C\(\frac{1}{1}-\frac{1}{2.3}+\frac{1}{3.4}-\frac{1}{4.5}+\frac{1}{5.6}\)-\(\frac{1}{6.7}\)+\(\frac{1}{7.8}\)-\(\frac{1}{8.9}+\frac{1}{9.10}\)

c=\(\frac{1}{1}-\frac{1}{10}\)

c=\(\frac{9}{10}\)

còn a và b rễ lắm mình ko thích làm bài rễ đâu bạn cố chờ lời giải khác nhé!

A =(1/2 +1)×(1/3 +1)×(1/4 +1)×....×(1/99 +1)

=3/2x4/3x...............x100/99

=2-1/99

=197/99

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

A=\(\frac{\left(3\cdot4\cdot5\cdot....\cdot99\right)\cdot100}{2\cdot\left(3\cdot4\cdot5\cdot...\cdot99\right)}\)

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

\(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{97\cdot99}\)

\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\)

=> \(\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)>\(\frac{32}{100}\)=32%

Câu đầu tiên:

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

\(A=\frac{3}{2}\cdot\frac{4}{3}\cdot...\cdot\frac{100}{99}=\frac{3\cdot4\cdot5\cdot...\cdot99\cdot100}{3\cdot4\cdot5\cdot...\cdot99\cdot2}=\frac{100}{2}=50\)

Câu thứ 2:

\(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{97.99}\)

\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}=\frac{1}{3}-\frac{1}{99}=\frac{32}{99}>\frac{32}{100}\)

tính giá trị biểu thức chứ còn cái gì nữa

a, \(A=\frac{22}{27}\)

b,\(B=\frac{1}{57}\)

C,\(C=\frac{1}{50}\)

d, \(D=0\)

A = (4+\(\frac{1}{5}\)) . \(\frac{18}{19}\)+ (2+\(\frac{8}{5}\)) . \(\frac{21}{5}\)

A= \(\frac{21}{5}\).18/19 + 18/5 . 21/5

A= 21/5 (18/19 + 18/5)

A= 21/5 . 432/95

A= 9288/95

b= 25/2. (3+2/7) - 23/7. (5 + 1/2)

b= 25/2 . 23/7 - 23/7 . 11/2

b= 23/7 (25/2 -11/2)

b=23/7 . 7

b= 23

1, =\(\frac{2\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}\right)}{4\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}\right)}=\frac{1}{2}\)

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

= \(\frac{1\cdot2\cdot3\cdot....\cdot99}{2\cdot3\cdot4\cdot...\cdot100}=\frac{1}{100}\)

Vậy ......

hok tốt

\(d=\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right).........\left(1+\frac{1}{99.101}\right)\)

    \(=\frac{4}{3}.\frac{9}{2.4}.............\frac{10000}{99.101}\)

    \(=\frac{2.2}{3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}............\frac{100.100}{99.101}\)

    \(=\frac{2.3.4..........100}{2.3.4............99}.\frac{2.3.4...........100}{3.4...........101}\)

     \(=100.\frac{2}{101}\)\(=\frac{200}{101}\)

\(C=\left(1-\frac{1}{2}\right)\times\left(1-\frac{1}{3}\right)\times...\times\left(1-\frac{1}{1994}\right)\)

    \(=\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times...\times\frac{1993}{1994}\)

    \(=\frac{1\times2\times3\times...\times1993}{2\times3\times4\times...\times1994}\)

    \(=\frac{1}{1994}\)                         (Giản ước còn lại như này)