Question

\(\left(1+\dfrac{x+\sqrt{x}}{1+\sqrt{x}}\right).\left(1+\dfrac{x-\sqrt{x}}{1-\sqrt{x}}\right)\) . Rút gọn

Answer 1

\(\left(1+\dfrac{x+\sqrt{x}}{1+\sqrt{x}}\right).\left(1+\dfrac{x-\sqrt{x}}{1-\sqrt{x}}\right)\left(đk:x\ge0,x\ne1\right)\)

\(=\dfrac{1+\sqrt{x}+x+\sqrt{x}}{1+\sqrt{x}}.\dfrac{1-\sqrt{x}+x-\sqrt{x}}{1-\sqrt{x}}\)

\(=\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}+1}.\dfrac{\left(1-\sqrt{x}\right)^2}{1-\sqrt{x}}=\left(\sqrt{x}+1\right)\left(1-\sqrt{x}\right)=1-x\)

Answer 2

\(\left(1+\dfrac{x+\sqrt{x}}{1+\sqrt{x}}\right).\left(1+\dfrac{x-\sqrt{x}}{1-\sqrt{x}}\right)\\ =\left(1+\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{1+\sqrt{x}}\right).\left(1+\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{1-\sqrt{x}}\right)\\ =\left(1+\sqrt{x}\right).\left(1-\sqrt{x}\right)\\ =1-x\)

Answer 3

\(\left(1+\dfrac{x+\sqrt{x}}{1+\sqrt{x}}\right).\left(1+\dfrac{x-\sqrt{x}}{1-\sqrt{x}}\right)\)

\(=\left(1+\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right).\left(1-\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\right)\)

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

Answer 4

Ta có: \(\left(1+\dfrac{x+\sqrt{x}}{\sqrt{x}+1}\right)\left(1-\dfrac{x-\sqrt{x}}{\sqrt{x}-1}\right)\)

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

=1-x