giúp mình với

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

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

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

nhớ bảo kê nha Duyên

thank you

nhanh nha mọi người. Ngày mai mình thi rồi. Ai giải nhanh nhất mình k cho!

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 \(A< B\)

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

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

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

Ta có: \(\frac{2}{2018^{2019}+1}>\frac{2}{2018^{2019}+2}\)

\(\Rightarrow1-\frac{2}{2018^{2019}+1}< 1-\frac{2}{2018^{2019}+2}\)

\(\Rightarrow A< B\)

Vậy .....

Be De no bi dien rui ak

VAY NO BI DIEN

Bọn mày ko giúp thì cút

A = 6cs + 7cs - 1 = 7cs              

B = 12cs - 2 = 12 cs

==>A>B

a, Vì  A, B < 1

\(A=\frac{15^{16}+1}{15^{17}+1}< \frac{15^{16}+1+14}{15^{17}+1+14}=\frac{15^{16}+15}{15^{17}+15}=\frac{15\left(15^{15}+1\right)}{15\left(15^{16}+1\right)}=\frac{15^{15}+1}{15^{16}+1}\)

b, \(B=\frac{2018^{2018}+1}{2018^{2019}+1}< 1< \frac{2018^{2019}+1}{2018^{2018}+1}=A\)