ĐKXĐ: \(x>0;x\ne1\)
\(A=\left(\dfrac{1+\sqrt{x}}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}+\dfrac{1-\sqrt{x}}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}\right):\left(\dfrac{1+\sqrt{x}}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}-\dfrac{1-\sqrt{x}}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}\right)+\dfrac{1}{1-\sqrt{x}}\)
\(=\dfrac{2}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}:\dfrac{2\sqrt{x}}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}+\dfrac{1}{1-\sqrt{x}}\)
\(=\dfrac{2}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}.\dfrac{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}{2\sqrt{x}}+\dfrac{1}{1-\sqrt{x}}\)
\(=\dfrac{1}{\sqrt{x}}+\dfrac{1}{1-\sqrt{x}}=\dfrac{1-\sqrt{x}+\sqrt{x}}{\sqrt{x}\left(1-\sqrt{x}\right)}\)
\(=\dfrac{1}{\sqrt{x}\left(1-\sqrt{x}\right)}=\dfrac{1}{\sqrt{x}-x}\)
b.
\(x=7+4\sqrt{3}=\left(2+\sqrt{3}\right)^2\Rightarrow\sqrt{x}=2+\sqrt{3}\)
\(A=\dfrac{1}{\sqrt{x}-x}=\dfrac{1}{2+\sqrt{3}-\left(7+4\sqrt{3}\right)}=\dfrac{1}{-5-3\sqrt{3}}=\dfrac{5-3\sqrt{3}}{2}\)
c.
Biểu thức ko có giá trị nhỏ nhất
a: ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x\ne1\end{matrix}\right.\)
\(A=\left(\dfrac{1}{1-\sqrt{x}}+\dfrac{1}{1+\sqrt{x}}\right):\left(\dfrac{1}{1-\sqrt{x}}-\dfrac{1}{1+\sqrt{x}}\right)+\dfrac{1}{1-\sqrt{x}}\)
\(=\dfrac{1+\sqrt{x}+1-\sqrt{x}}{1-x}:\dfrac{1+\sqrt{x}-1+\sqrt{x}}{1-x}+\dfrac{1}{1-\sqrt{x}}\)
\(=\dfrac{2}{1-x}\cdot\dfrac{1-x}{2\sqrt{x}}+\dfrac{1}{1-\sqrt{x}}=\dfrac{1}{\sqrt{x}}-\dfrac{1}{\sqrt{x}-1}\)
\(=\dfrac{\sqrt{x}-1-\sqrt{x}}{\sqrt{x}\cdot\left(\sqrt{x}-1\right)}=\dfrac{-1}{\sqrt{x}\cdot\left(\sqrt{x}-1\right)}\)
b: Thay \(x=7+4\sqrt{3}=\left(2+\sqrt{3}\right)^2\) vào A, ta được:
\(A=\dfrac{-1}{\sqrt{\left(2+\sqrt{3}\right)^2}\left(\sqrt{\left(2+\sqrt{3}\right)^2}-1\right)}\)
\(=\dfrac{-1}{\left(2+\sqrt{3}\right)\left(2+\sqrt{3}-1\right)}\)
\(=\dfrac{-1}{\left(2+\sqrt{3}\right)\left(\sqrt{3}+1\right)}=\dfrac{-1}{2\sqrt{3}+2+3+\sqrt{3}}\)
\(=\dfrac{-1}{3\sqrt{3}+5}=\dfrac{-\left(3\sqrt{3}-5\right)}{27-25}=\dfrac{-3\sqrt{3}+5}{2}\)
