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26 tháng 2 2020

Ta có \(E=\frac{2018^{99}-1}{2018^{100}-1}\)

\(\Leftrightarrow2018E=\frac{2018^{100}-2018}{2018^{100}-1}\)

\(\Leftrightarrow2018E=1-\frac{2017}{2018^{100}-1}\)   (2)

Lại có \(F=\frac{2018^{98}-1}{2018^{99}-1}\)

\(\Leftrightarrow2018F=\frac{2018^{99}-2018}{2018^{99}-1}\)

\(\Leftrightarrow2018F=1-\frac{2017}{2018^{99}-1}\)  (2)

Mà \(2018^{100}>2018^{99}>0\)

\(\Leftrightarrow2018^{100}-1>2018^{99}-1\)

\(\Leftrightarrow\frac{2017}{2018^{100}-1}< \frac{2017}{2018^{99}-1}\)

\(\Leftrightarrow-\frac{2017}{2018^{100}-1}>-\frac{2017}{2018^{99}-1}\)

\(\Leftrightarrow1-\frac{2017}{2018-1}>1-\frac{2017}{2018^{99}-1}\)   (3)

Từ (1) ;(2) và (3) <=> 2018E > 2018 F > 0

<=> E > F 

Vậy E > F

@@ Học tốt

Chiyuki Fujito

K cần tk

10 tháng 2 2019

giúp mình với

4 tháng 5 2018

1) Đặt dãy trên là \(A\)

Theo bài ra ta có :

\(A=\frac{1}{3.3}+\frac{1}{4.4}+\frac{1}{5.5}+\frac{1}{6.6}+...+\frac{1}{100.100}\)

\(\Rightarrow A< \frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)

\(\Rightarrow A< \frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)

\(\Rightarrow A< \frac{1}{2}-\frac{1}{100}< \frac{1}{2}\left(đpcm\right)\)

2) \(A=\frac{5^{2018}-2017+1}{5^{2018}-2017}=\frac{5^{2018}-2017}{5^{2018}-2017}+\frac{1}{5^{2018}-2017}=1+\frac{1}{5^{2018}-2017}\)( 1 )

\(B=\frac{5^{2018}-2019+1}{5^{2018}-2019}=\frac{5^{2018}-2019}{5^{2018}-2019}+\frac{1}{5^{2018}-2019}=1+\frac{1}{5^{2018}-2019}\)( 2 )

Từ ( 1 ) và ( 2 ) \(\Rightarrow\)\(A=1+\frac{1}{5^{2018}-2017}< 1+\frac{1}{5^{2018}-2019}=B\)

\(\Rightarrow A< B\)

Vậy \(A< B.\)

4 tháng 5 2018

1) Ta có B =

 \(\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\) < \(\frac{1}{1.3}+\frac{1}{3.4}+...+\frac{1}{99.100}=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)\(\frac{99}{100}\)

=> B < 1 ( chứ không phải \(\frac{1}{2}\) bạn nhé)

Sai thì thôi chứ mk chỉ làm rờ thôi

10 tháng 2 2019

\(\frac{2018^{100}+1}{2018^{90}+1}\)\(\frac{2018^{10}+1}{1+1}\)\(\frac{2018^{10}+1}{2}\)

\(\frac{2018^{99}+1}{2018^{89}+1}\)\(\frac{2018^{10}+1}{1+1}\)\(\frac{2018^{10}+1}{2}\)

=> \(\frac{2018^{100}+1}{2018^{90}+1}=\frac{2018^{99}+1}{2018^{89}+1}\)

nhớ bảo kê nha Duyên

10 tháng 2 2019

thank you

3 tháng 5 2019

\(A=\frac{100^{2017}+1}{100^{2018}+1}\)

\(\Rightarrow100A=\frac{100\cdot\left[100^{2017}+1\right]}{100^{2018}+1}\)

\(\Rightarrow100A=\frac{100^{2018}+100}{100^{2018}+1}\)

\(\Rightarrow100A=\frac{100^{2018}+1+99}{100^{2018}+1}\)

\(\Rightarrow100A=1+\frac{99}{100^{2018}+1}\)

\(B=\frac{100^{2018}+1}{100^{2019}+1}\)

\(\Rightarrow100B=\frac{100\cdot\left[100^{2018}+1\right]}{100^{2019}+1}\)

\(\Rightarrow100B=\frac{100^{2019}+100}{100^{2019}+1}\)

\(\Rightarrow100B=\frac{100^{2019}+1+99}{100^{2019}+1}\)

\(\Rightarrow100B=1+\frac{99}{100^{2019}+1}\)

Tự so sánh

3 tháng 5 2019

\(A=\frac{100^{2017}+1}{100^{2018}+1}\)

\(\Rightarrow100A=\frac{100^{2018}+100}{100^{2018}+1}\)

\(\Rightarrow100A=\frac{100^{2018}+1+99}{100^{2018}+1}\)

\(\Rightarrow100A=\frac{100^{2018}+1}{100^{2018}+1}+\frac{99}{100^{2018}+1}\)

\(\Rightarrow100A=1+\frac{99}{100^{2018}+1}\)(1)

\(B=\frac{100^{2018}+1}{100^{2019}+1}\)

\(\Rightarrow100B=\frac{100^{2019}+100}{100^{2019}+1}\)

\(\Rightarrow100B=\frac{100^{2019}+1+99}{100^{2019}+1}\)

\(\Rightarrow100B=\frac{100^{2019}+1}{100^{2019}+1}+\frac{99}{100^{2019}+1}\)

\(\Rightarrow100B=1+\frac{99}{100^{2019}+1}\)(2)

Từ (1) và (2) suy ra 100A > 100B hay A > B

21 tháng 6 2019

Bài toán : So sánh A và B

\(A=\frac{2018^{100}}{1+2018+2018^2+...+2018^{100}}\)

+) Ta có \(\frac{1}{A}=\frac{1+2018+2018^2+...+2018^{100}}{2018^{100}}\)

                     \(=\frac{1}{2018^{100}}+\frac{2018}{2018^{100}}+\frac{2018^2}{2018^{100}}+...+\frac{2018^{100}}{2018^{100}}\)

                      \(=\frac{1}{2018^{100}}+\frac{1}{2018^{99}}+\frac{1}{2018^{98}}+...+1\)

\(B=\frac{2019^{100}}{1+2019+2019^2+...+2019^{100}}\)

+) Ta có \(\frac{1}{B}=\frac{1+2019+2019^2+...+2019^{100}}{2019^{100}}\)

                     \(=\frac{1}{2019^{100}}+\frac{2019}{2019^{100}}+\frac{2019^2}{2019^{100}}+...+\frac{2019^{100}}{2019^{100}}\)

                      \(=\frac{1}{2019^{100}}+\frac{1}{2019^{99}}+\frac{1}{2019^{98}}+...+1\)

+) \(\frac{1}{2018^{100}}>\frac{1}{2019^{100}}\)

     \(\frac{1}{2018^{99}}>\frac{1}{2019^{99}}\)

     .....................................

     \(1=1\)

\(\Rightarrow\frac{1}{2018^{100}}+\frac{1}{2018^{99}}+\frac{1}{2018^{98}}+...+1>\frac{1}{2019^{100}}+\frac{1}{2019^{99}}+\frac{1}{2019^{98}}+...+1\)

\(\Rightarrow\frac{1}{A}>\frac{1}{B}\)

\(\Rightarrow A< B\)

Vậy \(A< B\)

4 tháng 10 2018

        \(2018^{100}+2018^{99}\)

\(=2018^{99}.\left(2018+1\right)\)

\(=2018^{99}.2019\)\(< 2019^{99}.2019=2019^{100}\)

\(\Rightarrow2018^{100}+2018^{99}< 2019^{100}\)

                          Vậy \(2018^{100}+2018^{99}< 2019^{100}\)

                                                      ~~Hok tốt~~