\(2^{101}+4=4\left(2^{99}+1\right)=4\left[\left(2^3\right)^{33}+1\right]=4\left(8^{33}+1\right)\)

Vì \(8^{33}=\left(9-1\right)^{33}\overline{=}\left(-1\right)^{33}=-1\left(mod9\right)\)

Do đó \(8^{33}+1\overline{=}\left(-1\right)+1=0\left(mod9\right)\)Hay \(8^{33}+1⋮9\)

\(\Rightarrow4\left(8^{33}+1\right)⋮9\)\(\Rightarrow2^{101}+4⋮9\)(đpcm)

đây không phải là toán

Ta có: 

  \(A=\frac{112}{13.20}+\frac{112}{20.27}+.........+\frac{112}{62.69}\)

\(\Rightarrow A=112.\left(\frac{1}{13.20}+\frac{1}{20.27}+..........+\frac{1}{62.69}\right)\)

\(\Rightarrow A=16.\left(\frac{7}{13.20}+\frac{7}{20.27}+.......+\frac{7}{62.69}\right)\)

\(\Rightarrow A=16.\left(\frac{1}{13}-\frac{1}{20}+\frac{1}{20}-\frac{1}{27}+........+\frac{1}{62}-\frac{1}{69}\right)\)

\(\Rightarrow A=16.\left(\frac{1}{13}-\frac{1}{69}\right)\)

\(\Rightarrow A=16.\frac{56}{897}\)

\(\Rightarrow A=\frac{896}{897}\)

Vậy: \(A=\frac{896}{897}\)

\(A=\frac{112}{13.20}+\frac{112}{20.27}+\frac{112}{27.34}+...+\frac{112}{62.69}\)

\(\Rightarrow A=112.\left(\frac{1}{13.20}+\frac{1}{20.27}+\frac{1}{27.34}+...+\frac{1}{62.69}\right)\)

\(\Rightarrow A=112.\frac{7}{7}.\left(\frac{1}{13.20}+\frac{1}{20.27}+\frac{1}{27.34}+...+\frac{1}{62.69}\right)\)

\(\Rightarrow A=112.\frac{1}{7}\left(\frac{7}{13.20}+\frac{7}{20.27}+\frac{7}{27.34}+...+\frac{7}{62.69}\right)\)

\(\Rightarrow A=16\left(\frac{1}{13}-\frac{1}{20}+\frac{1}{20}-\frac{1}{27}+\frac{1}{27}-\frac{1}{34}+...+\frac{1}{62}-\frac{1}{69}\right)\)

\(\Rightarrow A=16\left(\frac{1}{13}-\frac{1}{69}\right)\)

\(\Rightarrow A=16.\frac{56}{897}\)

\(\Rightarrow A=\frac{896}{897}\)

\(Can\)\(you\) \(k\) \(for\) \(me,everyone?\)

5x².3xy²=15x³y²

hệ số là 15

15 nha