Question

Giải giúp tui với :)) \(\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{4}}+...+\dfrac{1}{\sqrt{99}+\sqrt{100}}\)

Answer 1

bài này bn trục căn thức mẫu nhé

\(P=\dfrac{1-\sqrt{2}}{1-2}+\dfrac{\sqrt{2}-\sqrt{3}}{2-3}+...+\dfrac{\sqrt{99}-\sqrt{100}}{99-100}\)

\(P=\dfrac{1-\sqrt{2}+\sqrt{2}-\sqrt{3}+....+\sqrt{99}-\sqrt{100}}{-1}=\dfrac{1-\sqrt{100}}{-1}=\sqrt{100}-1\)

\(=10-1=9\)

Answer 2

`=(\sqrt1-\sqrt2)/(1-2)+(\sqrt2-\sqrt3)/(2-3)+.....+(\sqrt99-\sqrt100)/(99-100)`

`=\sqrt2-\sqrt1+\sqrt3-\sqrt2+.....+\sqrt100-\sqrt99`

`=-\sqrt1+\sqrt100=-1+10=9`

Answer 3

Ta có: \(\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{4}}+...+\dfrac{1}{\sqrt{99}+\sqrt{100}}\)

\(=-1+\sqrt{2}-\sqrt{2}+\sqrt{3}-\sqrt{3}+2-...-\sqrt{99}+10\)

=10-1=9