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Hello Triệu Mẫn điên .Tui là Nguyên 6n1^^
Tui đang suy nghĩ
Tui biết làm nhưng không nói
chỉ nói kết quả bằng 10
\(a)\) Ta có :
\(VP=\frac{2018}{1}+\frac{2017}{2}+\frac{2016}{3}+...+\frac{2}{2017}+\frac{1}{2018}\)
\(VP=\left(\frac{2018}{1}-1-...-1\right)+\left(\frac{2017}{2}+1\right)+\left(\frac{2016}{3}+1\right)+...+\left(\frac{2}{2017}+1\right)+\left(\frac{1}{2018}+1\right)\)
\(VP=1+\frac{2019}{2}+\frac{2019}{3}+...+\frac{2019}{2017}+\frac{2019}{2018}\)
\(VP=2019\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}\right)\)
Lại có :
\(VT=\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}\right).x\)
\(\Rightarrow\)\(x=2019\)
Vậy \(x=2019\)
Chúc bạn học tốt ~
\(a)\) Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\) ta có :
\(A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(A< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(A< 1-\frac{1}{100}=\frac{99}{100}< 1\)
Vậy \(A< 1\)
Chúc bạn học tốt ~
Đặt \(A=\frac{\frac{1}{2020}+\frac{2}{2019}+\frac{3}{2018}+...+\frac{2019}{2}+\frac{2020}{1}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2021}}\)
\(A=\frac{1+\left(\frac{1}{2020}+1\right)+\left(\frac{2}{2019}+1\right)+\left(\frac{3}{2018}+1\right)+...+\left(\frac{2019}{2}+1\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2021}}\)
\(A=\frac{\frac{2021}{2021}+\frac{2021}{2020}+\frac{2021}{2019}+...+\frac{2021}{2}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2021}}\)
\(A=\frac{2021\left(\frac{1}{2021}+\frac{1}{2020}+\frac{1}{2019}+...+\frac{1}{2}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2021}}=2021\)
mình nhầm \(A=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2^{2018}-1}\)
Ta có : \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2018^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2017.2018}\)
Xét B = \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2017.2018}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\)
=\(1-\frac{1}{2018}\)
Xét : \(\frac{2018}{2018}=1\)=) B < 1
khoan hình như sai đề
Đặt \(A=1+\frac{1}{3^2}+\frac{1}{3^4}+...+\frac{1}{3^{2018}}\)
\(\Rightarrow3^2A=3^2+1+\frac{1}{3^2}+...+\frac{1}{3^{2016}}\)
\(\Rightarrow3^2A-A=\frac{1}{3^{2016}}-3^2\)
\(8A=\frac{1}{3^{2016}}-9\)
\(A=\frac{\frac{1}{3^{2016}}-9}{8}\)
\(1+\frac{1}{3^2}+\frac{1}{3^4}+...+\frac{1}{3^{2018}}\)
Ta có :
Đặt A = \(1+\frac{1}{3^2}+\frac{1}{3^4}+...+\frac{1}{3^{2018}}\)
32A = \(3^2.\left(1+\frac{1}{3^2}+\frac{1}{3^4}+...+\frac{1}{3^{2018}}\right)\)
32A = \(3^2+1+\frac{1}{3^2}+....+\frac{1}{3^{2016}}\)
32A - A = \(\left(3^2+1+\frac{1}{3^2}+....+\frac{1}{3^{2016}}\right)\) - \(\left(1+\frac{1}{3^2}+\frac{1}{3^4}+...+\frac{1}{3^{2018}}\right)\)
8A = \(3^2-\frac{1}{3^{2018}}\)
A = \(\frac{9-\frac{1}{3^{2018}}}{8}\)