\(\sqrt{99}<\sqrt{100}=10\)

\(\sqrt{50}+\sqrt{10}\)\(>\sqrt{49}+\sqrt{9}=7+3=10\)

Vậy \(\sqrt{50}+\sqrt{10}>\sqrt{99}\)

Công thức tổng quát:

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

Vậy \(\dfrac{1}{1+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{99}+\sqrt{100}}=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+...+\sqrt{100}-\sqrt{99}=-1+\sqrt{100}=-1+10=9\)

Đề bài:
So sánh \(\sqrt{17}+\sqrt{26}\)và \(\sqrt{99}\)

b) Ta có:

\(\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{99}+\sqrt{100}}\)

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

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

\(=-\sqrt{1}+\sqrt{100}\)

\(=\left(-1\right)+10\)

\(=9.\)

Vì \(9=9.\)

\(\Rightarrow\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{99}+\sqrt{100}}=9\left(đpcm\right).\)

Chúc bạn học tốt!

ta co: gia tri tuyet doi cua 17+gia tri tuyet doi cua26+1=10,2(da lam tron)

gia tri tuyet doi cua 99=9,9(da lam tron)

ta co:9,9<10,2

Suy ra ..........................................................................................................

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