mk làm tắt dc ko

\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)

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

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

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

a, \(A=\frac{5}{11.16}+\frac{5}{16.21}+\frac{5}{21.26}+...+\frac{5}{61.66}\)

  \(A=\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+\frac{1}{21}-\frac{1}{26}+...+\frac{1}{61}-\frac{1}{66}\)

 \(A=\frac{1}{11}-\frac{1}{66}\)

\(A=\frac{5}{66}\)

b, \(B=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\)

\(B=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\)

\(B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\)

\(B=1-\frac{1}{7}\)

\(B=\frac{6}{7}\)

_Học tốt nha_

1) \(\frac{3^{2014}.8^{19}}{6^{60}.3^{1955}}=\frac{3^{2014}.\left(2^3\right)^{19}}{\left(2.3\right)^{60}.3^{1955}}=\frac{3^{2014}.2^{57}}{2^{60}.3^{2015}}=\frac{1}{2^3.3}=\frac{1}{24}\)

2) \(5^x+5^{x+1}=150\)

=> 5^x(1 + 5) = 150

=> 5^x.6 = 150

=> 5^x = 25

=> \(x=\pm2\)

3) \(\frac{3}{11.16}+\frac{3}{16.21}+...+\frac{3}{61.66}=\frac{3}{5}\left(\frac{5}{11.16}+\frac{5}{16.21}+...+\frac{5}{61.66}\right)\)

\(=\frac{3}{5}\left(\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+...+\frac{1}{61}-\frac{1}{66}\right)=\frac{3}{5}.\left(\frac{1}{11}-\frac{1}{66}\right)=\frac{3}{5}.\frac{5}{66}=\frac{1}{22}\)

cảm ơn bạn Xyz đã trả lời

Bài 1:

\(\frac{3^{2014}.8^{19}}{6^{60}.3^{1955}}=\frac{3^{59}.2^{57}}{2^{60}.3^{60}}=\frac{1}{24}\)

Bài 2:

\(5^x+5^{x+1}=150\\ 5^x+5^x.5=150\\ 5^x\left(1+5\right)=150\\ 5^x.6=150\\ 5^x=25=>x=2\)

Bài 3:

A = \(3\left(\frac{1}{11.16}+\frac{1}{16.21}+\frac{1}{21.26}+...+\frac{1}{61.66}\right)=\frac{5}{3}\left(\frac{5}{11.16}+\frac{5}{16.21}+...+\frac{5}{61.66}\right)=\frac{5}{3}.\frac{5}{66}=\frac{25}{198}\)

Ta có : 

\(S=\frac{5^2}{1.6}+\frac{5^2}{6.11}+\frac{5^2}{11.16}+\frac{5^2}{16.21}+\frac{5^2}{21.26}\)

\(S=5\left(\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+\frac{5}{16.21}+\frac{5}{21.26}\right)\)

\(S=5\left(\frac{1}{1}-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+\frac{1}{21}-\frac{1}{26}\right)\)

\(S=5\left(1-\frac{1}{26}\right)\)

\(S=5.\frac{25}{26}\)

\(S=\frac{125}{26}\)

Vậy \(S=\frac{125}{26}\)

Chúc bạn học tốt ~ 

S=125/26

\(\frac{5}{66}\)    nha

\(B=\frac{5}{11.13}+\frac{5}{16.21}+...+\frac{5}{61.66}\)

\(\Rightarrow B=\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+...+\frac{1}{61}-\frac{1}{66}\)

\(\Rightarrow B=\frac{1}{11}-\frac{1}{66}\)

\(\Rightarrow B=\frac{5}{66}\)

\(\frac{5}{66}\)\(nha\)\(b\text{ạn}\)

\(theo\)\(mk\)\(l\text{à}\)\(th\text{ế}\)

\(ch\text{úc}\)\(b\text{ạn}\)\(h\text{ọc}\)\(t\text{ốt}\)

^_^ !

\(A=5.\left(\frac{5}{11.16}+\frac{5}{16.21}+...+\frac{5}{56.61}\right)\))

\(A=5.\left(\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+...+\frac{1}{56}-\frac{1}{61}\right)\)

\(A=5.\left(\frac{1}{11}-\frac{1}{61}\right)\)

\(A=5.\frac{50}{671}\)

\(A=\frac{250}{671}\)

Chúc em học tốt^^

\(A=\frac{5^2}{11.16}+\frac{5^2}{16.21}+\frac{5^2}{21.26}+.....+\frac{5^2}{56.61}\)

\(=5.\left(\frac{5}{11.16}+\frac{5}{16.21}+.....+\frac{5}{56.61}\right)\)

 \(=5.\left(\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+....+\frac{1}{56}-\frac{1}{61}\right)\)

\(A=\frac{5^2}{11.16}+\frac{5^2}{16.21}+\frac{5^2}{21.26}+...+\frac{5^2}{56.61}\)

\(A=5\left(\frac{5}{11.16}+\frac{5}{16.21}+\frac{5}{21.26}+...+\frac{5}{56.61}\right)\)

\(A=5\left(\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+...+\frac{1}{56}-\frac{1}{61}\right)\)

\(A=5\left(\frac{1}{11}-\frac{1}{61}\right)\)

\(A=5.\frac{50}{671}\)

\(A=\frac{250}{671}\)

Bài 2:

a) \(\frac{5}{11.16}+\frac{5}{16.21}+...+\frac{5}{61.66}\)

\(=\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+...+\frac{1}{61}-\frac{1}{66}\)

\(=\frac{1}{11}-\frac{1}{66}\)

\(=\frac{5}{66}\)

b) \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{30}+\frac{1}{42}\)

\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\)

\(=1-\frac{1}{7}\)

\(=\frac{6}{7}\)

c) \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2006.2007}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2006}-\frac{1}{2007}\)

\(=1-\frac{1}{2007}\)

\(=\frac{2006}{2007}\)

Bài 2:

a) \(\frac{5}{11.16}\) + \(\frac{5}{16.21}\) + \(\frac{5}{21.26}\) + ... + \(\frac{5}{61.66}\)

= \(\frac{1}{11}\) - \(\frac{1}{16}\) + \(\frac{1}{16}\) - \(\frac{1}{21}\) + \(\frac{1}{21}\) - \(\frac{1}{26}\) + ... + \(\frac{1}{61}\) - \(\frac{1}{66}\)

= \(\frac{1}{11}\) - \(\frac{1}{66}\)

= \(\frac{5}{66}\)

b) \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\)

= \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\)

= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\)

= \(1-\frac{1}{7}\)

= \(\frac{6}{7}\)

c) \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{1989.1990}+...+\frac{1}{2006.2007}\)

= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{1989}-\frac{1}{1990}+...+\frac{1}{2006}-\frac{1}{2007}\)

= \(1-\frac{1}{2007}\)

= \(\frac{2006}{2007}\)

Chúc bạn học tốt!

\(A=\frac{5}{11.16}+\frac{5}{16.21}+...+\frac{5}{61.66}\)

\(=\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+...+\frac{1}{61}-\frac{1}{66}\)

\(=\frac{1}{11}-\frac{1}{66}\)

\(=\frac{5}{66}\)

Vậy \(A=\frac{5}{66}\)

\(A=\frac{5}{11.16}+\frac{5}{16.21}+...+\frac{5}{61.66}\)

\(=5.\left(\frac{1}{11.16}+\frac{1}{16.21}+...+\frac{1}{61.66}\right)\)

\(=5.\frac{1}{4}.\left(\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{24}+...+\frac{1}{61}-\frac{1}{66}\right)\)

\(=\frac{5}{4}.\left(\frac{1}{11}-\frac{1}{66}\right)\)

\(=\frac{5}{4}.\frac{5}{66}\)

\(=\frac{25}{264}\)

Mình sửa lại đề nhé :))

\(A=\frac{5}{11.16}+\frac{5}{16.21}+\frac{5}{21.26}+...+\frac{5}{61.66}\)

\(\Rightarrow A=\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+\frac{1}{21}-\frac{1}{26}+...+\frac{1}{61}-\frac{1}{66}\)

\(\Rightarrow A=\frac{1}{11}-\frac{1}{66}\)

\(\Rightarrow A=\frac{5}{66}\)

a) áp dụng dãy số cách đều đi

a, 1+6+11+16+...+46+51

Số số hạng là : (51-1):5+1 = 11 ( số )

Tổng là : (51+1).11:2=286

b, Đặt A = \(\dfrac{5^2}{1.6}+\dfrac{5^2}{6.11}+\dfrac{5^2}{11.16}+\dfrac{5^2}{16.21}+\dfrac{5^2}{21.26}+\dfrac{5^2}{26.31 } \)

\(\dfrac{1}{5}A=\) \(\dfrac{5}{1.6}+\dfrac{5}{6.11}+\dfrac{5}{11.16}+\dfrac{5}{16.21}+\dfrac{5}{21.26}+\dfrac{5}{26.31}\)

\(\dfrac{1}{5}A=\) \(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{21}+\dfrac{1}{21}-\dfrac{1}{26}+\dfrac{1}{26}-\dfrac{1}{31}\)

\(\dfrac{1}{5}A=1-\dfrac{1}{31}\)

\(\dfrac{1}{5}A=\dfrac{30}{31}\)

\(A=\dfrac{30}{31}:\dfrac{1}{5}=\dfrac{150}{31}\)

Vậy..