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 \(\frac{45\times16-17}{45\times15+28}\)

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

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

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

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

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

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

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

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

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

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

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

\(S=\frac{1}{3}+\frac{1}{9}+\frac{1}{18}+\frac{1}{30}+\frac{1}{45}+...+\frac{1}{14850}\)

\(\Rightarrow\frac{3}{2}S=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{9900}\)

               \(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)

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

\(S=\frac{1}{3}+\frac{1}{9}+\frac{1}{30}+\frac{1}{45}+...+\frac{1}{14850}\)

\(\Rightarrow\frac{3}{2}S=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{9900}\)

               \(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)

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

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

Vậy S = \(\frac{99}{100}:\frac{3}{2}\) = \(\frac{33}{50}\)

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

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

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

I'm sorry, I can't help you

\(\Rightarrow\left(\frac{\frac{31}{36}}{\frac{11}{9}:4-\frac{3}{4}}:31\right).x=\frac{-1}{16}\Rightarrow\left(\frac{\frac{31}{36}}{\frac{-4}{9}}:31\right)x=-\frac{1}{16}\Rightarrow\left(-\frac{31}{16}:31\right)x=-\frac{1}{16}\Rightarrow-\frac{1}{16}x=-\frac{1}{16}\Rightarrow x=1\)