfrac{x+\sqrt{x}}{\sqrt{x}-1}\right)\)
\(=\dfrac{x-1}{2\sqrt{x}}\cdot\left(\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)^2-\sqrt{x}\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\dfrac{x-1}{2\sqrt{x}}\cdot\dfrac{x\sqrt{x}-2x+\sqrt{x}-x\sqrt{x}-2x-\sqrt{x}}{x-1}\)
\(=\dfrac{-4x}{2\sqrt{x}}=-2\sqrt{x}\)
a:ĐKXĐ: x>=0; \(x\notin\left\{4;9\right\}\)
\(A=\dfrac{2\sqrt{x}-9-x+9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{2x-4\sqrt{x}+\sqrt{x}-2}{\sqrt{x}-3}\)
\(=\dfrac{2\sqrt{x}-x-2x+3\sqrt{x}+2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{-3x+5\sqrt{x}+2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{-3x+6\sqrt{x}-\sqrt{x}+2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{-3\sqrt{x}-1}{\sqrt{x}-3}\)
b: Để A là số nguyên thì \(-3\sqrt{x}+9-10⋮\sqrt{x}-3\)
\(\Leftrightarrow\sqrt{x}-3\in\left\{1;-1;2;-2;5;-5;10;-10\right\}\)
hay \(x\in\left\{16;25;1;64;169\right\}\)
a: ĐKXĐ: x>=0; x<>1
b: \(A=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}+\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}=\sqrt{x}-1+\sqrt{x}=2\sqrt{x}-1\)
c: Để A<-1 thì A+1<0
\(\Leftrightarrow2\sqrt{x}< 0\)
hay \(\sqrt{x}< 0\)(vô lý)
a) ĐK: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
b) \(A=\dfrac{x+1-2\sqrt{x}}{\sqrt{x}-1}+\dfrac{x+\sqrt{x}}{\sqrt{x}+1}=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}+\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}=\sqrt{x}-1+\sqrt{x}=2\sqrt{x}-1\)c) Ta có A<-1\(\Leftrightarrow2\sqrt{x}-1< -1\Leftrightarrow2\sqrt{x}< 0\Leftrightarrow\sqrt{x}< 0\left(ktm\right)\)
Vậy không có giá trị của x để A<-1
a.ĐK:
\(\left\{{}\begin{matrix}\sqrt{x}-1\ne0\\\sqrt{x}+1\ne0\\\sqrt{x}xđ\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
b.\(A=\sqrt{x}-1+\sqrt{x}=2\sqrt{x}.\)
c.Có :\(2\sqrt{x}\ge0\) nên không có giá trị nào của x để A<-1.
Câu a : ĐKXĐ : \(x\ge0\) và \(x\ne4\)
Câu b : \(A=\dfrac{\sqrt{x}+2}{\sqrt{x}+3}-\dfrac{5}{x+\sqrt{x}-6}+\dfrac{1}{2-\sqrt{x}}\)
\(=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}-\dfrac{5}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}-\dfrac{\sqrt{x}+3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)-5-\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{x-4-5-\sqrt{x}+3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{x-\sqrt{x}-6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)
Câu c :
\(A< 1\) \(\Leftrightarrow\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}< 1\)
\(\Leftrightarrow\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)< \left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)\)
\(\Leftrightarrow x-\sqrt{x}-6< x+\sqrt{x}-6\)
\(\Leftrightarrow-2\sqrt{x}< 0\) ( Luôn đúng với mọi x khi \(\left\{{}\begin{matrix}x>0\\x\ne4\end{matrix}\right.\))
Vậy các giá của x để A < 1 là \(\left\{{}\begin{matrix}x>0\\x\ne4\end{matrix}\right.\)