câu 2:

A = 1/1*2 + 1/2*3  + 1/3*4 + ... + 1/98*99 + 1/99*100

A = 2-1/1*2 + 3-2/2*3 + 4-3/3*4 + ... + 99-98/98*99 + 100-99/99*100

A = 2/1*2 - 1/ 1*2 + 3/2*3 - 2/2*3 + 4/ 3*4 -3/3*4 +...+ 99/98*99 - 98/98*99 + 100/99*100 - 99/99*100

A = 1 - 1/ 100

A = 99 / 100

phần 2 mk ko =bít

bài 1, 3 mk ko bít

a) x+ \(\frac{4}{5}\)= \(\frac{4}{5}\)+ ( \(\frac{3}{7}\)+ \(\frac{3}{5}\))

    x + \(\frac{4}{5}\)=  \(\frac{4}{5}\)+ \(\frac{36}{35}\)

    => x = \(\frac{36}{35}\)

Vậy : x = \(\frac{36}{35}\)

b) \(\frac{4}{9}\)+\(\frac{8}{9}\)+ \(\frac{12}{9}\)+ ... + \(\frac{56}{9}\)

    = \(\frac{4+8+12+...+56}{9}\)

    Ta có : 4+8+12 + ... +56

     Số các số hạng của dẫy số trên là : ( 56 -4) + 4 +1 = 14 ( số hạng )

      => Tổng trên = ( 4+56) x 14 : 2 = 420

      => \(\frac{4}{9}\)+ \(\frac{8}{9}\)+ ... + \(\frac{56}{9}\)= \(\frac{420}{9}\)= \(\frac{140}{3}\)

a)x=3/7+3/5=3x(1/7+1/5)=36/35

b)=(4+8+...+56)/9=(56+4)x14/9=280/3
 

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

\(A=\frac{3}{2}\times\frac{4}{3}\times\frac{5}{4}\times......\times\frac{2016}{2017}\times\frac{2018}{2017}\)

\(A=\frac{2018}{2}=1009\)

\(B=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+.......+\frac{2}{43.45}\)

\(B=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-......+\frac{1}{43}-\frac{1}{45}\)

\(B=\frac{1}{3}-\frac{1}{45}\)

\(B=\frac{14}{45}\)

2     \(\frac{2017}{2018}\times\frac{23}{47}+\frac{24}{2018}\times\frac{2017}{47}\)

\(=\frac{2017}{2018}\times\frac{23}{47}+\frac{24}{47}\times\frac{2017}{2018}\)

\(=\frac{2017}{2018}\times\left(\frac{23}{47}+\frac{24}{47}\right)\)

\(=\frac{2017}{2018}\times1\)

=\(\frac{2017}{2018}\)

bạn nào xem giải thế có đúng ko

= 3/2 + 4/3 + 5/4  ................................ 100/99

= 100/2 = 50

a)\(\dfrac{5}{7}+\dfrac{4}{9}=\dfrac{45}{63}+\dfrac{28}{63}=\dfrac{73}{63}\) ; \(\dfrac{9}{11}+\dfrac{3}{8}=\dfrac{72}{88}+\dfrac{33}{88}=\dfrac{105}{88}\)

\(\dfrac{4}{5}-\dfrac{2}{3}=\dfrac{12}{15}-\dfrac{10}{15}=\dfrac{2}{15}\); \(\dfrac{16}{25}-\dfrac{2}{5}=\dfrac{16}{25}-\dfrac{10}{25}=\dfrac{6}{25}\)

b)\(5+\dfrac{3}{5}=\dfrac{25}{5}+\dfrac{3}{5}=\dfrac{28}{5};10-\dfrac{9}{16}=\dfrac{160}{16}-\dfrac{9}{16}=\dfrac{151}{16}\)

\(\dfrac{2}{3}-\left(\dfrac{1}{6}+\dfrac{1}{8}\right)=\dfrac{2}{3}-\left(\dfrac{8}{48}+\dfrac{6}{48}\right)=\dfrac{2}{3}-\dfrac{14}{48}=\dfrac{32}{48}-\dfrac{14}{48}=\dfrac{3}{8}\)

\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right).....\left(1-\frac{1}{100}\right).200x=4036\)

\(\Leftrightarrow\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{99}{100}.200x=4036\)

\(\Leftrightarrow\frac{1.2.3...99}{2.3.4....100}.200x=4036\)

\(\Leftrightarrow\frac{1}{100}.200x=4036\)

\(\Leftrightarrow\frac{1}{100}.200x=4036\)

\(\Leftrightarrow2x=4036\)

\(\Leftrightarrow x=4036:2=2018\)

\(\left(1-\frac{1}{2}\right)\times\left(1-\frac{1}{3}\right)\times...\times\left(1-\frac{1}{100}\right)\times200\times x=4036\)

=> \(\frac{1}{2}\times\frac{2}{3}\times...\times\frac{99}{100}\times200\times x=4036\)

=> \(\frac{1\times2\times...\times99}{2\times3\times...\times100}\times200\times x=4036\)

\(\Rightarrow\frac{1}{100}\times200\times x=4036\)

\(\Rightarrow2\times x=4036\)

=> x = 2018 

\(a,\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)

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

\(b,\left(1-\frac{1}{2}\right)\times\left(1-\frac{1}{3}\right)\times\left(1-\frac{1}{4}\right)=\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\)

\(=\frac{1\times2\times3}{2\times3\times4}=\frac{1}{4}\)