1: \(A=\dfrac{2}{\sqrt{5}+\sqrt{3}}-\sqrt{\dfrac{2}{4-\sqrt{15}}}+6\sqrt{\dfrac{1}{3}}\)
\(=\dfrac{2\left(\sqrt{5}-\sqrt{3}\right)}{5-3}-\sqrt{\dfrac{2\left(4+\sqrt{15}\right)}{16-15}}+2\sqrt{3}\)
\(=\sqrt{5}-\sqrt{3}-\sqrt{8+2\sqrt{15}}+2\sqrt{3}\)
\(=\sqrt{5}+\sqrt{3}-\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}\)
\(=\sqrt{5}+\sqrt{3}-\left(\sqrt{5}+\sqrt{3}\right)=0\)
2:
a: \(B=\left(\dfrac{x-\sqrt{x}+2}{x-\sqrt{x}-2}-\dfrac{x}{x-2\sqrt{x}}\right):\dfrac{1-\sqrt{x}}{2-\sqrt{x}}\)
\(=\left(\dfrac{x-\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}}{\sqrt{x}-2}\right)\cdot\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\)
\(=\dfrac{x-\sqrt{x}+2-\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\)
\(=\dfrac{x-\sqrt{x}+2-x-\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\dfrac{-2\sqrt{x}+2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-2}{\sqrt{x}+1}\)
b: \(B< -\dfrac{1}{3}\)
=>\(B+\dfrac{1}{3}< 0\)
=>\(\dfrac{-2}{\sqrt{x}+1}+\dfrac{1}{3}< 0\)
=>\(\dfrac{-6+\sqrt{x}+1}{3\left(\sqrt{x}+1\right)}< 0\)
=>\(\sqrt{x}-5< 0\)
=>0<=x<25
Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}0< x< 25\\x\notin\left\{1;4\right\}\end{matrix}\right.\)
c: Để B là số nguyên thì \(-2⋮\sqrt{x}+1\)
=>\(\sqrt{x}+1\in\left\{1;-1;2;-2\right\}\)
=>\(\sqrt{x}\in\left\{0;-2;1;-3\right\}\)
=>\(\sqrt{x}\in\left\{0;1\right\}\)
=>\(x\in\left\{0;1\right\}\)
Kết hợp ĐKXĐ, ta được: \(x\in\varnothing\)
BÀi này làm như nào vậy mn?