S=\(\frac{1}{3}.\left(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+...+\frac{1}{4950}\right)\)

S=\(\frac{1}{3}.2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{99.100}\right)\)

S=\(\frac{2}{3}.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\)

S=\(\frac{2}{3}.\left(1-\frac{1}{100}\right)=\frac{2}{3}.\frac{99}{100}=\frac{33}{50}\)

\(\frac{33}{50}>\frac{30}{50}=\frac{3}{5}->S>\frac{3}{5}\)

\(F=\left(\frac{3}{1.8}+\frac{3}{8.15}+\frac{3}{15.22}+...+\frac{3}{106.113}\right)\)\(-\)\(\left(\frac{25}{50.55}+\frac{25}{55.60}+\frac{25}{60.65}+...+\frac{25}{95.100}\right)\)

\(=\frac{3}{7}\left(\frac{1}{3}-\frac{1}{8}+\frac{1}{8}-\frac{1}{15}+...+\frac{1}{106}-\frac{1}{113}\right)\) -  \(5\left(\frac{1}{50}-\frac{1}{55}+\frac{1}{55}-\frac{1}{60}+...+\frac{1}{95}-\frac{1}{100}\right)\)

\(=\frac{3}{7}\left(\frac{1}{3}-\frac{1}{113}\right)-5\left(\frac{1}{50}-\frac{1}{100}\right)\)

\(=\frac{3}{7}.\frac{110}{339}-5.\frac{1}{100}\)

\(=\frac{1}{7}-\frac{1}{20}=\frac{13}{140}\)

= \(\frac{3}{7}\left(\frac{7}{1.8}+\frac{7}{8.15}+...+\frac{7}{106.103}\right)-5\left(\frac{5}{50.55}+\frac{5}{55.60}+...+\frac{5}{95.100}\right)\)

=\(\frac{3}{7}\left(1-\frac{1}{8}+\frac{1}{8}-\frac{1}{15}+...+\frac{1}{106}-\frac{1}{113}\right)-5\left(\frac{1}{50}-\frac{1}{55}+\frac{1}{55}-\frac{1}{60}+...+\frac{1}{95}-\frac{1}{100}\right)\)

=\(\frac{3}{7}\left(1-\frac{1}{113}\right)-5\left(\frac{1}{50}-\frac{1}{100}\right)\)

=\(\frac{3}{7}\cdot\frac{112}{113}-5\cdot\frac{1}{100}\)

=\(\frac{48}{113}-\frac{1}{20}\)

=\(\frac{847}{2260}\)

\(A=\frac{-5}{7}+\frac{3}{4}+\frac{-1}{5}+\frac{-2}{7}+\frac{1}{4}\)

\(A=\left(\frac{-5}{7}+\frac{-2}{7}\right)+\left(\frac{3}{4}+\frac{1}{4}\right)+\frac{-1}{5}\)

\(A=-1+1+\frac{-1}{5}\)

\(A=\frac{-1}{5}\)

\(B=\frac{-4}{12}+\frac{18}{45}+\frac{-6}{9}+\frac{-21}{35}+\frac{6}{30}\)

\(B=\frac{-1}{3}+\frac{2}{5}+\frac{-2}{3}+\frac{-3}{5}+\frac{1}{5}\)

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

\(B=-1+0\)

\(B=-1\)

toan lop may nói lại

A=1/3 + 1/9 + 1/18 + 1/30 + 1/45 + ... + 1/570

= 1/3(1 + 1/3 + 1/6 + 1/10 + 1/15 + ... + 1/190)

=2/3(3/2 + 1/6 + 1/12 + 1/20 + 1/30 + .... + 1/380)

=2/3(3/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6 + ... + 1/19 + 1/20)

=2/3(3/2+1/2-1/20)

=2/3.39/20

=13/10

b) Đặt B = A : C ta có:

\(A=\frac{5^3}{6}+\frac{5^3}{12}+\frac{5^3}{20}+\frac{5^3}{42}+\frac{5^3}{56}+\frac{5^3}{72}+\frac{5^3}{90}\)

\(A=5^3.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\right)\)

\(A=5^3.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)

\(A=5^3.\left(\frac{1}{2}-\frac{1}{10}\right)\)

\(A=\frac{5^3.2}{5}\)

\(A=5^2.2\)

\(\Rightarrow A=50\)

\(C=\frac{1124.2247-1123}{1124+1123.2247}\)

\(C=\frac{\left(1123+1\right).2274-1123}{1123.2247+1124}\)

\(C=\frac{1123.2247-2247-1123}{1123.2247+1124}\)

\(C=\frac{1123.2247+1124}{1123.2247+1124}=1\)

\(\Rightarrow B=50:1=50\) 

Vậy B = 50

dễ mk nhìn là biết

Đặt A = \(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)

3A = \(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)

3A - A = (\(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)) - (\(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\))

2A = 1 - \(\frac{1}{729}\) = \(\frac{728}{729}\)

A = \(\frac{728}{729}:2=\frac{364}{729}\)

a, Gọi biểu thức đó là A

Ta có :

A = \(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)

A x 3 = \(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}-\frac{1}{729}\)

A x 3 = \(1+A-\frac{1}{729}\)

A x 3 = \(\frac{728}{729}+A\)

A x 2 + A = \(\frac{728}{729}+A\)

A x 2 = \(\frac{728}{729}\)(bỏ A ở cả 2 vế)

A = \(\frac{728}{729}\div2=\frac{364}{729}\)

Đáp án = \(\frac{364}{729}\)

b, Phần này mình nghĩ là bạn sai đề rồi. Phải là \(\frac{45\times16-17}{45\times15+28}\)

\(S=\frac{3}{4}-0,25-\left[\frac{7}{3}+\left(\frac{-9}{2}\right)\right]-\frac{5}{6}\)

\(S=\frac{3}{4}-\frac{1}{4}-\left[\frac{14}{6}+\left(\frac{-27}{6}\right)\right]-\frac{5}{6}\)

\(S=\frac{1}{2}-\left(\frac{-13}{6}\right)-\frac{5}{6}\)

\(S=\frac{3}{6}-\left(\frac{-13}{6}\right)-\frac{5}{6}\)

\(S=\frac{11}{6}\)