\(=\frac{1}{1.3}.\frac{1}{2.4}...\frac{1}{9.11}=\frac{1}{1.2.3^2...9^2.10.11}\)

xí nha!đây ko phải toán lớp 5 nha bạn !

cô giáo mình ra đề như vậy mà

A={x|x=3ⁿ;n\(\in\)N;\(1\le n\le4\)}
B={x|x=(2k)²;k\(\in\)N;\(1\le k\le5\)}
C={x|x=\(\frac{1}{\frac{1}{2}n^3-\frac{5}{2}n^2+7n-3}\);\(n\in N\);\(1\le n\le4\)}
Cái C tui làm bừa đấy

Làm lại đề cho:

\(\left(\frac{1}{4}-1\right)\cdot\left(\frac{1}{9}-1\right)\cdot\left(\frac{1}{16}-1\right)...\left(\frac{1}{81}-1\right)\cdot\left(\frac{1}{100}-1\right)\)

Tính nhẩm

TOI KO BIET

\(S=\left(\frac{1}{4}-1\right).\left(\frac{1}{9}-1\right).\left(\frac{1}{16}-1\right)...\left(\frac{1}{81}-1\right).\left(\frac{1}{100}-1\right)\)

\(S=\frac{-3}{4}.\frac{-8}{9}.\frac{-15}{16}........\frac{-80}{81}.\frac{-99}{100}\)

\(-S=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}......\frac{80}{81}.\frac{99}{100}\)

\(-S=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}........\frac{8.10}{9.9}.\frac{9.11}{10.10}\)

\(-S=\frac{1.3.2.4.3.5........8.10.9.11}{2.2.3.3.4.4.......9.9.10.10}\)

\(-S=\frac{\left(1.2.3......8.9\right).\left(3.4.5.......10.11\right)}{\left(2.3.4.......9.10\right).\left(2.3.4........9.10\right)}\)\(=\frac{1}{10}.\frac{11}{2}=\frac{11}{20}=>S=\frac{-11}{20}\)

sao câu này thầy ko chọn ạ

Ngân Hoàng Xuân: bị 2 lần rồi, ức chế v~~~

\(\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)\left(\frac{1}{16}-1\right)....\left(\frac{1}{81}-1\right)\left(\frac{1}{100}-1\right)\)

\(=\frac{-3}{4}.\frac{-8}{9}.\frac{-15}{16}....\frac{-80}{81}.\frac{-99}{100}\)

\(=\left[\left(-1\right).\left(-1\right)...\left(-1\right)\left(9\text{số (-1)}\right)\right].\frac{3}{4}.\frac{8}{9}....\frac{99}{100}\)

\(=\left(-1\right).\frac{1.3}{2.2}.\frac{2.4}{3.3}....\frac{9.11}{10.10}\)

\(=-\frac{1.11}{2.10}=-\frac{11}{10}\)

Cảm ơn bạn nha !