so sánh\(\frac{2017^6+5}{2017^6+3}va\frac{2017^6+7}{2017^6+5}\)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\frac{x-2017}{5}-\frac{x-2017}{6}=\frac{x-2017}{7}-\frac{x-2017}{8}\)
\(\frac{x-2017}{5}-\frac{x-2017}{6}-\frac{x-2017}{7}+\frac{x-2017}{8}=0\)
\(\left(x-2017\right)\left(\frac{1}{5}-\frac{1}{6}-\frac{1}{7}+\frac{1}{8}\right)=0\)
mà \(\frac{1}{5}-\frac{1}{6}-\frac{1}{7}+\frac{1}{8}\ne0\)
\(\Rightarrow x-2017=0\)
\(\Rightarrow x=2017\)
Vậy x = 2017
\(\Rightarrow\frac{x-2017}{5}-\frac{x-2017}{6}-\frac{x-2017}{7}+\frac{x-2017}{8}=0\)
\(\Rightarrow\left(x-2017\right)\left(\frac{1}{5}-\frac{1}{6}-\frac{1}{7}+\frac{1}{8}\right)=0\)
\(\Rightarrow x-2017=0\)(vì \(\frac{1}{5}-\frac{1}{6}-\frac{1}{7}+\frac{1}{8}\ne0\))
=>x=2017
Chúc bạn học tốt
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 = \(\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6.8^4.3^5}-\frac{5^{10}.7^3:25^5.49}{\left(125.7\right)^3+5^9.14^3}\)
\(=\frac{2^{12}.3^5-\left(2^2\right)^6.\left(3^2\right)^2}{2^{12}.3^6+\left(2^3\right)^4.3^5}-\frac{5^{10}.7^3-\left(5^2\right)^5.7^2}{\left(5^3\right)^3.7^3+5^9.\left(7.2\right)^3}\)
\(=\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}-\frac{5^{10}.7^3-5^{10}-7^2}{5^9.7^3+5^9.7^3.2^3}\)
\(=\frac{2^{12}.3^4.\left(3-1\right)}{2^{12}.3^5\left(3+1\right)}-\frac{5^{10}.7^2.\left(7-1\right)}{5^9.7^3\left(1+2^3\right)}\)
\(=\frac{1}{3.2}-\frac{5.2}{7.3}\)
\(=\frac{7}{3.2.7}-\frac{5.2.2}{7.3.2}\)
\(=\frac{7}{42}-\frac{20}{42}\)
\(=-\frac{13}{42}\)
`a,`
`5/6=1-1/6`
`7/8=1-1/8`
Mà `1/6>1/8 -> 5/6<7/8`
`b,`
`9/5=(9 \times 2)/(5 \times 2)=18/10`
`3/2=(3 \times 5)/(2 \times 5)=15/10`
`18/10 > 15/10 -> 9/5 > 3/2`
`c,`
`2017/2018 = 1-1/2018`
`2019/2020=1-1/2020`
`1/2018 > 1/2020 -> 2017/2018 < 2019/2020`
`d,`
`2018/2017 = 1+1/2017`
`2020/2019 = 1+1/2019`
`1/2017 > 1/2019 -> 2018/2017>2020/2019`
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 \(A=\frac{1001}{501}\)
\(\frac{2017^6+5}{2017^6+3}=1+\frac{2}{2017^6+3}>1+\frac{2}{2017^6+5}.\)=\(\frac{2017^6+7}{2017^6+5}\)
bạn Eihwaz ơi vậy kết quả