Question

Rút gọn bt ; \((2+\dfrac{x+\sqrt{x}}{\sqrt{x}+1})\) . \((2-\dfrac{x-\sqrt{x}}{\sqrt{x}-1})\) với x\(\ge\) 0 , x\(\ne\) 1

Answer 1

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

= \(\dfrac{x+3\sqrt{x}+2}{\sqrt{x}+1}.\dfrac{-x+3\sqrt{x}-2}{\sqrt{x}-1}\)

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

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

Answer 2

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

= \(\dfrac{x+3\sqrt{x}+2}{\sqrt{x}+1}:\dfrac{-x+3\sqrt{x}-2}{\sqrt{x}-1}\)

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

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

Answer 3

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

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

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

=\(\dfrac{3\sqrt{x}+2+x}{\sqrt{x}+1}\times\dfrac{3\sqrt{x}-2-x}{\sqrt{x}-1}\)

=\(\dfrac{5x-4-x^2}{x-1}\)

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

=\(\dfrac{-x\left(x-1\right)+4\left(x-1\right)}{x-1}\)

=\(\dfrac{\left(x-1\right)\left(4-x\right)}{x-1}\)

=\(4-x\)