a) 29^57 < 29^75

b) 1011^22 < 1101^22

c) ( 2021 + 2018 )^2019 < ( 2020 + 2019 )^2020

d) 2^5000 > 7^2000

Chúc bạn học tốt @!!!

Nếu có thể thì t.i.c.k cho mình nha ! Thank

Đại ca toàn học giỏi nhất hành tinh đây

a) 29⁵⁷và 29⁷⁵

 ^=> 29⁵⁷<29⁷⁵

ko biết

`a,`

`5/6=1-1/6`

`7/8=1-1/8`

Mà `1/6>1/8 -> 5/6<7/8`

`b,`

`9/5=(9 \times 2)/(5 \times 2)=18/10`

`3/2=(3 \times 5)/(2 \times 5)=15/10`

`18/10 > 15/10 -> 9/5 > 3/2`

`c,`

`2017/2018 = 1-1/2018`

`2019/2020=1-1/2020`

`1/2018 > 1/2020 -> 2017/2018 < 2019/2020`

`d,`

`2018/2017 = 1+1/2017`

`2020/2019 = 1+1/2019`

`1/2017 > 1/2019 -> 2018/2017>2020/2019`

\(-\frac{2018}{2019}.\frac{2}{7}-\frac{2018}{2019}.\frac{5}{7}+1\frac{2018}{2019}=\frac{2018}{2019}\left(\frac{-2-5}{7}\right)+1\frac{2018}{2019}=\frac{2018}{2019}.\left(-1\right)+1\frac{2018}{2019}=\frac{-2018}{2019}+1\frac{2018}{2019}=1\)

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

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

\(\Leftrightarrow C< D\)