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

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

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

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

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

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

Đặt \(\left\{{}\begin{matrix}2018-x=a\\x-2019=b\end{matrix}\right.\) \(\Rightarrow a+b=-1\Rightarrow b=-1-a\)

\(\frac{a^2+ab+b^2}{a^2-ab+b^2}=\frac{19}{49}\Leftrightarrow49\left(a^2+ab+b^2\right)=19\left(a^2-ab+b^2\right)\)

\(\Leftrightarrow15a^2+34ab+15b^2=0\)

\(\Leftrightarrow\left(5a+3b\right)\left(3a+5b\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}5a=-3b\\3a=-5b\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}5a=-3\left(-1-a\right)\\3a=-5\left(-1-a\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2a=3\\2a=-5\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}a=\frac{3}{2}\\a=-\frac{5}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2018-x=\frac{3}{2}\\2018-x=-\frac{5}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{4033}{2}\\x=\frac{4041}{2}\end{matrix}\right.\)

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

\(x\left(\sqrt{2019}+\sqrt{2018}\right)+y\left(\sqrt{2019}-\sqrt{2018}\right)=2019\sqrt{2019}+2018\sqrt{2018}\)

\(\Leftrightarrow x\left(\sqrt{2019}+\sqrt{2018}\right)+y\left(\sqrt{2019}-\sqrt{2018}\right)=2018\left(\sqrt{2019}+\sqrt{2018}\right)+\sqrt{2019}\)

\(\Leftrightarrow x+y.\left(\sqrt{2019}-\sqrt{2018}\right)^2=2018+\sqrt{2019}\left(\sqrt{2019}-\sqrt{2018}\right)\)

\(\Leftrightarrow x+y\left(4037-2\sqrt{2019.2018}\right)=4037-\sqrt{2019.2018}\)

\(\Leftrightarrow x+4037.y-4037=2y\sqrt{2019.2018}-\sqrt{2019.2018}\)

\(\Leftrightarrow x+4037y-4037=\left(2y-1\right).\sqrt{2019.2018}\)(1)

Do \(x;y\) hữu tỉ \(\Rightarrow x+4037y-4037\) và \(2y-1\) đều là số hữu tỉ

Mà \(\sqrt{2019.2018}\) là số vô tỉ

\(\Rightarrow\)đẳng thức (1) xảy ra khi và chỉ khi: \(\left\{{}\begin{matrix}2y-1=0\\x+4037y-4037=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}y=\dfrac{1}{2}\\x=\dfrac{4037}{2}\end{matrix}\right.\)