= 9/10

k nha

\(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+...+\frac{1}{8x9}\)

=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}\)

=\(1-\frac{1}{9}\)

=\(\frac{8}{9}\)

OK XONG NHỚ CHO MIK NHA

\(\frac{1}{1\times2}+\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+.......+\frac{1}{7x8}+\)\(\frac{1}{8x9}\)

=1-\(\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{8}-\frac{1}{9}\)

=1-\(\frac{1}{9}\)

=\(\frac{8}{9}\)

a) \(\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\)

\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)

\(=\frac{1}{5}-\frac{1}{10}\)

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

b) \(\frac{2}{10.12}+\frac{2}{12.14}+\frac{2}{14.16}+...+\frac{2}{998.1000}\)

\(=\frac{1}{10}-\frac{1}{12}+\frac{1}{12}-\frac{1}{14}+\frac{1}{14}-\frac{1}{16}+...+\frac{1}{998}-\frac{1}{1000}\)

\(=\frac{1}{10}-\frac{1}{1000}\)

\(=\frac{99}{1000}\)

c) \(\frac{4}{1.2}+\frac{4}{2.3}+\frac{4}{3.4}+...+\frac{4}{69.90}\)

\(=4.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{89.90}\right)\)

\(=4.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{89}-\frac{1}{90}\right)\)

\(=4.\left(1-\frac{1}{90}\right)\)

\(=4.\frac{89}{90}\)

\(=\frac{178}{45}\)

_Chúc bạn học tốt_

a, \(=\frac{1}{10}\)

\(6xy+\left(\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+...+\frac{1}{7x8}\right)=\frac{29}{8}\)

Đăt \(A=\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+...+\frac{1}{7x8}\)

\(\Rightarrow A=\frac{3-2}{2x3}+\frac{4-3}{3x4}+\frac{5-4}{4x5}+...+\frac{8-7}{7x8}\)

\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{7}-\frac{1}{8}=\frac{1}{2}-\frac{1}{8}=\frac{3}{8}\)

\(\Rightarrow6xy+A=6xy+\frac{3}{8}=\frac{29}{8}\Rightarrow6xy=\frac{26}{8}\Rightarrow y=\frac{26}{8x6}\)

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

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+

A = \(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+...+\frac{1}{8x9}\)

A = \(\frac{1}{1}-\frac{1}{9}=\frac{9}{9}-\frac{1}{9}=\frac{8}{9}\)

Mk đầu tiên nha

\(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)

\(=\frac{1}{2}-\frac{1}{10}\)

\(=\frac{2}{5}\)

\(\frac{1}{1x2}+\frac{1}{2x3}+...+\frac{1}{9x10}\)

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

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

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

1/2x3+1/3x4+......+1/38x39+1/39x40

=1/2-1/3+1/3-1/4+.....+1/39-1/40

=1/2-1/40

=19/40