-25*72+25*21-49*25

=(-25).(72-21+49)

=(-25).100

=(-2500)

bn nhớ k nha

\(A=1+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2020}}\)

\(2A=2+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2019}}\)

\(\Rightarrow A=2A-A=2+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2019}}-\left(1+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2020}}\right)\)

\(=2+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2019}}-1-\frac{1}{2^2}-\frac{1}{2^3}-...-\frac{1}{2^{2020}}\)

\(=1+\frac{1}{2}-\frac{1}{2^{2020}}=\frac{2^{2020}+2^{2019}-1}{2^{2020}}\)

\(A=\frac{7}{4}.\)\(\left(\frac{3333}{1212}+\frac{3333}{2020}+\frac{3333}{3030}+\frac{3333}{4242}\right)\)

\(A=\frac{7}{4}.\)\(\left[3333.\left(\frac{1}{1212}+\frac{1}{2020}+\frac{1}{3030}+\frac{1}{4242}\right)\right]\)

\(A=\frac{7}{4}.\)\(\left[3333.\left(\frac{1}{12.101}+\frac{1}{20.101}+\frac{1}{30.101}+\frac{1}{42.101}\right)\right]\)

\(A=\frac{7}{4}.\)\(\left[3333.\frac{1}{101}\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)\right]\)

\(A=\frac{7}{4}.\)\(33.\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\)

\(A=33.\left[\frac{7}{4}.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\right]\)

\(A=33.\left[\frac{7}{4}.\left(\frac{1}{3}-\frac{1}{7}\right)\right]\)

\(A=33.\left(\frac{7}{4}.\frac{4}{21}\right)\)

\(A=33.\frac{1}{3}\)

\(A=11\)

 Vậy \(A=11\)

a) \(\frac{7}{12}+\frac{x}{15}=\frac{1}{20}\)

=> \(\frac{35}{60}+\frac{4x}{60}=\frac{3}{60}\)

=> 35 + 4x = 3

=> 4x = -32

=> x = -8

b) \(\frac{-7}{x}+\frac{8}{15}=\frac{-1}{20}\)(ĐK \(x\ne0\))

=> \(\frac{-7}{x}=\frac{-1}{20}-\frac{8}{15}\)

=> \(\frac{-7}{x}=\frac{-3}{60}-\frac{32}{60}\)

=> \(\frac{-7}{x}=\frac{-35}{60}\)

=> \(\frac{-7}{x}=\frac{-7}{12}\)

=> x = 12(TM)

có ai cjoiw Phai Phai ko