Ta có: \(\left(26^{2018}+3^{2018}\right)^{2019}=26^{2018\cdot2019}+3^{2018\cdot2019}\left(1\right)\)

          \(\left(26^{2019}+3^{2019}\right)^{2018}=26^{2019\cdot2018}+3^{2019\cdot2018}\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left(26^{2018}+3^{2018}\right)^{2019}=\left(26^{2019}+3^{2019}\right)^{2018}\)

\(A=\left(26^{2018}+3^{2018}\right)^{2019}\)

\(B=\left(26^{2019}+3^{2019}\right)^{2018}\)

\(B=\left(26^{2018}.26+3.3^{2018}\right)^{2018}< \left(26^{2018}.26+3^{2018}.26\right)^{2018}\)

\(B< \left(26^{2018}+3^{2018}\right)^{2018}.26^{2018}< \left(26^{2018}+3^{2018}\right)^{2018}.\left(26^{2018}+3^{2018}\right)\)

\(\Rightarrow B< \left(26^{2018}+3^{2018}\right)^{2019}\Rightarrow B< A\)

Ta có: \(\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ac\)

\(\Rightarrow a^2+b^2+c^2=\left(a+b+c\right)^2-2\left(ab+bc+ac\right)\)

hay \(a^2+b^2+c^2=0\Rightarrow a=b=c=0\)

Thay a = b = c = 0 vào M rồi tính như bình thường nha bạn!

Ta có : 

\(a+b+c=0\)

\(\Leftrightarrow\)\(\left(a+b+c\right)^2=0\)

\(\Leftrightarrow\)\(a^2+b^2+c^2+2ab+2bc+2ca=0\)

\(\Leftrightarrow\)\(a^2+b^2+c^2+2\left(ab+bc+ca\right)=0\)

\(\Leftrightarrow\)\(a^2+b^2+c^2=0\)

\(\Leftrightarrow\)\(\hept{\begin{cases}a^2=0\\b^2=0\\c^2=0\end{cases}\Leftrightarrow a=b=c=0}\)

\(\Rightarrow\)\(M=\left(a-2018\right)^{2019}+\left(b-2018\right)^{2019}-\left(c+2018\right)^{2019}\)

\(\Rightarrow\)\(M=-2018^{2019}-2018^{2019}-2018^{2019}\)

\(\Rightarrow\)\(M=-3.2018^{2019}\)

Chúc bạn học tốt ~ 

Cứu mình với 9:00 sáng nay mình nộp bài rùi

bạn ơi bạn có câu trả lời chưa, cho mik xin vs

<=> \(2a^2+2b^2+2c^2=2ab+2bc+2ca< =>\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0< =>\)

a=b=c => 3²⁰²⁰ = 3.a²⁰¹⁹ <=> 3²⁰¹⁹ = a²⁰¹⁹ => a=b=c=3

A= 1²⁰¹⁷ + 0²⁰¹⁸ + (-1)²⁰¹⁹ = 0

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