Đặt \(B=\frac{2}{3}.\frac{4}{5}.\frac{6}{7}.\frac{8}{9}....\frac{100}{101}\)

Nhận xét: Nếu \(\frac{a}{b}<1\) thì \(\frac{a+m}{b+m}<1\) với số m> 0 bất kì

=> \(\frac{1}{2}<\frac{2}{3}\)

   \(\frac{3}{4}<\frac{4}{5}\)

.......

\(\frac{99}{100}<\frac{100}{101}\)

=> \(\frac{1}{2}.\frac{3}{4}....\frac{99}{100}<\frac{2}{3}.\frac{4}{5}....\frac{100}{101}\)=> A < B

=> A . A < A.B <=> A² <  \(\left(\frac{1}{2}.\frac{3}{4}....\frac{99}{100}\right).\left(\frac{2}{3}.\frac{4}{5}....\frac{100}{101}\right)=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}...\frac{99}{100}.\frac{100}{101}=\frac{1}{101}\)

=> \(A<\frac{1}{\sqrt{101}}\) (ĐPCM)

a.4^7

b.8^5 

 c.cho x mk sẻ tính kết quả nhưng tìm xmk ko tính đâu

b) \(\frac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5}.\frac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5}=\frac{4^5.\left(1+1+1+1\right)}{3^5.\left(1+1+1\right)}.\frac{6^5.\left(1+1+1+1+1+1\right)}{2^5.\left(1+1\right)}\)

                                                                                                       \(=\frac{4^5.4}{3^5.3}.\frac{6^5.6}{2^5.2}=\frac{4^6}{3^6}.\frac{6^6}{2^6}=\frac{2^{12}.2^6.3^6}{3^6.2^6}=2^{12}\)

   Ta có: \(2^{12}=\left(2^3\right)^4=8^4\)                                                    

Vậy x= 4

a, \(A=\frac{12}{3.7}+\frac{12}{7.11}+...+\frac{12}{195.199}\)

       \(=3.\left(\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{195.199}\right)\)

       \(=3.\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{195}-\frac{1}{199}\right)\) 

       \(=3.\left(\frac{1}{3}-\frac{1}{199}\right)\)

       \(=3.\left(\frac{199}{597}-\frac{3}{597}\right)\)

       \(=3.\frac{196}{597}\)

       \(=\frac{196}{199}\)