Bạn ơi, mk sử đề lại chút, chả bt bn có phải đánh nhầm không ( chỗ dấu "+" với dấu "." này ý)

sửa: 1+3+3^2+...+3^18.3^19 thành 1+3+3^2+...+3^18+3^19

Ta có: S = 1+3+3^2+...+3^18+3^19

\(\Rightarrow\) 3S = 3.(1+3+3^2+...+3^18+3^19)

             3S = 3+3^2+...+3^19+3^20

\(\Rightarrow\)3S - S = (3+3^2+...+3^19+3^20) - (1+3+3^2+...+3^18+3^19)

            2S = 3+3^2+...+3^19+3^20 - 1- 3- 3^2- ...- 3^18- 3^19

           2S = 3^20 - 1 

             S = (3^20 - 1) : 2

( Kquả ta nên để dưới dạng phân số cung đc )

a) \(\left(4\frac{1}{2}-2x\right)\cdot3\frac{2}{3}=\frac{11}{5}\)

\(\left(\frac{9}{2}-2x\right)=\frac{11}{5}\cdot\frac{3}{11}\)

\(2x=\frac{45-6}{10}\)

\(2x=\frac{39}{10}\)

\(x=\frac{39}{10\cdot2}=\frac{39}{20}\)

b) \(\frac{3}{4}\cdot x+\frac{4}{7}\cdot x=-\frac{15}{8}\)

\(x\cdot\left(\frac{21+16}{28}\right)=-\frac{15}{8}\)

\(x=-\frac{15}{8}\cdot\frac{28}{37}\)

\(x=-\frac{105}{74}\)

Chứng minh rằng tổng phân số đó <1/4 nha các bạn

a, Ta có: \(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};...;\frac{1}{2017^2}< \frac{1}{2016.2017}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2017^2}>\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2016.2017}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2016}-\frac{1}{2017}=1-\frac{1}{2017}< 1\)Vậy...

b, Đặt A = \(\frac{1}{4}+\frac{1}{16}+\frac{1}{36}+...+\frac{1}{10000}\)

\(A=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{100^2}\)

\(A=\frac{1}{2^2}\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)\)

Đặt B = \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\)

Ta có: \(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};.....;\frac{1}{50^2}< \frac{1}{49.50}\)

\(\Rightarrow B< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}=1-\frac{1}{50}< 1\)

Thay B vào A ta được:

\(A< \frac{1}{4}\left(1+1\right)=\frac{1}{4}.2=\frac{1}{2}\)

Vậy....

c, Ta có: \(\frac{1}{2^2}>\frac{1}{2.3};\frac{1}{3^2}>\frac{1}{3.4};....;\frac{1}{9^2}>\frac{1}{9.10}\)

\(\Rightarrow A>\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}=\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\)(1)

Lại có: \(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};....;\frac{1}{9^2}< \frac{1}{8.9}\)

\(\Rightarrow A< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{8.9}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{8}-\frac{1}{9}=1-\frac{1}{9}=\frac{8}{9}\)(2)

Từ (1) và (2) suy ra \(\frac{2}{5}< A< \frac{8}{9}\)(đpcm)

d, chắc là đề sai

e, giống câu a

a>

\(\frac{1}{2^2}+\frac{1}{100^2}\)=1/4+1/10000

ta có 1/4<1/2(vì 2 đề bài muốn chứng minh tổng đó nhỏ 1 thì chúng ta phải xét xem có bao nhiêu lũy thừa hoặc sht thì ta sẽ lấy 1 : cho số số hạng )

1/100^2<1/2

=>A<1

oke

C.-1/2<3/2<5/2

A=\(\frac{1}{30}\)+\(\frac{1}{42}\)+\(\frac{1}{56}\)+\(\frac{1}{72}\)+\(\frac{1}{90}\)+\(\frac{1}{110}\)+\(\frac{1}{132}\)

A=\(\frac{1}{5.6}\)+\(\frac{1}{6.7}\)+\(\frac{1}{7.8}\)+\(\frac{1}{8.9}\)+\(\frac{1}{9.10}\)+\(\frac{1}{10.11}\)+\(\frac{1}{11.12}\)

A= \(\frac{1}{5}\)-\(\frac{1}{6}\)+\(\frac{1}{6}\)-\(\frac{1}{7}\)+\(\frac{1}{7}\)-\(\frac{1}{8}\)+\(\frac{1}{8}\)-\(\frac{1}{9}\)+\(\frac{1}{9}\)-\(\frac{1}{10}\)+\(\frac{1}{10}\)-\(\frac{1}{11}\)+\(\frac{1}{11}\)-\(\frac{1}{12}\)

A= \(\frac{1}{5}\)-\(\frac{1}{12}\)=\(\frac{7}{60}\)

A = 1/2 + 1/2^2 + 1/2^3 + ... + 1/2^n < 1

A = 1/2 + 1/2^2 + 1/2^3 + ... + 1/2^n

2A = 1 + 1/2 + 1/2^2+ ..+ 1/2^n-1

2A - A = 1 + 1/2 + 1/2^2+ ..+ 1/2^n-1 - (1/2 + 1/2^2 + 1/2^3 + ... + 1/2^n)

A = (1 - 1/2^n) + (1/2 - 1/2) + ..+ (1/2^n-1 -1/2^n-1)

A = 1 - 1/2^n

A < 1 (đpcm)

Câu b:

B = 1/3 + 1/3^2 + 1/3^3 + ...+ 1/3^n < 1/2

3B = 1 + 1/3 + 1/3^2 + ..+ 1/3^n - 1

3B - B = 1 + 1/3 + 1/3^2 + ..+ 1/3^n - 1 - (1/3 + 1/3^2 + 1/3^3 + ...+ 1/3^n)

2B = 1 + 1/3 + 1/3^2 + ..+ 1/3^n - 1 - 1/3 - 1/3^2 - 1/3^3 -..- 1/3^n-1 - 1/3^n

2B = (1 - 1/3^n) + (1/3 - 1/3) +..+(1/3^n-1-1/3^n-1)

2B = 1 - 1/3^n

B = 1/2 - 1/2.3^n < 1/2 (đpcm)