\(\frac{2017^8-2}{2017^8+1}\)so sánh với \(\frac{2017^8}{2017^8-3}\)
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Khách
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Những câu hỏi liên quan
XO
26 tháng 10 2019
A = \(\frac{\frac{3}{4}-\frac{3}{11}+\frac{3}{13}}{\frac{5}{4}-\frac{5}{11}+\frac{5}{13}}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{\frac{5}{4}-\frac{5}{6}+\frac{5}{8}}\)
\(=\frac{3.\left(\frac{1}{4}-\frac{1}{11}+\frac{1}{13}\right)}{5.\left(\frac{1}{4}-\frac{1}{11}+\frac{1}{13}\right)}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}\right)}\)
\(=\frac{3}{5}+\frac{1}{\frac{5}{2}}\)
\(=\frac{3}{5}+\frac{2}{5}=1\)
XO
26 tháng 10 2019
b) B = \(\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6.8^4.3^5}-\frac{5^{10}.7^3:25^5.49}{\left(125.7\right)^3+5^9.14^3}\)
\(=\frac{2^{12}.3^5-\left(2^2\right)^6.\left(3^2\right)^2}{2^{12}.3^6+\left(2^3\right)^4.3^5}-\frac{5^{10}.7^3-\left(5^2\right)^5.7^2}{\left(5^3\right)^3.7^3+5^9.\left(7.2\right)^3}\)
\(=\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}-\frac{5^{10}.7^3-5^{10}-7^2}{5^9.7^3+5^9.7^3.2^3}\)
\(=\frac{2^{12}.3^4.\left(3-1\right)}{2^{12}.3^5\left(3+1\right)}-\frac{5^{10}.7^2.\left(7-1\right)}{5^9.7^3\left(1+2^3\right)}\)
\(=\frac{1}{3.2}-\frac{5.2}{7.3}\)
\(=\frac{7}{3.2.7}-\frac{5.2.2}{7.3.2}\)
\(=\frac{7}{42}-\frac{20}{42}\)
\(=-\frac{13}{42}\)
4 tháng 4 2018
\(B=\frac{18}{37}-\frac{8}{2017}+\frac{19}{37}-1\frac{2009}{2017}+\frac{2017}{2018}\)
\(B=\left(\frac{18}{37}+\frac{19}{37}\right)-\left(\frac{8}{2017}+1\frac{2009}{2017}\right)+\frac{2017}{2018}\)
\(B=1-\left(\frac{8}{2017}+\frac{4026}{2017}\right)+\frac{2017}{2018}\)
\(B=1-2+\frac{2017}{2018}\)
\(B=-1+\frac{2017}{2018}=\frac{-2018}{2018}+\frac{2017}{2018}\)
\(B=\frac{-1}{2018}\)
CHÚC BN HỌC TỐT!!!!!!
mk nha!!
30 tháng 3 2018
Bài 1 : dễ bạn tự làm được :)
Bài 2 :
Ta có :
\(B=\frac{2015+2016+2017}{2016+2017+2018}=\frac{2015}{2016+2017+2018}+\frac{2016}{2016+2017+2018}+\frac{2017}{2016+2017+2018}\)
Vì :
\(\frac{2015}{2016}>\frac{2015}{2016+2017+2018}\)
\(\frac{2016}{2017}>\frac{2016}{2016+2017+2018}\)
\(\frac{2017}{2018}>\frac{2017}{2016+2017+2018}\)
Nên \(\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}>\frac{2015}{2016+2017+2018}+\frac{2016}{2016+2017+2018}+\frac{2017}{2016+2017+2018}\)
\(\Leftrightarrow\)\(\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}>\frac{2015+2016+2017}{2016+2017+2018}\)
\(\Leftrightarrow\)\(A>B\)
Vậy \(A>B\)
Chúc bạn học tốt ~
30 tháng 3 2018
Ta có : B = 2016 + 2017 + 2018 2015 + 2016 + 2017 = 2016 + 2017 + 2018 2015 + 2016 + 2017 + 2018 2016 + 2016 + 2017 + 2018 2017 Vì : 2016 2015 > 2016 + 2017 + 2018 2015 2017 2016 > 2016 + 2017 + 2018 2016 2018 2017 > 2016 + 2017 + 2018 2017 Nên 2016 2015 + 2017 2016 + 2018 2017 > 2016 + 2017 + 2018 2015 + 2016 + 2017 + 2018 2016 + 2016 + 2017 + 2018 2017 ⇔ 2016 2015 + 2017 2016 + 2018 2017 > 2016 + 2017 + 2018 2015 + 2016 + 2017 ⇔A > B Vậy A > B Chúc bạn học tốt ~
16 tháng 9 2018
\(\frac{x-2017}{5}-\frac{x-2017}{6}=\frac{x-2017}{7}-\frac{x-2017}{8}\)
\(\frac{x-2017}{5}-\frac{x-2017}{6}-\frac{x-2017}{7}+\frac{x-2017}{8}=0\)
\(\left(x-2017\right)\left(\frac{1}{5}-\frac{1}{6}-\frac{1}{7}+\frac{1}{8}\right)=0\)
mà \(\frac{1}{5}-\frac{1}{6}-\frac{1}{7}+\frac{1}{8}\ne0\)
\(\Rightarrow x-2017=0\)
\(\Rightarrow x=2017\)
Vậy x = 2017
16 tháng 9 2018
\(\Rightarrow\frac{x-2017}{5}-\frac{x-2017}{6}-\frac{x-2017}{7}+\frac{x-2017}{8}=0\)
\(\Rightarrow\left(x-2017\right)\left(\frac{1}{5}-\frac{1}{6}-\frac{1}{7}+\frac{1}{8}\right)=0\)
\(\Rightarrow x-2017=0\)(vì \(\frac{1}{5}-\frac{1}{6}-\frac{1}{7}+\frac{1}{8}\ne0\))
=>x=2017
Chúc bạn học tốt
24 tháng 6 2017
A=\(\frac{2017^{2017}+2}{2017^{2017}-1}\)=\(\frac{\left(2017^{2017}-1\right)+3}{2017^{2017}-1}\)=\(1\)+\(\frac{3}{2017^{2017}-1}\)
B=\(\frac{2017^{2017}}{2017^{2017}-3}\)=\(\frac{\left(2017^{2017}-3\right)+3}{2017^{2017}-3}\)=\(1\)+\(\frac{3}{2017^{2017}-3}\)
Vì \(2017^{2017}-1\)\(>\)\(2017^{2017}-3\)nên \(\frac{3}{2017^{2017}-1}\)\(< \)\(\frac{3}{2017^{2017}-3}\)=> A<B
vậy A<B
chúc bạn học giỏi
k giùm mk nhé
Ta có:
\(\frac{2017^8-2}{2017^8+1}=\frac{2017^8+1-3}{2017^8+1}=1-\frac{3}{2017^8+1}\)
\(\frac{2017^8}{2017^8-3}=\frac{2017^8-3+3}{2017^8-3}=1+\frac{3}{2017^8-3}\)
Hiển nhiên: \(\frac{3}{2017^8-3}+\frac{3}{2017^8+1}>0\)
\(\Rightarrow\left(\frac{3}{2017^8-3}\right)>-\frac{3}{2017^8+1}\)
\(\Rightarrow1+\frac{3}{2017^8-3}>1-\frac{3}{2017^8+1}\)
\(\Rightarrow\left(\frac{2017^8-2}{2017^8+1}\right)< \frac{2017^8}{2017^8-3}\)
ta có:2017^8-2/2017^8+1=2017^8+1-3/2017^8+1=1-3/2017^8+1
2017^8/2017^8-3=2017^8-3+3/2017^8-3=1+3/2017^8-3
Vì 1-3/2017^8+1<1+3/2017^8-3
=>2017^8-2/2017^8+1<2017^8/2017^8-3