=\(\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}...\cdot\frac{899}{900}=\frac{1.3}{2.2}\cdot\frac{2.4}{3.3}\cdot\frac{3.5}{4.4}...\cdot\frac{29\cdot31}{30\cdot30}=\frac{1.2.3.4...29\cdot3.4.5...30.31}{2.2.3.3.4.4...30.30}=\frac{1.31}{2.30}=\frac{31}{60}\)

\(A=\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}....\frac{899}{30^2}=\frac{\left(1.3\right).\left(2.4\right).\left(3.5\right)...\left(29.31\right)}{\left(2.3.4...30\right).\left(2.3.4...30\right)}=\frac{\left(1.2....29\right).\left(3.4.5...31\right)}{\left(2.3.4...30\right).\left(2.3.4..30\right)}=\frac{1.31}{30.2}=\frac{31}{60}\)

= \(\frac{3}{4}.\frac{8}{9}.\frac{15}{16}....\frac{899}{900}\)

= \(\frac{3}{2.2}.\frac{8}{3.3}.\frac{15}{4.4}.....\frac{899}{9.100}\)

= (bạn tự giải tiếp nhé)

\(1-\dfrac{1}{n^2}=\dfrac{n^2-1}{n^2}=\dfrac{\left(n-1\right)\left(n+1\right)}{n^2}\)

Do đó:

\(M=\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\left(1-\dfrac{1}{4^2}\right)...\left(1-\dfrac{1}{30^2}\right)\)

\(=\dfrac{\left(2-1\right)\left(2+1\right)}{2^2}.\dfrac{\left(3-1\right)\left(3+1\right)}{3^2}.\dfrac{\left(4-1\right)\left(4+1\right)}{4^2}...\dfrac{\left(30-1\right)\left(30+1\right)}{30^2}\)

\(=\dfrac{1.3}{2^2}.\dfrac{2.4}{3^2}.\dfrac{3.5}{4^2}...\dfrac{29.31}{30^2}=\dfrac{1.2.3...29}{2.3.4...30}.\dfrac{3.4.5...31}{2.3.4...30}\)

\(=\dfrac{1}{30}.\dfrac{31}{2}=\dfrac{31}{60}\)

a,

\(A=\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)...\left(1-\frac{1}{900}\right)\\ =\left(1-\frac{1}{2}\right)\left(1+\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1+\frac{1}{3}\right)...\left(1-\frac{1}{30}\right)\left(1+\frac{1}{30}\right)\\ =\frac{1}{2}\cdot\frac{3}{2}\cdot\frac{2}{3}\cdot\frac{4}{3}\cdot...\cdot\frac{29}{30}\cdot\frac{31}{30}\\ =\frac{1}{2}\cdot\frac{2}{3}\cdot...\cdot\frac{29}{30}\cdot\frac{3}{2}\cdot\frac{4}{3}\cdot...\cdot\frac{31}{30}\\ =\frac{1\cdot2\cdot...\cdot29}{2\cdot3\cdot...\cdot30}\cdot\frac{3\cdot4\cdot...\cdot31}{2\cdot3\cdot...\cdot30}\\ =\frac{1}{30}\cdot\frac{31}{2}=\frac{31}{60}\)

b,

\(B=\frac{1}{2}\left(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+...+\frac{2}{98\cdot99\cdot100}\right)\\ =\frac{1}{2}\left(\frac{3-1}{1\cdot2\cdot3}+\frac{4-2}{2\cdot3\cdot4}+...+\frac{100-98}{98\cdot99\cdot100}\right)\\ =\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{98\cdot99}-\frac{1}{99\cdot100}\right)\\ =\frac{1}{2}\left(\frac{1}{2}-\frac{1}{9900}\right)\\ =\frac{1}{2}\cdot\frac{4450-1}{9900}=\frac{1}{2}\cdot\frac{4449}{9900}=\frac{4449}{19800}=\frac{1483}{6600}\)

c, (Chịu :V)

d,

\(D=\frac{1}{3}\left(\frac{3}{1\cdot2\cdot3\cdot4}+\frac{3}{2\cdot3\cdot4\cdot5}+...+\frac{3}{27\cdot28\cdot29\cdot30}\right)\\ =\frac{1}{3}\left(\frac{4-1}{1\cdot2\cdot3\cdot4}+\frac{5-2}{2\cdot3\cdot4\cdot5}+...+\frac{30-27}{27\cdot28\cdot29\cdot30}\right)\\ =\frac{1}{3}\left(\frac{1}{1\cdot2\cdot3}-\frac{1}{2\cdot3\cdot4}+\frac{1}{2\cdot3\cdot4}-\frac{1}{3\cdot4\cdot5}+...+\frac{1}{27\cdot28\cdot29}-\frac{1}{28\cdot29\cdot30}\right)\\ =\frac{1}{3}\left(\frac{1}{6}-\frac{1}{24630}\right)\\ =\frac{228}{4105}\)

Chúc bạn học tốt nha.

\(A=\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)...\left(1-\frac{1}{900}\right)\)

\(A=\frac{3}{4}\cdot\frac{8}{9}\cdot...\cdot\frac{899}{900}\)

\(A=\frac{\left(1\cdot3\right)\left(2\cdot4\right)...\left(29\cdot31\right)}{\left(2\cdot2\right)\left(3\cdot3\right)...\left(30\cdot30\right)}\)

\(A=\frac{\left(1\cdot2\cdot..\cdot29\right)\left(3\cdot4\cdot...\cdot31\right)}{\left(2\cdot3\cdot...\cdot30\right)\left(2\cdot3\cdot...\cdot30\right)}\)

\(A=\frac{1\cdot31}{30\cdot2}\)

\(A=\frac{31}{60}\)

\(\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)\left(\frac{1}{16}-1\right)...\left(\frac{1}{900}-1\right)\)

\(=\left(\frac{1}{2}-1\right)\left(\frac{1}{2}+1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{3}+1\right)\left(\frac{1}{4}-1\right)\left(\frac{1}{4}+1\right)...\left(\frac{1}{30}-1\right)\left(\frac{1}{30}+1\right)\)

\(=\frac{-1}{2}.\frac{3}{2}.\frac{-2}{3}.\frac{4}{3}.\frac{-3}{4}.\frac{5}{4}...\frac{-29}{30}.\frac{31}{30}=-\frac{31}{60}\)

$\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)\left(\frac{1}{16}-1\right)...\left(\frac{1}{900}-1\right)$

$=\left(\frac{1}{2}-1\right)\left(\frac{1}{2}+1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{3}+1\right)\left(\frac{1}{4}-1\right)\left(\frac{1}{4}+1\right)...\left(\frac{1}{30}-1\right)\left(\frac{1}{30}+1\right)$

$=\frac{-1}{2}.\frac{3}{2}.\frac{-2}{3}.\frac{4}{3}.\frac{-3}{4}.\frac{5}{4}...\frac{-29}{30}.\frac{31}{30}=-\frac{31}{60}$

3. \(M=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{10.11.12}\)

\(\Leftrightarrow2M=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{10.11.12}\)

\(\Leftrightarrow2M=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{10.11}-\frac{1}{11.12}\)

\(\Leftrightarrow2M=\frac{1}{1.2}-\frac{1}{11.12}\)

\(\Leftrightarrow2M=\frac{1}{2}-\frac{1}{132}\)

\(\Leftrightarrow2M=\frac{65}{132}\)

\(\Leftrightarrow M=\frac{65}{132}\div2\)

\(\Leftrightarrow M=\frac{65}{264}\)

1\(A=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}...\frac{899}{900}\)

\(\Leftrightarrow A=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{29.31}{30.30}\)

\(\Leftrightarrow A=\frac{1.3.2.4.3.5...29.31}{2.2.3.3.4.4...30.30}\)

\(\Leftrightarrow A=\frac{\left(1.2.3....29\right)\left(3.4.5...31\right)}{\left(2.3.4...30\right)\left(2.3.4...30\right)}\)

\(\Leftrightarrow A=\frac{1.31}{30.2}\)

\(\Leftrightarrow A=\frac{31}{60}\)

2. Đặt \(A=\frac{1}{5}+\frac{1}{6}+...+\frac{1}{17}\)

\(\Rightarrow A< \frac{1}{5}+\frac{1}{5}+...+\frac{1}{5}+\frac{1}{8}+\frac{1}{8}+...+\frac{1}{8}\)

\(\Rightarrow A< 1+1=2\)

Vậy a < 2 (đpcm)

Tìm y:

-y:1/2-5/2=4+1/2

-y:1/2 = 4+1/2+5/2

-y:1/2 = 7

-y = 7.2

y = -14

Vậy y = -14