\(\left(-2\frac{3}{4}+\frac{1}{2}\right)^2\)

\(=\left(-\frac{11}{4}+\frac{1}{2}\right)^2\)

     \(=\left(-\frac{9}{4}\right)^2\)

     \(=\frac{81}{16}\)

tíc mình nha

\(\left(-2\frac{3}{4}+\frac{1}{2}\right)^2=\left(\frac{-11}{4}+\frac{2}{4}\right)^2=\left(\frac{-9}{4}\right)^2=\frac{-81}{16}\)

\(K=\frac{-3}{4}.\frac{-8}{9}.\frac{-15}{16}...\frac{-9999}{10000}=\left(-1\right)^{99}.\frac{1.3.2.4...99.101}{2.2.3.3.4.4...100.100}=-\frac{1.2...99}{2.3...100}.\frac{3.4...101}{2.3...100}=-\frac{1}{100}.\frac{101}{2}=-\frac{101}{200}< -\frac{100}{200}=-\frac{1}{2}\)

Đặt A = \(\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-\frac{1}{16}-\)...\(-\frac{1}{1024}\)

A= \(\frac{1}{2^1}-\frac{1}{2^2}-\frac{1}{2^3}-\frac{1}{2^4}-\)....\(-\frac{1}{2^{10}}\)

2A=\(\frac{1}{1}\)\(-\frac{1}{2^1}-\frac{1}{2^2}-\frac{1}{2^3}-\)...\(-\frac{1}{2^9}\)

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

A=\(1+\frac{1}{2^{10}}\)

A= \(\frac{1025}{1024}\)

Mình làm như thế này nek

\(\frac{\frac{1}{4}+\frac{3}{7}-\frac{4}{5}}{0,75+\frac{9}{7}-2\frac{2}{5}}+\frac{\frac{3}{14}-\frac{2}{10}+\frac{5}{18}+\frac{7}{66}}{\frac{6}{7}-\frac{4}{5}+\frac{10}{9}+\frac{14}{33}}\)

\(=\frac{\frac{1}{4}+\frac{3}{7}-\frac{4}{5}}{\frac{2}{4}+\frac{9}{7}-\frac{12}{5}}+\frac{\frac{1}{2}\cdot\left(\frac{3}{7}-\frac{2}{5}+\frac{5}{9}+\frac{7}{33}\right)}{2\cdot\left(\frac{3}{7}-\frac{2}{5}+\frac{5}{9}+\frac{7}{33}\right)}\)

\(=\frac{\frac{1}{4}+\frac{3}{7}-\frac{4}{5}}{3\cdot\left(\frac{1}{4}+\frac{3}{7}-\frac{4}{5}\right)}+\frac{\frac{1}{2}}{2}\)

\(=\frac{1}{3}+\frac{1}{4}=\frac{7}{12}\)