Xét hạng tổng quát:

\(\frac{1}{\sqrt{n-1}+\sqrt{n}}=\frac{1}{\sqrt{n}+\sqrt{n-1}}=\frac{\sqrt{n}-\sqrt{n-1}}{n-n+1}=\sqrt{n}-\sqrt{n-1}\)

Áp dụng vào bài, ta có:

\(\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{n-1}+\sqrt{n}}\)

\(=\left(\sqrt{2}-1\right)+\left(\sqrt{3}-\sqrt{2}\right)+\left(\sqrt{4}-\sqrt{3}\right)+\left(\sqrt{n}-\sqrt{n-1}\right)\)

\(=\sqrt{n}-1\)

\(\frac{1}{2}:0,5-\frac{1}{4}:0,25+\frac{1}{8}:0,125-\frac{1}{10}:0,1\)

\(=\frac{1}{2}\times2-\frac{1}{4}\times4+\frac{1}{8}\times8-\frac{1}{10}\times10\)

\(=1-1+1-1\)

\(=0\)

\(1h45p=1,75h\)

quãng đường mà xe đạp đi đc là:

\(27\times1,75=47,25\left(km\right)\)

đs,,,,,,,,,,,,,,,,,,,,,

2/3 + 1/6 -7/12

= 5/6 - 7/12

=      1/4 

Học tối nha !!!

\(\frac{2}{3}+\frac{1}{6}-\frac{7}{12}\)

\(=\frac{5}{6}-\frac{7}{12}\)

\(=\frac{1}{4}\)

_HT_

ta có:

\(\sqrt{2+\sqrt{2}+\sqrt{2}+....+\sqrt{2}}>\sqrt{1}=1\)

lại có: \(\sqrt{2+\sqrt{2}+\sqrt{2}+....+\sqrt{2}}< \sqrt{2+\sqrt{2+\sqrt{2+...+\sqrt{4}}}}=\sqrt{2+\sqrt{2+\sqrt{2+...+2}}=2}\)\(\Rightarrow1< \sqrt{2+\sqrt{2+\sqrt{2+....+\sqrt{2}}}}< 2\)

\(\Rightarrow\sqrt{2+\sqrt{2+\sqrt{2+...+\sqrt{2}}}}\) ko phải là STN

Nhìn vào bài dễ thấy, \(A>1\)hay ta chứng minh \(A< 2\)

Vậy: \(\sqrt{2+\sqrt{2}}< \sqrt{2+2}=\sqrt{4}=2\)

\(\sqrt{2+\sqrt{2+\sqrt{2}}}< \sqrt{2+2}=\sqrt{4}=2\)

Nên:

\(A=\sqrt{2+\sqrt{2+\sqrt{2+...+\sqrt{2}}}}< \sqrt{2+2}=\sqrt{4}=2\)

\(\Rightarrow1< A< 2\)hay \(A\neℕ\left(đpcm\right)\)

Bài 8:

\(\frac{1}{2}+\frac{2}{4}+\frac{3}{6}+\frac{4}{8}+\frac{5}{10}+\frac{6}{12}\)

\(=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}\)

\(=6\times\frac{1}{2}=3\)

= 3 nhé

tích mik nha

500 + 200 +68 +90 =858

858 bn nhé

HT