= 9/10

k nha

A= 5.(1/5.6+1/6.7+...+1/10.11)

A=5.(1/5-1/6+1/6-1/7+.....+1/10-1/11)

A=5.(1/5-1/11)

A=5.6/55=6/11

A=6/11<1

A<1

Ta có: \(A=\frac{5}{5x6}+\frac{5}{6x7}+\frac{5}{7x8}+\frac{5}{8x9}+\frac{5}{9x10}+\frac{5}{10x11}.\)

\(=5\left(\frac{1}{5x6}+\frac{1}{6x7}+\frac{1}{7x8}+...+\frac{1}{10x11}\right)\)

\(=5\left(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{10}-\frac{1}{11}\right)\)

\(=5\left(\frac{1}{5}-\frac{1}{11}\right)=5x\frac{6}{55}=\frac{6}{11}\)

a) \(\frac{2^{12}x3^5-4^6.9^2}{\left(2^2x3\right)^6+8^4x3^5}=\frac{2^{12}x3^5+\left(2^2\right)^6x\left(3^2\right)^2}{2^{12}x3^6+\left(2^3\right)^4x3^5}\)

\(=\frac{2^{12}x3^5-2^{12}x3^4}{2^{12}x3^6+2^{12}x3^5}=\frac{2^{12}x3^4x\left(3-1\right)}{2^{12}x3^5x\left(3+1\right)}\)

\(=\frac{2}{3.4}=\frac{1}{3.2}=\frac{1}{6}\)

b) \(\frac{1}{9x10}-\frac{1}{8x9}-\frac{1}{7x8}-\frac{1}{6x7}-\frac{1}{5x6}-\frac{1}{4x5}-\frac{1}{3x4}-\frac{1}{2x3}-\frac{1}{1x2}\)

\(=-\left(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+\frac{1}{5x6}+\frac{1}{6x7}+\frac{1}{7x8}+\frac{1}{8x9}+\frac{1}{9x10}\right)\)

\(=-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)

\(=-\left(1-\frac{1}{10}\right)\)

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

sorry bn nha! mk ko bk lm phần c

trả lời :

a) 1/6

b)-9/10

Đặng Thị Dung cả 2 phần bạn CÔNG CHÚA ÔRI làm sai hết rồi. Nếu muốn biết sai ở đâu thì kb rồi nhắn tin mk chỉ cho

c,\(\frac{1}{2}\div\left(-1\frac{1}{2}\right)\div1\frac{1}{3}:\left(-1\frac{1}{4}\right)\div...\div1\frac{1}{99}\div\left(-1\frac{1}{100}\right)\)

\(=\frac{1}{2}\div\left(-\frac{3}{2}\right)\div\frac{4}{3}\div\left(-\frac{5}{4}\right)\div...\div\frac{100}{99}\div\left(-\frac{101}{100}\right)\)

\(=\frac{1}{2}\times\left(-\frac{2}{3}\right)\times\frac{3}{4}\times...\times\left(-\frac{100}{101}\right)\)

=\(=\frac{1.\left(-2\right).3.....\left(-100\right)}{2.3.4.....101}=\frac{1}{101}\)

\(A=\frac{5}{1.2}+\frac{5}{2.3}+...+\frac{5}{7.8}\)

\(\Rightarrow5A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{7.8}\)

\(\Rightarrow5A=1.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...-\frac{1}{8}\right)\)

\(\Rightarrow5A=1-\frac{1}{8}\)

\(\Rightarrow A=\left(1-\frac{1}{8}\right).\frac{1}{5}=\frac{7}{40}\)

\(A=\frac{5}{1.2}+\frac{5}{2.3}+...+\frac{5}{7.8}\)

\(A=5\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{5}{7.8}\right)\)

\(A=5\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{8}\right)\)

\(A=5\left(1-\frac{1}{8}\right)\)

\(A=5.\frac{7}{8}\)

\(A=\frac{38}{8}\)

nhầm rồi :)

\(A=5.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-....-\frac{1}{8}\right)\)

\(\Rightarrow A=\frac{5.7}{8}\)

\(\Rightarrow A=\frac{35}{8}\)

Mình không thể giải thích được nhưng kết quả chắc chắn là : \(\frac{8}{9}\)

đặt A=1/1×2+1/2×3+1/3×4+

bn đặt ra đi sau đó rồi tính vd 1/1x2=1/1-1/2

\(\frac{1}{6x7}+\frac{1}{7x8}+\frac{1}{8x9}+\frac{1}{9x10}+\frac{1}{10x11}\)

\(=\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\)

\(=\frac{1}{6}-\frac{1}{11}=\frac{5}{66}\)

\(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{13.15}+\frac{2}{1.2}+\frac{2}{2.3}+...+\frac{2}{9.10}\)

\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}+2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)

\(=\frac{1}{3}-\frac{1}{15}+2\left(1-\frac{1}{10}\right)\)

\(=\frac{4}{15}+\frac{9}{5}\)

\(=\frac{31}{15}\)

              Bài làm :

Ta có :

\(\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{13\times15}+\frac{2}{1\times2}+\frac{2}{2\times3}+...+\frac{2}{9\times10}\)

\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}+2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)

\(=\frac{1}{3}-\frac{1}{15}+2\left(1-\frac{1}{10}\right)\)

\(=\frac{31}{15}\)