\(\frac{2016\times2017+4034}{2018\times2019-4034}\) Tính
Những câu hỏi liên quan
Tính nhanh :
2016 x 2017 + 4034/2018 x 2019 + 4036
Tính nhanh
A= 2016 x 2017 +4034 / 2018 x 2019 - 4036
2016.x2017+2018=4034
2016.x2017 + 2018 = 4034
2016.x2017 = 4034-2018
2016.x2017 = 2016
x2017 = 1
=>x =1
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Tính gía trị của biểu thức
\(\frac{\left(2015^2\times2025+31\times2016-1\right)\times\left(2015\times2020+4\right)}{2016^2\times2017\times2018\times2019\times2020}\)
Chứng minh rằng :
\(\frac{2015}{4034}< \frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2016^2}< \frac{2015}{2016}\)
Giúp vs ak
\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2016^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2015.2016}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2015}-\frac{1}{2016}\)
\(=1-\frac{1}{2016}=\frac{2015}{2016}\)
=> \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2016^2}< \frac{2015}{2016}\)
\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2016^2}>\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2016.2017}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}...+\frac{1}{2016}-\frac{1}{2017}\)
\(=\frac{1}{2}-\frac{1}{2017}=\frac{2015}{4024}\)
=> \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2016^2}>\frac{2015}{4034}\)
vậy ta có điều cần chứng minh
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2017\(^2\)-4034 x 2016+2016\(^2\)
\(2017^2-4034.2016+2016^2\)
\(=2017^2+2.2017.2016+2016^2\)
\(=\left(2017+2016\right)^2\)
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Chứng minh rằng:
\(\frac{2015}{4034}\)< \(\frac{1}{^{2^2}}\)+\(\frac{1}{3^2}\)+.....+\(\frac{1}{2016^2}\)<\(\frac{2015}{2016}\)
Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2016^2}\) ta có :
\(A>\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2016.2017}\)
\(A>\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2016}-\frac{1}{2017}\)
\(A>\frac{1}{2}-\frac{1}{2017}\)
\(A>\frac{2015}{4034}\) \(\left(1\right)\)
Lại có :
\(A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2015.2016}\)
\(A< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2015}-\frac{1}{2016}\)
\(A< 1-\frac{1}{2016}\)
\(A< \frac{2015}{2016}\) \(\left(2\right)\)
Từ (1) và (2) suy ra : \(\frac{2015}{4034}< A< \frac{2015}{2016}\) ( đpcm )
Vậy \(\frac{2015}{4034}< \frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2016^2}< \frac{2015}{2016}\)
Chúc bạn học tốt ~
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cam on ban rat nhieu PHUNG MINH QUAN !!!!!!!!!!
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Tính nhanh : \(\frac{2018\times2017-1}{2016\times2018+2017}\)
\(\frac{2018\times2017-1}{2016\times2018+2017}\)
\(=\frac{2018\times\left(2016+1\right)-1}{2016\times2018+2017}\)
\(=\frac{2018\times2016+2018-1}{2016\times2018+2017}\)
\(=\frac{2018\times2016+2017}{2016\times2018+2017}\)
\(=1\)
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Xem thêm câu trả lời
Cho A=\(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{4034},B=1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+...+\frac{1}{4033}\). So sánh A/B với \(1\frac{2017}{2018}\)