\(1,\sqrt[3]{-64}+\sqrt[3]{27}=-4+3=-1\)
\(2,\\ a,A=\left(1+\dfrac{\sqrt{x}}{x+1}\right):\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2\sqrt{x}}{x\sqrt{x}+\sqrt{x}-x-1}\right)\left(x\ge0;x\ne1\right)\\ A=\dfrac{x+\sqrt{x}+1}{x+1}:\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+1\right)}\right)\\ A=\dfrac{x+\sqrt{x}+1}{x+1}:\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+1\right)}\\ A=\dfrac{x+\sqrt{x}+1}{x+1}\cdot\dfrac{\left(\sqrt{x}-1\right)\left(x+1\right)}{\left(\sqrt{x}-1\right)^2}=\dfrac{x+\sqrt{x}+1}{\sqrt{x}-1}\)
\(b,x=4+2\sqrt{3}=\left(\sqrt{3}+1\right)^2\\ A=\dfrac{4+2\sqrt{3}+\sqrt{3}+1+1}{\sqrt{3}+1-1}\\ A=\dfrac{3\left(\sqrt{3}+2\right)}{\sqrt{3}}=\sqrt{3}\left(\sqrt{3}+2\right)\)
\(c,A>1\Leftrightarrow A-1>0\Leftrightarrow\dfrac{x+\sqrt{x}+1}{\sqrt{x}-1}>0\\ \Leftrightarrow\sqrt{x}-1>0\left[x+\sqrt{x}+1=\left(\sqrt{x}+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\right]\\ \Leftrightarrow\sqrt{x}>1\Leftrightarrow x>1\)
Bài 1:
Ta có: \(\sqrt[3]{-64}+\sqrt[3]{27}\)
\(=-4+3\)
=-1
Bài 2:
a: Ta có: \(A=\left(1+\dfrac{\sqrt{x}}{x+1}\right):\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2\sqrt{x}}{x\sqrt{x}+\sqrt{x}-x-1}\right)\)
\(=\dfrac{x+\sqrt{x}+1}{x+1}:\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+1\right)}\)
\(=\dfrac{x+\sqrt{x}+1}{x+1}\cdot\dfrac{x+1}{\sqrt{x}-1}\)
\(=\dfrac{x+\sqrt{x}+1}{\sqrt{x}-1}\)
