Question

Answer 1

a: \(B=\left(\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{1}{x-\sqrt{x}}\right):\left(\frac{1}{\sqrt{x}+1}+\frac{2}{x-1}\right)\)

\(=\left(\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\frac{1}{\sqrt{x}+1}+\frac{2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)

\(=\frac{x+1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\frac{\sqrt{x}-1+2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)

\(=\frac{x+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}+1}=\frac{x+1}{\sqrt{x}}\)

c: B<0

=>\(\frac{x+1}{\sqrt{x}}<0\)

mà x+1>0 và \(\sqrt{x}>0\forall x\) thỏa mãn ĐKXĐ

nên x∈∅