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 \(A< B\)

1

\(A=\frac{2019^{2019}+1}{2019^{2020}+1}< \frac{2019^{2019}+1+2018}{2019^{2020}+1+2018}=\frac{2019^{2019}+2019}{2019^{2020}+2019}=\frac{2019\left(2019^{2018}+1\right)}{2019\left(2019^{2019}+1\right)}\)

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

2

\(M=\frac{100^{101}+1}{100^{100}+1}< \frac{100^{101}+1+99}{100^{100}+1+99}=\frac{100^{101}+100}{100^{100}+100}=\frac{100\left(100^{100}+1\right)}{100\left(100^{99}+1\right)}\)

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

giúp mình với

help Khai phong voi

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

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

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

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

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

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

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

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

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

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

Tự so sánh

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

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

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

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

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

\(B=\frac{100^{2018}+1}{100^{2019}+1}< \frac{100^{2018}+1+99}{100^{2019}+1+99}=\frac{100^{2018}+100}{100^{2019}+100}=\frac{100\left(100^{2017}+1\right)}{100\left(100^{2018}+1\right)}=\frac{100^{2017}+1}{100^{2018}+1}=A\)

\(\Rightarrow B< A\)

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

Bạn nên mua sách bồi dưỡng năg lực tự học toán 6 thì sẽ có dạng này đó, chúc bn hc tốt (•~•)

Ta có: B = (2018 + 2019)/(2019 + 2020) = (2018 + 2019)/4039 = 2018/4039 + 2019/4039
Ta thấy : 2018/2019 > 2018/4039
            2019/2020 > 2019/4039
=> 2018/2019 + 2019/2020 > 2018/4039 > 2019/4039
=> 2018/2019 + 2019/2020 > (2018 + 2019)/(2019 + 2020)
=> A  > B

Ta có: \(C=\dfrac{2019-2018}{2019+2018}\)

\(\Leftrightarrow C=\dfrac{\left(2019-2018\right)\left(2019+2018\right)}{\left(2019+2018\right)^2}\)

\(\Leftrightarrow C=\dfrac{2019^2-2018^2}{\left(2019+2018\right)^2}\)

Ta có: \(\left(2019+2018\right)^2=2019^2+2018^2+2\cdot2019\cdot2018\)

\(2019^2+2018^2=2019^2+2018^2+0\)

Do đó: \(\left(2019+2018\right)^2>2019^2+2018^2\)

\(\Leftrightarrow\dfrac{2019^2-2018^2}{\left(2019+2018\right)^2}< \dfrac{2019^2-2018^2}{2019^2+2018^2}\)

\(\Leftrightarrow C< D\)