Rút gọn biểu thức sau và tìm đkxđ: P=\(\left(\dfrac{\sqrt{a}-2}{\sqrt{a}+2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-2}\right).\left(\sqrt{a}-\dfrac{4}{\sqrt{a}}\right)\)
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10 tháng 7 2021
Sửa đề: \(Q=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}-1}{\sqrt{a}+2}\right)\)
a) ĐKXĐ: \(\left\{{}\begin{matrix}a\ge0\\a\notin\left\{1;4\right\}\end{matrix}\right.\)
Ta có: \(Q=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}-1}{\sqrt{a}+2}\right)\)
\(=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a+3\sqrt{a}+2-a+3\sqrt{a}-2}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\)
\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}{6\sqrt{a}}\)
\(=\dfrac{a-4}{6a\left(\sqrt{a}-1\right)}\)
c) Thay \(a=9-4\sqrt{5}\) vào Q, ta được:
\(Q=\dfrac{5-4\sqrt{5}}{6\left(9-4\sqrt{5}\right)\left(\sqrt{5}-3\right)}\)
\(=\dfrac{5-4\sqrt{5}}{6\left(9\sqrt{5}-27-20+12\sqrt{5}\right)}\)
\(=\dfrac{5-4\sqrt{5}}{6\left(21\sqrt{5}-47\right)}\)
\(=\dfrac{\left(5-4\sqrt{5}\right)\left(21\sqrt{5}+47\right)}{-24}\)
\(=\dfrac{105\sqrt{5}+235-420-188\sqrt{5}}{-24}\)
\(=\dfrac{-83\sqrt{5}-185}{-24}=\dfrac{83\sqrt{5}+185}{24}\)
10 tháng 7 2021
a) ĐKXĐ: \(\left\{{}\begin{matrix}a>0\\a\ne1\end{matrix}\right.\)
b) Ta có: \(M=\left(\dfrac{\sqrt{a}}{2}-\dfrac{1}{2\sqrt{a}}\right)\left(\dfrac{a-\sqrt{a}}{\sqrt{a}+1}-\dfrac{a+\sqrt{a}}{\sqrt{a}-1}\right)\)
\(=\dfrac{a-1}{2\sqrt{a}}\cdot\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)^2-\sqrt{a}\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{\sqrt{a}\left[\left(\sqrt{a}-1\right)^2-\left(\sqrt{a}+1\right)^2\right]}{2\sqrt{a}}\)
\(=\dfrac{a-2\sqrt{a}+1-a-2\sqrt{a}-1}{2}\)
\(=\dfrac{-4\sqrt{a}}{2}=-2\sqrt{a}\)
c) Để M=-4 thì \(-2\sqrt{a}=-4\)
\(\Leftrightarrow\sqrt{a}=2\)
hay a=4(thỏa ĐK)
13 tháng 8 2021
a,\(ĐK:x>0,x\ne1,x\ne4\)
\(A=\left[\dfrac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right]:\left[\dfrac{x-1-x+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\right]\)
\(A=\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{3}=\dfrac{\sqrt{x}-2}{3\sqrt{x}}\)
b,\(x=3-2\sqrt{2}=2-2\sqrt{2}+1=\left(\sqrt{2}-1\right)^2\)
\(=>A=\dfrac{\sqrt{2}-3}{3\sqrt{2}-3}\)
NT
13 tháng 8 2021
a) ĐKXĐ: \(\left\{{}\begin{matrix}\sqrt{x}\ge0\\\sqrt{x}-1>0\\\sqrt{x}-2>0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x>1\\x>4\end{matrix}\right.\) \(\Leftrightarrow x>4\)
\(A=\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}}\right):\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\right)\)
\(=\dfrac{\sqrt{x}-\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\left(x-1\right)-\left(x-4\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{3}\)
\(=\dfrac{\sqrt{x}-2}{3\sqrt{x}}\)
b) Ta có \(x=3-2\sqrt{2}=2-2\sqrt{2}+1=\left(2-1\right)^2=1\)
Thay \(x=1\) vào \(A\), ta được:
\(A=\dfrac{\sqrt{1}-2}{3\sqrt{1}}=\dfrac{1-2}{3}=-\dfrac{1}{3}\)
28 tháng 5 2021
ĐKXĐ: \(x\ge0;x\ne1\)
Ta có: \(A=\left(2+\dfrac{2x+\sqrt{x}}{2\sqrt{x}+1}\right)\left(2-\dfrac{x-\sqrt{x}}{\sqrt{x}-1}\right)\)
\(A=\left(2+\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{2\sqrt{x}+1}\right)\left(2-\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\right)\)
\(A=\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)=4-x\)
ĐKXĐ :
\(\left\{{}\begin{matrix}\sqrt{a}+2\ne0\\\sqrt{a}-2\ne0\\\sqrt{a}\ne0\\\sqrt{a}x\text{đ}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{a}\ne2\\a\ne0\\a\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a\ne4\\a>0\end{matrix}\right.\)
Rút gọn :
\(\left(\dfrac{\sqrt{a}-2}{\sqrt{a}+2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-2}\right).\left(\sqrt{a}-\dfrac{4}{\sqrt{a}}\right)\)
\(=\dfrac{\left(\sqrt{a}-2\right)^2-\left(\sqrt{a}+2\right)^2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}.\dfrac{a-4}{\sqrt{a}}\)
\(=\dfrac{\left(\sqrt{a}-2+\sqrt{a}+2\right)\left(\sqrt{a}-2-\sqrt{a}-2\right)}{\left(\sqrt{a}\right)^2-2^2}.\dfrac{a-4}{\sqrt{a}}\)
\(=\dfrac{2\sqrt{a}.\left(-4\right)}{\sqrt{a}}\)
\(=2.\left(-4\right)=-8\)