Câu 1:
a: \(A=\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}\)
\(=\sqrt{\left(\sqrt{5}+1\right)^2}-\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=\sqrt{5}+1-\sqrt{5}+1=2\)
b: \(B=a+1-\sqrt{a^2-2a+1}\)
\(=a+1-\sqrt{\left(a-1\right)^2}\)
Vì a<1 nên a-1<0
=>\(B=a+1-\left(1-a\right)=a+1-1+a=2a\)
Câu 2:
a: \(P=\left(\dfrac{2x+1}{\sqrt{x^3}-1}-\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\right):\left(\dfrac{3}{\sqrt{x}-1}+\dfrac{2\sqrt{x}+5}{1-x}\right)\)
\(=\left(\dfrac{2x+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\right):\dfrac{3\left(\sqrt{x}+1\right)-2\sqrt{x}-5}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{2x+1-\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{3\sqrt{x}+3-2\sqrt{x}-5}\)
\(=\dfrac{2x+1-x+\sqrt{x}}{\left(x+\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-2}=\dfrac{\sqrt{x}+1}{\sqrt{x}-2}\)
b: \(x=\dfrac{8}{3-\sqrt{5}}=\dfrac{8\left(3+\sqrt{5}\right)}{9-5}=2\left(3+\sqrt{5}\right)=6+2\sqrt{5}\)
Thay x\(=6+2\sqrt{5}=\left(\sqrt{5}+1\right)^2\) vào P, ta được:
\(P=\dfrac{\sqrt{\left(6+2\sqrt{5}\right)}+1}{\sqrt{6+2\sqrt{5}}-2}=\dfrac{\sqrt{5}+1+1}{\sqrt{5}+1-2}=\dfrac{\sqrt{5}+2}{\sqrt{5}-1}=\dfrac{7+3\sqrt{5}}{4}\)
c: Để P là số tự nhiên thì \(\left\{{}\begin{matrix}\sqrt{x}+1⋮\sqrt{x}-2\\P>=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\sqrt{x}-2+3⋮\sqrt{x}-2\\\sqrt{x}-2>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3⋮\sqrt{x}-2\\\sqrt{x}-2>0\end{matrix}\right.\)
=>\(\sqrt{x}-2\in\left\{1;3\right\}\)
=>\(\sqrt{x}\in\left\{3;5\right\}\)
=>\(x\in\left\{9;25\right\}\)
