Từ đề bài suy ra: A=3/4 x 8/9 x ...x 9800/9801 x 9999/10000

                       =>A=<1x3/2x2> x <2x4/3x3> x ... x <99x101/100x100>

                       =>A=(1x2x...x99)/(2x3x...x100) x (3x4x...x101)/(2x3x...x100) 

                       =>A=1/100 x 101/2 = 101/200

BẠn hãy tính ra rồi phân tích tử và mẫu là ra

a)

\(\begin{array}{l}\frac{1}{9} - 0,3.\frac{5}{9} + \frac{1}{3}\\ = \frac{1}{9} - \frac{3}{{10}}.\frac{5}{9} + \frac{1}{3}\\ = \frac{1}{9} - \frac{3}{{2.5}}.\frac{5}{{3.3}} + \frac{1}{3}\\ = \frac{1}{9} - \frac{1}{6} + \frac{1}{3}\\ = \frac{2}{{18}} - \frac{3}{{18}} + \frac{6}{{18}}\\ = \frac{5}{{18}}\end{array}\)

b)

\(\begin{array}{l}{\left( {\frac{{ - 2}}{3}} \right)^2} + \frac{1}{6} - {\left( { - 0,5} \right)^3}\\ = \frac{4}{9} + \frac{1}{6} - \left( {\frac{{ - 1}}{2}} \right)^3\\  = \frac{4}{9} + \frac{1}{6} - \left( {\frac{{ - 1}}{8}} \right)\\ = \frac{4}{9} + \frac{1}{6} + \frac{1}{8}\\ = \frac{{32}}{{72}} + \frac{{12}}{{72}} + \frac{9}{{72}}\\ = \frac{{53}}{{72}}\end{array}\)

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

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

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

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

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

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

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

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

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

mình làm được nhưng đánh lâu lắm

x. (x^2)^3 = x^5 
x^7 ≠ x^5 
Nếu, 
x^7 - x^5 = 0 
mủ lẻ nên phương trình có 3 nghiệm 
Đáp số: 
x = -1 
hoặc 
x = 0 
hoặc 
x = 1 

a, \(\left(1-\frac{1}{4}\right)\cdot\left(1-\frac{1}{9}\right)\cdot\left(1-\frac{1}{16}\right)\cdot\left(1-\frac{1}{25}\right)\cdot\left(1-\frac{1}{36}\right)\)

\(=\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}\cdot\frac{24}{25}\cdot\frac{35}{36}\)

\(=\frac{1.3}{2.2}\cdot\frac{2.4}{3.3}\cdot\frac{3.5}{4.4}\cdot\frac{4.6}{5.5}\cdot\frac{5.7}{6.6}\)

\(=\frac{1.2.3.4.5}{2.3.4.5.6}\cdot\frac{3.4.5.6.7}{2.3.4.5.6}=\frac{1}{6}\cdot\frac{7}{2}\)

\(=\frac{7}{12}\)

b, \(\left(2-\frac{3}{2}\right)\cdot\left(2-\frac{4}{3}\right)\cdot\left(2-\frac{5}{4}\right)\cdot\left(2-\frac{6}{5}\right)\)

\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{4}{5}=\frac{1.2.3.4}{2.3.4.5}\)

\(=\frac{1}{5}\)

bạn tự chép lại đề nha

=(1-1/4).(1-1/9)(1-1/16)........(1-1/9801)

=3/4.8/9.15/16......9800/9801

=1.3.2.4.3.5........98.100/2.2.3.3.4.4.......99.99

=1.2.3.4......98/2.3........99      .     3.4.5........100/2.3.......99

rút gọn,ta có :1/99 .100/2=50/99

chổ nào bạn ko hiểu có thể hỏi minh

\(1.\)\(M=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{42}\)

\(M=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{6.7}\)

\(M=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{6}-\frac{1}{7}\)

\(M=1-\frac{1}{7}=\frac{6}{7}\)

Mình làm câu 1 thoi nha!

1.

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

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

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

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

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