Cho x= \(\dfrac{\sqrt{13-4\sqrt{3}}}{2}.Tính\) A = \(\dfrac{\sqrt{x}+2}{2\sqrt{x}-3}\)
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18 tháng 12 2021
\(a,ĐK:x\ge0;x\ne9\\ A=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\\ A=\dfrac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}=\dfrac{-3}{\sqrt{x}+3}\\ b,x=13-4\sqrt{3}=\left(2\sqrt{3}-1\right)^2\\ \Leftrightarrow A=\dfrac{-3}{2\sqrt{3}-1+3}=\dfrac{-3}{2\sqrt{3}+2}=\dfrac{-3\left(2\sqrt{3}-2\right)}{8}\)
\(c,A< -\dfrac{1}{2}\Leftrightarrow\dfrac{-3}{\sqrt{x}+3}+\dfrac{1}{2}< 0\Leftrightarrow\dfrac{\sqrt{x}-3}{2\left(\sqrt{x}+3\right)}< 0\\ \Leftrightarrow\sqrt{x}-3< 0\left(\sqrt{x}+3>0\right)\\ \Leftrightarrow\sqrt{x}< 3\Leftrightarrow0\le x< 9\\ d,A=-\dfrac{2}{3}\Leftrightarrow\dfrac{3}{\sqrt{x}+3}=\dfrac{2}{3}\\ \Leftrightarrow2\sqrt{x}+6=9\\ \Leftrightarrow\sqrt{x}=\dfrac{3}{2}\Leftrightarrow x=\dfrac{9}{4}\left(tm\right)\\ e,\Leftrightarrow\sqrt{x}+3\inƯ\left(-3\right)=\left\{-3;-1;1;3\right\}\\ \Leftrightarrow\sqrt{x}=0\left(\sqrt{x}\ge0\right)\\ \Leftrightarrow x=0\left(tm\right)\\ f,\sqrt{x}+3\ge3\\ \Leftrightarrow A=-\dfrac{3}{\sqrt{x}+3}\ge-\dfrac{3}{3}=-1\\ A_{min}=-1\Leftrightarrow x=0\)
17 tháng 10 2021
a: Ta có: \(A=\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}-\dfrac{3\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)
\(=\dfrac{15\sqrt{x}-11-\left(3x+7\sqrt{x}-6\right)-\left(2x+\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{15\sqrt{x}-11-3x-7\sqrt{x}+6-2x-\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-5\sqrt{x}+2}{\sqrt{x}+3}\)
5 tháng 2 2022
a: \(=6+2\sqrt{11}-4+\sqrt{11}=2+3\sqrt{11}\)
b: \(=\dfrac{3x+9\sqrt{x}-2x+4\sqrt{x}}{\left(\sqrt{x}+3\right)\left(x-2\sqrt{x}\right)}\cdot\dfrac{\left(\sqrt{x}+3\right)^2}{\sqrt{x}+13}=\dfrac{\sqrt{x}+3}{x-2\sqrt{x}}\)
16 tháng 10 2021
a: Ta có: \(x=\sqrt{28-16\sqrt{3}}+2\sqrt{3}\)
\(=4-2\sqrt{3}+2\sqrt{3}\)
=4
Thay x=4 vào B, ta được:
\(B=\dfrac{2-4}{2}=-1\)
Lời giải:
\(x=\frac{\sqrt{13-4\sqrt{3}}}{2}=\frac{\sqrt{13-2\sqrt{12}}}{2}=\frac{\sqrt{12+1-2\sqrt{12}}}{2}=\frac{\sqrt{(\sqrt{12}-1)^2}}{2}=\frac{\sqrt{12}-1}{2}\)
\(2A=1+\frac{7}{2\sqrt{x}-3}=1+\frac{7}{\sqrt{2\sqrt{12}-2}}\)
\(A=\frac{1}{2}+\frac{7}{2\sqrt{4\sqrt{3}-2}}\)