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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: \(\left\{{}\begin{matrix}0\le x< 9\\x\ne1\end{matrix}\right.\)
\(a,ĐK:x>0;x\ne9\\ b,A=\dfrac{\sqrt{x}+3+\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}}\\ A=\dfrac{2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+3\right)}=\dfrac{2}{\sqrt{x}+3}\\ c,A>\dfrac{2}{5}\Leftrightarrow\dfrac{2}{\sqrt{x}+3}-\dfrac{2}{5}>0\\ \Leftrightarrow\dfrac{1}{\sqrt{x}+3}-\dfrac{1}{5}>0\\ \Leftrightarrow\dfrac{2-\sqrt{x}}{5\left(\sqrt{x}+3\right)}>0\\ \Leftrightarrow2-\sqrt{x}>0\left(\sqrt{x}+3>0\right)\\ \Leftrightarrow\sqrt{x}< 2\Leftrightarrow0< x< 4\)
a) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
b) Thay x=0 vào A, ta được:
\(A=\dfrac{15\cdot\sqrt{0}-11}{0+2\sqrt{0}-3}-\dfrac{3\sqrt{0}-2}{\sqrt{0}-1}-\dfrac{2\sqrt{0}+3}{\sqrt{0}+3}\)
\(=\dfrac{-11}{-3}-\dfrac{-2}{-1}-\dfrac{3}{3}\)
\(=\dfrac{11}{3}-2-1\)
\(=\dfrac{11}{3}-\dfrac{9}{3}=\dfrac{2}{3}\)
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: x>=0; \(x\notin\left\{1;\dfrac{9}{4};\dfrac{441}{256}\right\}\)
b: \(A=\left(\dfrac{2\sqrt{x}+3-6\sqrt{x}+9}{4x-9}+\dfrac{1}{\sqrt{x}-1}\right)\cdot\dfrac{\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{16\sqrt{x}-21}\)
\(=\left(\dfrac{-4\sqrt{x}+12}{4x-9}+\dfrac{1}{\sqrt{x}-1}\right)\cdot\dfrac{\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{16\sqrt{x}-21}\)
\(=\dfrac{-4x+4\sqrt{x}+12\sqrt{x}-12+4x-9}{\left(2\sqrt{x}+3\right)\left(2\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{16\sqrt{x}-21}\)
\(=\dfrac{16\sqrt{x}-21}{2\sqrt{x}-3}\cdot\dfrac{1}{16\sqrt{x}-21}=\dfrac{1}{2\sqrt{x}-3}\)
c: Để A<0 thì \(2\sqrt{x}-3< 0\)
=>0<x<9/4
a: ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)
Ta có: \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{x-\sqrt{x}}\right):\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{x-1}\right)\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x-1}{\sqrt{x}}\)
b: Thay \(x=3+2\sqrt{2}\) vào P, ta được:
\(P=\dfrac{2\sqrt{2}+2}{\sqrt{2}+1}=2\)
1) ĐKXĐ: \(x\notin\left\{0;1\right\}\)
2) Ta có: \(A=\left(\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\right):\left(1-\dfrac{3-\sqrt{x}}{\sqrt{x}+1}\right)\)
\(=\dfrac{x+\sqrt{x}+1-\left(x-\sqrt{x}+1\right)}{\sqrt{x}}:\dfrac{\sqrt{x}+1-3+\sqrt{x}}{\sqrt{x}+1}\)
\(=2\cdot\dfrac{\sqrt{x}+1}{2\sqrt{x}-2}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
Sửa đề: \(A=\left(\dfrac{1}{2\sqrt{x}-3}-\dfrac{3}{2\sqrt{x}+3}+\dfrac{1}{\sqrt{x}-1}\right):\dfrac{16\sqrt{x}-21}{2x+\sqrt{x}-3}\)
a: ĐKXĐ: x>=0; \(x\notin\left\{1;\dfrac{9}{4};\dfrac{441}{256}\right\}\)
b: \(A=\left(\dfrac{2\sqrt{x}+3-6\sqrt{x}+9}{4x-9}+\dfrac{1}{\sqrt{x}-1}\right)\cdot\dfrac{\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{16\sqrt{x}-21}\)
\(=\left(\dfrac{-4\sqrt{x}+12}{4x-9}+\dfrac{1}{\sqrt{x}-1}\right)\cdot\dfrac{\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{16\sqrt{x}-21}\)
\(=\dfrac{-4x+4\sqrt{x}+12\sqrt{x}-12+4x-9}{\left(2\sqrt{x}+3\right)\left(2\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{16\sqrt{x}-21}\)
\(=\dfrac{16\sqrt{x}-21}{2\sqrt{x}-3}\cdot\dfrac{1}{16\sqrt{x}-21}=\dfrac{1}{2\sqrt{x}-3}\)
c: Để A<0 thì \(2\sqrt{x}-3< 0\)
=>0<x<9/4