Question

Rút gọn:

\(B=\left(2+\frac{x-2\sqrt{x}+1}{1-\sqrt{x}}\right).\left(2+\frac{x+2\sqrt{x}+1}{\sqrt{x}+1}\right)\)với \(x\ge0;x\ne1\)

Answer 1

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

=\(\left[2-\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}\right]\left[2+\frac{\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right]\)

=\(\left[2-\left(\sqrt{x}-1\right)\right]\left[2+\left(\sqrt{x}+1\right)\right]\)

=4 - \(\left(\sqrt{x}-1\right)^2\)

=4-x+2\(\sqrt{x}\) -1

=4-x+2\(\sqrt{x}\)