giúp mình với

Mình giải theo lớp 6

  \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2016.2017}\)

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

  Ta loại các cặp số đổi của nhau như : \(-\frac{1}{2}\)và \(\frac{1}{2}\)thì còn

\(\frac{1}{1}-\frac{1}{2017}\)

\(=\frac{2016}{2017}\)

=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+.......+1/2016-1/2017

=1-1/2017

=2016/2017

xong rồi bạn ạ

=1/1-1/2+1/2-1/3+...+1/2016-1/2017

= 1/1-1/2017

=2016/2017

k cho mk nhé mk ít điểm lắm . mk học lớp 5 rồi đúng 100 % đấy

\(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+...+\dfrac{1}{\left(x-1\right)\times x}=\dfrac{15}{16}\)

\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{x-1}-\dfrac{1}{x}=\dfrac{15}{16}\)

\(1-\dfrac{1}{x}=\dfrac{15}{16}\)

\(\dfrac{1}{x}=1-\dfrac{15}{16}=\dfrac{16}{16}-\dfrac{15}{16}\)

\(\dfrac{1}{x}=\dfrac{1}{16}\)

\(\Rightarrow x=16\)

?

A=1²/1.2 .2²/2.3 .3²/3.4 .4²/4.5

=1/2. 2.2/2.3 .3.3/3.4 .4.4/4.5

=1/2.2/3.3.4.4./5

=1/5

\(\frac{1313}{1212}:x=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5}\)

\(\frac{1313}{1212}:x=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}\)

\(\frac{1313}{1212}:x=\frac{1}{1}-\frac{1}{5}+\frac{1}{5}\)

\(\frac{1313}{1212}:x=\frac{4}{5}+\frac{1}{5}\)

\(\frac{1313}{1212}:x=1\)

\(x=\frac{1313}{1212}:1\)

\(x=\frac{13}{12}\)

Lời giải 

\(\frac{1313}{1212}:x=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}\)

\(\frac{1313}{1212}:x=\frac{1}{1}-\frac{1}{5}+\frac{1}{5}\)

\(\frac{1313}{1212}:x=\frac{4}{5}+\frac{1}{5}\)

\(x=\frac{1313}{1212}:1\)

\(x=\frac{13}{12}\)

\(\frac{1313}{1212}\)\(\div x\)\(=\)\(\frac{1}{1}\)\(-\)\(\frac{1}{2}\)\(+\)\(\frac{1}{2}\)\(-\)\(\frac{1}{3}\)\(+\)\(\frac{1}{3}\)\(-\)\(\frac{1}{4}\)\(+\)\(\frac{1}{4}\)\(-\)\(\frac{1}{5}\)\(+\)\(\frac{1}{5}\)

\(\frac{1313}{1212}\)\(\div x\)\(=\)\(\frac{1}{1}\)\(-\)\(\frac{1}{5}\)\(+\)\(\frac{1}{5}\)

\(\frac{1313}{1212}\)\(\div x\)\(=\)\(\frac{4}{5}\)\(+\)\(\frac{1}{5}\)

\(\frac{1313}{1212}\)\(\div x\)\(=\)\(1\)

\(x=\)\(\frac{1313}{1212}\)\(\div1\)

\(x=\)\(\frac{13}{12}\)

1/1x2 + 1/2x3 + 1/3x4 + 1/4x5 + 1/5x6

Đặt A = 1/1x2 + 1/2x3 + 1/3x4 + 1/4x5 + 1/5x6

A = 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6

A = 1/1 - 1/6

A = 5/6

Vậy A = 5/6

C=2+3/2+4/3+5/4+6/5

C=437/60

\(C=\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+\frac{1}{5x6}\)

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

\(C=1-\frac{1}{6}\)

\(C=\frac{5}{6}\)

0 nha bn

Tận cùng các số của từng nhóm : 

        2 + 6 + 2 + 0 + 0 + 2 + 6 + 2 + 0 + 0.....

Vậy cứ 5 nhóm thì tận cùng là :

        2 + 6 + 2 = 10 ( tận cùng là 0 )

Có số nhóm 2016

Vậy :     2016 : 5 =  dư 1

Vậy tận cùng là :

        0 + 2 = 2

            Đ/s : 2

Hok tốt !

1-2+2-3+3-4+4-5+.....+2015-2016=

2016-1=2015

chú ý những số ko có trong dòng thứ hai thì phải gạch chéo nha.

\(=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{5.6}\)

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

\(=1-\frac{1}{6}\)

\(=\frac{5}{6}\)

\(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+\frac{1}{5x6}\)

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

=> 1-\(\frac{1}{6}\)

=\(\frac{6}{6}-\frac{1}{6}=\frac{6}{6}+\frac{-1}{6}=\frac{5}{6}\)

Bài 1:

Đặt \(A=\frac{2}{1x2}+\frac{2}{2x3}+\frac{2}{3x4}+...+\frac{2}{18x19}+\frac{2}{19x20}\)

\(\frac{A}{2}=\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+...+\frac{1}{18x19}+\frac{1}{19x20}\)

\(\frac{A}{2}=\frac{2-1}{1x2}+\frac{3-2}{2x3}+\frac{4-3}{3x4}+...+\frac{19-18}{18x19}+\frac{20-19}{19x20}\)

\(\frac{A}{2}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}=1-\frac{1}{20}=\frac{19}{20}\)

\(A=\frac{2x19}{20}=\frac{19}{10}\)

Bài 2:

Đặt \(B=\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+...+\frac{1}{8x9}+\frac{1}{9x10}\)

Làm tương tự câu 1 có \(B=1-\frac{1}{10}=\frac{9}{10}\)

\(Bx100=\frac{9}{10}x100=90\)

=> \(\left[\frac{5}{2}:\left(x+\frac{206}{100}\right)\right]:\frac{1}{2}=1\)

=> \(\left[\frac{5}{2}:\left(x+\frac{206}{100}\right)\right]=\frac{1}{2}\)

=>  \(x+\frac{206}{100}=\frac{5}{2}:\frac{1}{2}=5\Rightarrow x=5-\frac{206}{100}=\frac{294}{100}=\frac{147}{50}\)

bài 1 đáp án là:19/10

2:147/50

đáp án bài 1 là: 19/10

đáp án bài 2 là 147/50