Bài 1. 

b) \(\frac{5+55+555+5555}{9+99+999+9999}\)

= \(\frac{5\left(1+11+111+1111\right)}{9\left(1+11+111+1111\right)}=\frac{5}{9}\)

c) \(39,2\cdot27+39,2\cdot43+78,4\cdot15\)

= \(39,2\cdot27+39,2\cdot43+39,2\cdot2\cdot15\)

= \(39,2\left(27+43+30\right)=39,2\cdot100=3920\)

d) \(\frac{4}{17}\cdot\frac{3}{11}+\frac{8}{11}\cdot\frac{4}{17}-\frac{4}{17}\)

= \(\frac{4}{17}\left(\frac{3}{11}+\frac{8}{11}-1\right)=\frac{4}{17}\cdot0=0\)

Bài 2.

a) \(\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+\frac{1}{9\cdot11}+...+\frac{1}{57\cdot59}\)

= \(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{57}-\frac{1}{59}\)

= \(\frac{1}{5}-\frac{1}{59}=\frac{54}{295}\)

b) \(\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}\right)-\left(\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\right)\)

= \(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}-\frac{1}{3}-\frac{1}{4}-\frac{1}{5}-\frac{1}{6}-\frac{1}{7}-\frac{1}{8}\)

= \(\frac{1}{2}-\frac{1}{8}=\frac{3}{8}\)

c) \(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)\left(1-\frac{1}{5}\right)...\left(1-\frac{1}{2012}\right)\)

= \(\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{4}{5}\cdot...\cdot\frac{2011}{2012}=\frac{1}{2012}\)

1, A=\(\frac{3}{2}.\frac{4}{3}.\frac{5}{4}....\frac{100}{99}\)

   A= \(\frac{100}{2}\)

   A=50

2, B=\(\frac{-1}{2}.\frac{-2}{3}....\frac{-98}{99}\)

     B= \(\frac{1}{99}\)

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

     \(=\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}......\frac{99}{98}\cdot\frac{100}{99}\)

     \(=\frac{100}{2}\)

       \(=50\)

\(B=\left(\frac{1}{2}-1\right)\cdot\left(\frac{1}{3}-1\right)\cdot\left(\frac{1}{4}-1\right)......\left(\frac{1}{99}-1\right)\)

     \(=\left(-\frac{1}{2}\right)\cdot\left(-\frac{2}{3}\right)\cdot\left(-\frac{3}{4}\right).....\left(-\frac{97}{98}\right)\cdot\left(-\frac{98}{99}\right)\)

       \(=-\frac{1}{99}\)

P=1/2.2/3.3/4.4/5.5/6.6/7...98/99

P=1/99

giải hẵn ra đi bạn

\(P=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)....\left(1-\frac{1}{99}\right)\)

\(P=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}....\frac{98}{99}\)

\(P=\frac{1.2.3...98}{2.3.4...99}\)

\(P=\frac{1}{99}\)

ai trả lời đúng k

có cách làm nữa nha

\(P=\left(1+\frac{1}{2}\right).\left(1+\frac{1}{3}\right).\left(1+\frac{1}{4}\right).....\left(1+\frac{1}{99}\right)\)

     \(=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}.....\frac{100}{99}\)

     \(=\frac{100}{2}\)

     =500

1+1=3 :)))

\(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1+\frac{1}{4}\right)\cdot\cdot\left(1+\frac{1}{99}\right)\)

\(=\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot\cdot\cdot\cdot\frac{100}{99}\)

\(=\frac{100}{2}=50\)

\(\left(1+\frac{1}{2}\right).\left(1+\frac{1}{3}\right).\left(1+\frac{1}{4}\right)...\left(1+\frac{1}{99}\right)\)

\(=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}...\frac{100}{99}\)

\(=\frac{3.4.5...100}{2.3.4...99}\)

\(=\frac{100}{2}\)

\(=50\)

hình như thiếu 1/98
đề đó đúng ko ( 1+ 1/2) x (1+ 1/3) x (1+ 1/4) x...x (1+ 1/98) x (1+ 1/99)

                                   Giải:

( 1+ 1/2) x (1+ 1/3) x (1+ 1/4) x...x (1+ 1/98) x (1+ 1/99)

= 3/2 x 4/3 x 5/4 x … x 99/98 x 100/99

= (3 x 4 x 5 x … x 99 x 100) / 2 x 3 x 4 x … 98 x 99

Giản ước ta được:

= 100/2

= 50