dễ thôi bn, tính về trên đầu , vế dưới

\(a.\)

\(\frac{15}{33}+\frac{7}{20}+\frac{18}{33}+\frac{13}{20}\)

\(=\left(\frac{15}{33}+\frac{18}{33}\right)+\left(\frac{13}{20}+\frac{7}{20}\right)\)

\(=\frac{33}{33}+\frac{20}{20}\)

\(=1+1=2\)

\(b.\)

\(2\frac{1}{2}+\frac{4}{7}:\left(-\frac{8}{21}\right)\)

\(=\frac{5}{2}+\frac{4}{7}:\left(-\frac{8}{21}\right)\)

\(=\frac{5}{2}+\frac{4}{7}.\left(-\frac{21}{8}\right)\)

\(=\frac{5}{2}+\frac{1}{1}.\left(-\frac{3}{2}\right)\)

\(=\frac{5}{2}-\frac{3}{2}\)

\(=1\)

\(c.\)

\(\left(-\frac{1}{2}\right)^3+\frac{1}{2}:5\)

\(=-\frac{1}{8}+\frac{1}{2}.\frac{1}{5}\)

\(=-\frac{1}{8}+\frac{1}{10}\)

\(=-\frac{1}{40}\)

a) \(\frac{15}{33}+\frac{7}{20}+\frac{18}{33}+\frac{13}{20}=\left(\frac{15}{33}+\frac{18}{33}\right)+\left(\frac{7}{20}+\frac{13}{20}\right)\) = 1 + 1 = 2

b) \(2\frac{1}{2}+\frac{4}{7}:\left(\frac{-8}{21}\right)=\frac{5}{2}+\frac{4}{7}:\left(\frac{-8}{21}\right)=\frac{5}{2}+\frac{-3}{2}\) = 1

c) \(\left(\frac{-1}{2}\right)\)³ + \(\frac{1}{2}\) : 5 = \(\frac{-1}{8}+\frac{1}{10}\) = \(\frac{-1}{40}\)

Chúc bạn học tốt!

Ta có: \(2\frac{1}{3}+\frac{11}{5}:33-\frac{1}{50}.\left(-5\right)^2\)

\(=\frac{7}{3}+\frac{11}{5}.\frac{1}{33}-\frac{1}{50}.25\)

\(=\frac{7}{3}+\frac{11}{165}-\frac{25}{50}\)

\(=\frac{7}{3}+\frac{1}{15}-\frac{1}{2}\)

\(=\frac{70}{30}+\frac{2}{30}-\frac{15}{30}\)

\(=\frac{56}{30}=\frac{28}{15}\)

b/ Ta có: \(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n}.\sqrt{n+1}.\left(\sqrt{n+1}+\sqrt{n}\right)}\)

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

Áp dụng vào bài toán ta được

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

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

\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{100}}=1-\frac{1}{10}=\frac{9}{10}\)

Cả 2 câu là n tự nhiên khác 0 hết nhé

a/ Ta có: \(\frac{1}{\sqrt{n}+\sqrt{n+1}}=\frac{\sqrt{n+1}-\sqrt{n}}{n+1-n}=\sqrt{n+1}-\sqrt{n}\)

Áp đụng vào bài toán được

\(\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{1680}+\sqrt{1681}}\)

\(=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{1681}-\sqrt{1680}\)

\(=\sqrt{1681}-\sqrt{1}=41-1=40\)

đặt A=...

ta có 

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

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

Ta có:

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

Lại có:

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

Do đó:

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

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

\(\Leftrightarrow\sqrt{100}-1=10-1=9\)

1 : 29 x ( 19 -13 ) - 19 x ( 29 - 13 )

= 29 x 6 - 19 x 16

= 174 - 304

=  - 130

2 : 1 - \(\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

⁼ 1 - \(\frac{1}{100}\)

= \(\frac{99}{100}\)

\(\frac{1}{100.99}-\left(\frac{1}{99.98}+\frac{1}{98.97}+...+\frac{1}{2.1}\right)\)

\(=\frac{1}{100}-\frac{1}{99}-\left(\frac{1}{99}-\frac{1}{98}+\frac{1}{98}-\frac{1}{97}+...+\frac{1}{2}-1\right)\)

\(=\frac{1}{100}-\frac{1}{99}-\left(\frac{1}{99}-1\right)\)

\(=\frac{1}{100}-\frac{1}{99}-\frac{1}{99}+1\)

\(=\frac{9799}{9900}\)

A=\(33.\dfrac{3-2}{3}.\dfrac{5-2}{5}....\dfrac{99-2}{99}\)

A=\(33.\dfrac{1}{3}.\dfrac{3}{5}.\dfrac{5}{7}...\dfrac{95}{97}.\dfrac{97}{99}\)

A=\(33.\dfrac{1}{99}=\dfrac{1}{3}\)