1: \(A=\sqrt{13+4\sqrt{3}}-\dfrac{11}{2\sqrt{3}+1}\)
\(=\sqrt{12+2\cdot2\sqrt{3}\cdot1+1}-\dfrac{11\left(2\sqrt{3}-1\right)}{12-1}\)
\(=\sqrt{\left(2\sqrt{3}+1\right)^2}-\left(2\sqrt{3}-1\right)\)
\(=2\sqrt{3}+1-2\sqrt{3}+1=2\)
2:
a:
ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x\ne1\end{matrix}\right.\)
\(B=\left(\dfrac{x\sqrt{x}+x\sqrt{x}}{x\sqrt{x}-1}-\dfrac{\sqrt{x}+3}{1-\sqrt{x}}\right)\cdot\dfrac{x-1}{2x+\sqrt{x}-1}\)
\(=\left(\dfrac{2x\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\sqrt{x}+3}{\left(\sqrt{x}-1\right)}\right)\cdot\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)}\)
\(=\dfrac{2x\sqrt{x}+\left(\sqrt{x}+3\right)\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}-1}{2\sqrt{x}-1}\)
\(=\dfrac{2x\sqrt{x}+x\sqrt{x}+4x+4\sqrt{x}+3}{\left(2\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\dfrac{3x\sqrt{x}+4x+4\sqrt{x}+3}{\left(2\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{3\left(x\sqrt{x}+1\right)+4\sqrt{x}\left(\sqrt{x}+1\right)}{\left(2\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{\left(\sqrt{x}+1\right)\left(3x-3\sqrt{x}+3+4\sqrt{x}\right)}{\left(2\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{\left(\sqrt{x}+1\right)\left(3x+\sqrt{x}+3\right)}{\left(2\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
b: Để B>0 thì \(\dfrac{\left(\sqrt{x}+1\right)\left(3x+\sqrt{x}+3\right)}{\left(2\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}>0\)
=>\(2\sqrt{x}-1>0\)
=>\(x>\dfrac{1}{4}\)
Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}x>\dfrac{1}{4}\\x\ne1\end{matrix}\right.\)
