b) \(\frac{\frac{-6}{5}+\frac{6}{19}-\frac{6}{23}}{\frac{9}{5}-\frac{9}{19}+\frac{9}{23}}=\frac{\left(-6\right).\left(\frac{1}{5}-\frac{1}{19}+\frac{1}{23}\right)}{9.\left(\frac{1}{5}-\frac{1}{19}+\frac{1}{23}\right)}=\frac{-6}{9}=\frac{-2}{3}\)

d) \(\frac{\frac{2}{3}-\frac{2}{5}-\frac{2}{7}+\frac{2}{11}}{\frac{13}{3}-\frac{13}{5}-\frac{13}{7}+\frac{13}{11}}=\frac{2\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}+\frac{1}{11}\right)}{13\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}+\frac{1}{11}\right)}=\frac{2}{13}\)

Làm tiếp:

\(=\left(1+\frac{1}{2}+.....+\frac{1}{2017}\right)-\left(1+\frac{1}{2}+....+\frac{1}{1008}\right)\)

\(=\frac{1}{1009}+\frac{1}{1010}+.........+\frac{1}{2017}\)

\(\Rightarrow\frac{\frac{1}{1009}+....+\frac{1}{2017}}{1-\frac{1}{2}+.....+\frac{1}{2015}-\frac{1}{2016}+\frac{1}{2017}}=1\)

Bài 2:

Đặt \(A=\frac{1}{2^2}+.......+\frac{1}{2^{800}}\)

\(4A=1+\frac{1}{2^2}+.....+\frac{1}{2^{798}}\)

\(\Rightarrow4A-A=1-\frac{1}{2^{800}}\)

\(\Rightarrow3A=1-\frac{1}{2^{800}}< 1\Rightarrow A< \frac{1}{3}\)

Vậy \(\frac{1}{2^2}+\frac{1}{2^4}+........+\frac{1}{2^{800}}< \frac{1}{3}\)

Bài 1:Tính

a,   Xét biểu thức \(\frac{\left(1+\frac{n}{1}\right)\left(1+\frac{n}{2}\right).........\left(1+\frac{n}{n+2}\right)}{\left(1+\frac{n+2}{1}\right)\left(1+\frac{n+2}{2}\right)..........\left(1+\frac{n+2}{n}\right)}\) với\(n\in N\)

Ta có:\(\frac{\left(1+\frac{n}{1}\right)\left(1+\frac{n}{2}\right).......\left(1+\frac{n}{n+2}\right)}{\left(1+\frac{n+2}{1}\right)\left(1+\frac{n+2}{2}\right)......\left(1+\frac{n+2}{n}\right)}\)

\(=\frac{\frac{n+1}{1}.\frac{n+2}{2}........\frac{2n+2}{n+2}}{\frac{n+3}{1}.\frac{n+4}{2}.........\frac{2n+2}{n}}\)

\(=\frac{\frac{\left(n+1\right)\left(n+2\right).......\left(2n+2\right)}{1.2.3.........\left(n+2\right)}}{\frac{\left(n+3\right)\left(n+4\right)........\left(2n+2\right)}{1.2.3.........n}}\)

\(=\frac{\left(n+1\right)\left(n+2\right).......\left(2n+2\right).1.2.3.......n}{\left(n+3\right)\left(n+4\right)........\left(2n+2\right).1.2.3......\left(n+2\right)}\)

\(=\frac{\left(n+1\right)\left(n+2\right)}{\left(n+1\right)\left(n+2\right)}=1\)

Áp dụng vào bài toán ta có đáp số là:1

b, \(\frac{\frac{-6}{5}+\frac{6}{19}-\frac{6}{23}}{\frac{9}{5}-\frac{9}{19}+\frac{9}{23}}=\frac{\left(-6\right).\left(\frac{1}{5}-\frac{1}{19}+\frac{1}{23}\right)}{9.\left(\frac{1}{5}-\frac{1}{19}+\frac{1}{23}\right)}=\frac{-6}{9}=-\frac{2}{3}\)

c,\(\frac{\frac{1}{6}-\frac{1}{39}+\frac{1}{51}}{\frac{1}{8}-\frac{1}{52}+\frac{1}{68}}=\frac{\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{13}+\frac{1}{17}\right)}{\frac{1}{4}.\left(\frac{1}{2}-\frac{1}{13}+\frac{1}{17}\right)}=\frac{\frac{1}{3}}{\frac{1}{4}}=12\)

d,\(\frac{\frac{2}{3}-\frac{2}{5}-\frac{2}{7}}{\frac{13}{3}-\frac{13}{5}-\frac{13}{7}}=\frac{2\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}\right)}{13\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}\right)}=\frac{2}{13}\)

e,Xét mẫu số ta có:

\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+..........+\frac{1}{2015}-\frac{1}{2016}+\frac{1}{2017}\)

\(=1+\frac{1}{2}-2.\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-2.\frac{1}{4}+.....+\frac{1}{2015}+\frac{1}{2016}-2.\frac{1}{2016}+\frac{1}{2017}\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+.......+\frac{1}{2017}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+.........+\frac{1}{2016}\right)\)

2/

S = 2 + 2² + 2³ +...+ 2⁹⁹

= (2+2²+2³) +...+ (2⁹⁷+2⁹⁸+2⁹⁹)

= 2(1+2+2²)+...+2⁹⁷(1+2+2²)

= 2.7 +...+ 2⁹⁷.7

= 7(2+...+2⁹⁷) chia hết cho 7

S = 2+2²+2³+...+2⁹⁹

= (2+2²+2³+2⁴+2⁵)+...+(2⁹⁵+2⁹⁶+2⁹⁷+2⁹⁸+2⁹⁹)

= 2(1+2+2²+2³+2⁴)+...+2⁹⁵(1+2+2²+2³+2⁴)

= 2.31+...+2⁹⁵.31

= 31(2+...+2⁹⁵) chia hết cho 31

3/

A = 1+5+5²+....+5¹⁰⁰ (1)

5A = 5+5²+5³+...+5101 (2)

Lấy (2) - (1) ta được

4A = 5¹⁰¹ - 1

A = \(\frac{5^{101}-1}{4}\)

4/

Đặt A là tên của biểu thức trên

Ta có: \(\frac{1}{2^2}< \frac{1}{1.2}=\frac{1}{1}-\frac{1}{2}\)

\(\frac{1}{3^2}< \frac{1}{2.3}=\frac{1}{2}-\frac{1}{3}\)

........

\(\frac{1}{8^2}< \frac{1}{7.8}=\frac{1}{7}-\frac{1}{8}\)

\(\Rightarrow A< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{8}=\frac{1}{1}-\frac{1}{8}=\frac{7}{8}< 1\)

Vậy...

5/

a, Gọi UCLN(n+1,2n+3) = d

Ta có : n+1 chia hết cho d => 2(n+1) chia hết cho d => 2n+2 chia hết cho d

           2n+3 chia hết cho d

=> 2n+2 - (2n+3) chia hết cho d

=> -1 chia hết cho d => d = {-1;1}

Vậy...

b, Gọi UCLN(2n+3,4n+8) = d

Ta có: 2n+3 chia hết cho d => 2(2n+3) chia hết cho d => 4n+6 chia hết cho d

          4n+8 chia hết cho d 

=> 4n+6 - (4n+8) chia hết cho d

=> -2 chia hết cho d => d = {1;-1;2;-2}

Mà 2n+3 lẻ => d lẻ => d khác 2;-2 => d = {1;-1}

Vậy...

6/

a,Vì B > 1

\(\Rightarrow B=\frac{2016^{2016}}{2016^{2016}-3}>\frac{2016^{2016}+2}{2016^{2016}-3+2}=\frac{2016^{2016}+2}{2016^{2016}-1}=A\) 

Vậy A < B

b, C = \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{9900}\)

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

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

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

1\(\frac{1}{2}\)+2\(\frac{2}{3}\)+3\(\frac{3}{4}\)+4\(\frac{4}{5}\)+.......+50\(\frac{50}{51}\)+\(\frac{1}{2}\)+\(\frac{1}{3}\)+\(\frac{1}{4}\)+\(\frac{1}{5}\)+....+\(\frac{1}{51}\)

=(1\(\frac{1}{2}\)+\(\frac{1}{2}\))+(2\(\frac{2}{3}\)+\(\frac{1}{3}\))+(3\(\frac{3}{4}\)+\(\frac{1}{4}\))+.......+(50\(\frac{50}{51}\)+\(\frac{1}{51}\))

=2+3+4+.....+51

=1325

Vậy:1\(\frac{1}{2}\)+2\(\frac{2}{3}\)+3\(\frac{3}{4}\)+4\(\frac{4}{5}\)+.......+50\(\frac{50}{51}\)+\(\frac{1}{2}\)+\(\frac{1}{3}\)+\(\frac{1}{4}\)+\(\frac{1}{5}\)+....+\(\frac{1}{51}\)=1325

Học Tốt!

\(1\frac{1}{2}+2\frac{2}{3}+3\frac{3}{4}+4\frac{4}{5}+...+50\frac{50}{51}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{51}\)

\(=1+\frac{1}{2}+2+\frac{2}{3}+3+\frac{3}{4}+...+50+\frac{50}{51}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{51}\)

\(=\left(1+2+3+...+50\right)+\left(\frac{1}{2}+\frac{1}{2}\right)+\left(\frac{2}{3}+\frac{1}{3}\right)+...+\left(\frac{50}{51}+\frac{1}{51}\right)\)

\(=\frac{50.51}{2}+1+1+1+...+1\) ( có 50 số 1 )

\(=1275+50\)

\(=1325\)

cứ nhóm vào ta được

2+3+......+50+51

suy ra biểu thức trên bằng 1325

help me

sao nhiều dữ vậy

Quá dài, quá khó, quá nguy hiểm.

a: ~12.58

b: 10

c:-1

d:55/4

dkr68

a)~12.58

b)10

c)-1

d)55/4

Từ dãy trên ta có:

(\(\frac{3}{2}\)+\(\frac{1}{2}\))+(\(\frac{8}{3}\)+\(\frac{2}{3}\))+......+(\(\frac{2600}{51}\)+\(\frac{1}{51}\))                  < vì không có cách nhập hỗn số nên mình đổi ra phân số >

= 2 + 3 + 4 + 5 + 6 + ..........................+ 51

Từ 2 -> 51 có :( 51 - 2 ) : 1 + 1 = 50 số 

Chia ra : 50 : 2 = 25 cặp 

ta có( 51 + 2 ) x 25 =1325

Vậy tổng trên có kết quả bằng 1325       (tớ chỉ nghĩ thế thôi chứ sai đừng trách nhá.Đùa thôi,đúng đấy )

A = 1 + 1/2 + 1/3 + 1/4 + 1/5 + ... + 1/100

Ta đổi A = 2-1+1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+...+1/99-1/100

A= 2 - 1 - 1/100 =200/100 -100/100 - 1/100

A= 99/100

Cảm ơn bạn Kudo Shinichi, nhưng 

1=2-1 ->ok

1/2=1-1/2 ->ok

1/3=1/2-1/3 -> sai 

vì 1/2-1/3=1/6