1) Đặt dãy trên là \(A\)

Theo bài ra ta có :

\(A=\frac{1}{3.3}+\frac{1}{4.4}+\frac{1}{5.5}+\frac{1}{6.6}+...+\frac{1}{100.100}\)

\(\Rightarrow A< \frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)

\(\Rightarrow A< \frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)

\(\Rightarrow A< \frac{1}{2}-\frac{1}{100}< \frac{1}{2}\left(đpcm\right)\)

2) \(A=\frac{5^{2018}-2017+1}{5^{2018}-2017}=\frac{5^{2018}-2017}{5^{2018}-2017}+\frac{1}{5^{2018}-2017}=1+\frac{1}{5^{2018}-2017}\)( 1 )

\(B=\frac{5^{2018}-2019+1}{5^{2018}-2019}=\frac{5^{2018}-2019}{5^{2018}-2019}+\frac{1}{5^{2018}-2019}=1+\frac{1}{5^{2018}-2019}\)( 2 )

Từ ( 1 ) và ( 2 ) \(\Rightarrow\)\(A=1+\frac{1}{5^{2018}-2017}< 1+\frac{1}{5^{2018}-2019}=B\)

\(\Rightarrow A< B\)

Vậy \(A< B.\)

1) Ta có B =

 \(\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\) < \(\frac{1}{1.3}+\frac{1}{3.4}+...+\frac{1}{99.100}=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)= \(\frac{99}{100}\)

=> B < 1 ( chứ không phải \(\frac{1}{2}\) bạn nhé)

Sai thì thôi chứ mk chỉ làm rờ thôi

\(A=5+5^2+5^3+5^4+...+5^{2004}\)

\(5A=5^2+5^3+5^4+5^5+...+5^{2005}\)

\(5A-A=\left(5^2+5^3+5^4+5^5+...+5^{2005}\right)-\left(5+5^2+5^3+5^4+...+5^{2004}\right)\)

\(4A=5^{2005}-5\)

\(A=\dfrac{5^{2005}-5}{4}\)

\(B=7^1+7^2+7^3+....+7^{2015}\)

\(7B=7^2+7^3+7^4+....+7^{2016}\)

\(7B-B=\left(7^2+7^3+7^4+...+7^{2016}\right)-\left(7+7^2+7^3+....+7^{2015}\right)\)

\(6B=7^{2016}-7\)

\(B=\dfrac{7^{2016}-7}{6}\)

\(C=4^5+4^6+4^7+...+4^{2016}\)

\(4C=4^6+4^7+4^8+...+4^{2017}\)

\(4C-C=\left(4^6+4^7+4^8+...+4^{2017}\right)-\left(4^5+4^6+4^7+...+4^{2016}\right)\)

\(3C=4^{2017}-4^5\)

\(C=\dfrac{4^{2017}-4^5}{3}\)

A = 5 + 5² + 5³ + 5⁴ + ... + 5²⁰⁰⁴

5A = 5² + 5³ + 5⁴ + 5⁵ + ... + 5²⁰⁰⁵

5A - A = 5²⁰⁰⁵ - 5

4A = 5²⁰⁰⁵ - 5

A = (5²⁰⁰⁵ - 5) : 4

B = 7¹ + 7² + 7³ + ... + 7²⁰¹⁵

7B = 7² + 7³ + 7⁴ + ... + 7²⁰¹⁶

7B - B = 7²⁰¹⁶ - 7

6B = 7²⁰¹⁶ - 7

B = (7²⁰¹⁶ - 7) : 6

C = 4⁵ + 4⁶ + 4⁷ + ... + 4²⁰¹⁶

4C = 4⁶ + 4⁷ + 4⁸ + ... + 4²⁰¹⁷

4C - C = 4²⁰¹⁷ - 4⁵

3C = 4²⁰¹⁷ - 4⁵

C = (4²⁰¹⁷ - 4⁵) : 3

Ô...mai..gót

Thế này ko ai giải cho bn đâu vì họ ko dại gì làm tất cả chỉ để lấy cái T.I.C.K

Hãy đăng từng câu một 

Ai đồng quan điểm

Bạn lấy mấy bài này từ mấy cái đề học sinh giỏi vậy ?

\(\frac{19}{37}+\left(1-\frac{19}{37}\right)\)

\(=\frac{19}{37}+1-\frac{19}{37}\)
\(=\left(\frac{19}{37}-\frac{19}{37}\right)+1\)

\(=0+1=1\)

@@ dùng máy tính mà tính 

Anh làm mẫu 1 phần 

\(\frac{\frac{2}{2017}+\frac{2}{2018}}{\frac{5}{2017}+\frac{5}{2018}}=\frac{2.\left(\frac{1}{2017}+\frac{1}{2018}\right)}{5.\left(\frac{1}{2017}+\frac{1}{2018}\right)}=\frac{2}{5}\)

Thanks!

\(A=1+7+7^2+7^3+...+7^{2016}\)

\(\Rightarrow7A=7\left(1+7+7^2+7^3+...+7^{2016}\right)\)

\(7A=7+7^2+7^3+7^4+...+7^{2017}\)

\(\Rightarrow7A-A=\left(7+7^2+7^3+...+7^{2017}\right)-\left(1+7+7^2+...+7^{2016}\right)\)

\(\Rightarrow6A=7^{2017}-1\)

\(\Rightarrow A=\dfrac{7^{2017}-1}{6}\)

mn giúp mk vs, mk cần rất gấp òi. cảm ơn mn nhìu

$A=1-$$3+5-7+...+2001-2003+2005$

$=\left(1-3\right)+\left(5-7\right)+...+\left(2001-2003\right)+2005$(Có 1002 cặp)

$=\left(-2\right).1002+2005$

$=-2004+2005$

$=1$

\(2.THPT\)

\(A=\frac{9}{1.2}+\frac{9}{2.3}+\frac{9}{3.4}+...+\frac{9}{98.99}+\frac{9}{99.100}\)

\(A=9\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)

\(A=9\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)

\(A=9\left(1-\frac{1}{100}\right)\)

\(A=9.\frac{99}{100}\)

\(A=\frac{891}{100}\)

\(B=\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{2}{93.95}\)

\(B=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{93}-\frac{1}{95}\)

\(B=\frac{1}{5}-\frac{1}{95}\)

\(B=\frac{18}{95}\)

\(D=\frac{5}{2.7}+\frac{4}{7.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\)

\(D=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{28}\)

\(D=\frac{1}{2}-\frac{1}{28}\)

\(D=\frac{13}{28}\)