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a, ĐK: \(x\ge0,x\ne1\)
\(A=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{3\sqrt{x}+1}{x-1}\)
\(=\dfrac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{3\sqrt{x}+1}{x-1}\)
\(=\dfrac{x+1+2\sqrt{x}+x+1-2\sqrt{x}-3\sqrt{x}-1}{x-1}\)
\(=\dfrac{2x-3\sqrt{x}+1}{x-1}\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(2\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}\)
Lời giải:
ĐKXĐ: $x\geq 0; x\neq 1$
a)
\(A=\frac{x+\sqrt{x}+1}{x+1}:\left[\frac{1}{\sqrt{x}-1}-\frac{2\sqrt{x}}{(\sqrt{x}-1)(x+1)}\right]\)
\(=\frac{x+\sqrt{x}+1}{x+1}:\frac{x+1-2\sqrt{x}}{(\sqrt{x}-1)(x+1)}=\frac{x+\sqrt{x}+1}{x+1}.\frac{(\sqrt{x}-1)(x+1)}{(\sqrt{x}-1)^2}=\frac{x+\sqrt{x}+1}{\sqrt{x}-1}\)
b)
\(A=7\Leftrightarrow x+\sqrt{x}+1=7(\sqrt{x}-1)\)
\(\Leftrightarrow x-6\sqrt{x}+8=0\Leftrightarrow (\sqrt{x}-2)(\sqrt{x}-4)=0\)
\(\Leftrightarrow \left[\begin{matrix} x=4\\ x=16\end{matrix}\right.\) (đều thỏa mãn)
c)
\(x=2(2+\sqrt{3})=4+2\sqrt{3}=3+1+2\sqrt{3.1}=(\sqrt{3}+1)^2\Rightarrow \sqrt{x}=\sqrt{3}+1\)
\(\Rightarrow A=\frac{4+2\sqrt{3}+\sqrt{3}+1+1}{\sqrt{3}}=\frac{6+3\sqrt{3}}{\sqrt{3}}=3+2\sqrt{3}\)
d)
\(A< 1\Leftrightarrow \frac{x+\sqrt{x}+1}{\sqrt{x}-1}-1<0\Leftrightarrow \frac{x-2\sqrt{x}+2}{\sqrt{x}-1}<0\)
\(\Leftrightarrow \frac{(\sqrt{x}-1)^2+1}{\sqrt{x}-1}<0\Leftrightarrow \sqrt{x}-1< 0\Leftrightarrow 0\leq x< 1\)
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: \(\left\{{}\begin{matrix}0\le x< 9\\x\ne1\end{matrix}\right.\)
\(a,ĐK:x\ge0;x\ne9\\ A=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\\ A=\dfrac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}=\dfrac{-3}{\sqrt{x}+3}\\ b,x=13-4\sqrt{3}=\left(2\sqrt{3}-1\right)^2\\ \Leftrightarrow A=\dfrac{-3}{2\sqrt{3}-1+3}=\dfrac{-3}{2\sqrt{3}+2}=\dfrac{-3\left(2\sqrt{3}-2\right)}{8}\)
\(c,A< -\dfrac{1}{2}\Leftrightarrow\dfrac{-3}{\sqrt{x}+3}+\dfrac{1}{2}< 0\Leftrightarrow\dfrac{\sqrt{x}-3}{2\left(\sqrt{x}+3\right)}< 0\\ \Leftrightarrow\sqrt{x}-3< 0\left(\sqrt{x}+3>0\right)\\ \Leftrightarrow\sqrt{x}< 3\Leftrightarrow0\le x< 9\\ d,A=-\dfrac{2}{3}\Leftrightarrow\dfrac{3}{\sqrt{x}+3}=\dfrac{2}{3}\\ \Leftrightarrow2\sqrt{x}+6=9\\ \Leftrightarrow\sqrt{x}=\dfrac{3}{2}\Leftrightarrow x=\dfrac{9}{4}\left(tm\right)\\ e,\Leftrightarrow\sqrt{x}+3\inƯ\left(-3\right)=\left\{-3;-1;1;3\right\}\\ \Leftrightarrow\sqrt{x}=0\left(\sqrt{x}\ge0\right)\\ \Leftrightarrow x=0\left(tm\right)\\ f,\sqrt{x}+3\ge3\\ \Leftrightarrow A=-\dfrac{3}{\sqrt{x}+3}\ge-\dfrac{3}{3}=-1\\ A_{min}=-1\Leftrightarrow x=0\)
a) ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\ne4\end{matrix}\right.\)
b) Ta có: \(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{4}{x-2\sqrt{x}}\right)\left(\dfrac{1}{\sqrt{x}+2}+\dfrac{4}{x-4}\right)\)
\(=\dfrac{x-4}{\sqrt{x}\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}-2+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
d) Để A>0 thì \(\sqrt{x}-2>0\)
hay x>4
\(a,A=\dfrac{2\sqrt{x}-2+2\sqrt{x}+x-3\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}\left(x\ge0;x\ne1;x\ne9\right)\\ A=\dfrac{x+\sqrt{x}-2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}+2}{\sqrt{x}-3}\)
\(b,A\in Z\Leftrightarrow\dfrac{\sqrt{x}-3+5}{\sqrt{x}-3}\in Z\Leftrightarrow1+\dfrac{5}{\sqrt{x}-3}\in Z\\ \Leftrightarrow\sqrt{x}-3\inƯ\left(5\right)=\left\{-5;-1;1;5\right\}\\ Mà.x\ge0\\ \Leftrightarrow\sqrt{x}\in\left\{2;4;8\right\}\\ \Leftrightarrow x\in\left\{4;16;64\right\}\)
a) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne9\\x\ne1\end{matrix}\right.\)
\(A=\dfrac{2\sqrt{x}-2+2\sqrt{x}+x-3\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}=\dfrac{x+\sqrt{x}-2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}=\dfrac{\sqrt{x}+2}{\sqrt{x}-3}\)
b) \(A=\dfrac{\sqrt{x}+2}{\sqrt{x}-3}=1+\dfrac{5}{\sqrt{x}-3}\in Z\)
\(\Rightarrow\sqrt{x}-3\inƯ\left(5\right)=\left\{-5;-1;1;5\right\}\)
Kết hợp đk
\(\Rightarrow x\in\left\{4;16;64\right\}\)
a: Thay x=36 vào B, ta được:
\(B=\dfrac{6}{6-3}=\dfrac{6}{3}=2\)
