1+2=3

2+3=5

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Vẫn còn 1324 món ăn đúng ko. Hjhj !

Ta có: \(P=\left\{1-\frac{1}{2}\right\}.\left\{1-\frac{1}{3}\right\}....\left\{1-\frac{1}{99}\right\}\)

\(\Rightarrow P=\frac{1}{2}.\frac{2}{3}.........\frac{98}{99}\)\(=\frac{1.2.3...98}{2.3.4...99}=\frac{1}{99}\)

Vậy \(P=\frac{1}{99}\)

\(A=\left[\frac{5^3}{6}+\frac{5^3}{12}+\frac{5^3}{20}+\frac{5^3}{42}+\frac{5^3}{56}+\frac{5^3}{72}+\frac{5^3}{90}\right]:\frac{1124.2247-1123}{1124+1123.2247}\)

\(A=5^3\left[\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\right]:\frac{1124.2247-1123}{1124+1123.2247}\)

\(A=5^3\left[\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\right]:\frac{1124.2247-1123}{1124+1123.2247}\)

\(A=5^3\left[\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right]:\)\(\frac{1124.2247-1124+1}{1123.2247+1123+1}\)

\(A=5^3.\left(\frac{1}{2}-\frac{1}{10}\right):\frac{1224.\left(2247-1\right)+1}{1223.\left(2247+1\right)-1}\)

\(A=5^3.\frac{2}{5}:1\)

\(A=5^2.2\)

\(A=50\)

Là sao vậy Khải Nhi