\(N=\frac{6}{10^{2015}}+\frac{8}{10^{2016}}=M=\frac{8}{10^{2015}}+\frac{6}{10^{2016}}\)

Hk tốt

k nhé

Ta có :N= \(\frac{6}{10^{2015}}+\frac{8}{10^{2016}}=\frac{6}{10^{2015}}+\frac{6}{10^{2016}}+\frac{2}{10^{2016}}\)

          M=\(\frac{8}{10^{2015}}+\frac{6}{10^{2016}}=\frac{6}{10^{2015}}+\frac{6}{10^{2016}}+\frac{2}{10^{2015}}\)

         Ta Xét:                   \(\frac{2}{10^{2016}},\frac{2}{10^{2015}}\)

          Vì   10²⁰¹⁶>10²⁰¹⁵

          Nên:     \(\frac{2}{10^{2016}}< \frac{2}{10^{2015}}\)

          Do đó :                                 N<M

A = 1/2.3/4.....2015/2016

= 1.3.5.....2015/2.4.6......2016

= 1.3.5.....2015/(1.2).(2.2).....(2.1008)

= 1.3.5.....2015/2^1008 . 1.2....1008

A = \(\frac{\frac{3}{4}-\frac{3}{11}+\frac{3}{13}}{\frac{5}{4}-\frac{5}{11}+\frac{5}{13}}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{\frac{5}{4}-\frac{5}{6}+\frac{5}{8}}\)

\(=\frac{3.\left(\frac{1}{4}-\frac{1}{11}+\frac{1}{13}\right)}{5.\left(\frac{1}{4}-\frac{1}{11}+\frac{1}{13}\right)}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}\right)}\)

\(=\frac{3}{5}+\frac{1}{\frac{5}{2}}\)

\(=\frac{3}{5}+\frac{2}{5}=1\)

b) B = \(\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6.8^4.3^5}-\frac{5^{10}.7^3:25^5.49}{\left(125.7\right)^3+5^9.14^3}\)

\(=\frac{2^{12}.3^5-\left(2^2\right)^6.\left(3^2\right)^2}{2^{12}.3^6+\left(2^3\right)^4.3^5}-\frac{5^{10}.7^3-\left(5^2\right)^5.7^2}{\left(5^3\right)^3.7^3+5^9.\left(7.2\right)^3}\)

\(=\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}-\frac{5^{10}.7^3-5^{10}-7^2}{5^9.7^3+5^9.7^3.2^3}\)

\(=\frac{2^{12}.3^4.\left(3-1\right)}{2^{12}.3^5\left(3+1\right)}-\frac{5^{10}.7^2.\left(7-1\right)}{5^9.7^3\left(1+2^3\right)}\)

 \(=\frac{1}{3.2}-\frac{5.2}{7.3}\)

\(=\frac{7}{3.2.7}-\frac{5.2.2}{7.3.2}\)

\(=\frac{7}{42}-\frac{20}{42}\)

\(=-\frac{13}{42}\)

Ta có :

\(P=\dfrac{6^{2016}+4}{6^{2016}-1}=\dfrac{6^{2016}-1+5}{6^{2016}-1}=\dfrac{6^{2016}-1}{6^{2016}-1}+\dfrac{5}{6^{2016}-1}\)\(=1+\dfrac{5}{6^{2016}-1}\)

\(Q=\dfrac{6^{2016}}{6^{2016}-5}=\dfrac{6^{2016}-5+5}{6^{2016}-5}=\dfrac{6^{2015}-5}{6^{2016}-5}+\dfrac{5}{6^{2016}-5}=1+\dfrac{5}{6^{2016}-5}\)

Vì \(1+\dfrac{5}{6^{2016}-1}< 1+\dfrac{5}{6^{2016}-5}\Rightarrow P< Q\)

Ta có:

\(P-Q=\dfrac{6^{2016}+4}{6^{2016}-1}-\dfrac{6^{2016}}{6^{2016}-5}=1+\dfrac{5}{6^{2016}-1}-1-\dfrac{5}{6^{2016}-5}\)

\(=\dfrac{5}{6^{2016}-1}-\dfrac{5}{6^{2016}-5}=5\left(\dfrac{1}{6^{2016}-1}-\dfrac{1}{6^{2016}-5}\right)< 0\)

Vậy A < B

1. Bài giải:

Đặt \(A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{1000}\)

\(\Rightarrow\frac{1}{2}A=\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{1002}\)

\(\Rightarrow\frac{1}{2}A=A-\frac{1}{2}A=\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{1000}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{1002}\right)\)

\(\Rightarrow\frac{1}{2}A=1-\frac{1}{1002}=\frac{1001}{1002}\Rightarrow A=\frac{2002}{1002}=\frac{1001}{501}\)

Vậy \(A=\frac{1001}{501}\)