Ta có :

\(\frac{a+b-b-c}{2018-2019}=\frac{a-c}{-1}\)

\(\frac{b+c-c-a}{2019-2020}=\frac{b-a}{-1}\)

\(\frac{b-c}{2018-2020}=\frac{b-c}{-2}\)     

Đặt \(\frac{a-c}{-1}=\frac{b-a}{-1}=\frac{b-c}{-2}=k\left(k\ne0\right)\)

\(\Rightarrow\hept{\begin{cases}\frac{a-c}{-1}=k\\\frac{b-a}{-1}=k\\\frac{b-c}{-2}=k\end{cases}\Rightarrow\hept{\begin{cases}a-c=-k\\b-a=-k\\b-c=k.\left(-2\right)\end{cases}}}\)

\(\Rightarrowđpcm\)

Đặt \(\frac{a}{2018}=\frac{b}{2019}=\frac{c}{2020}=k\)

\(\Rightarrow a=2018k\), \(b=2019k\), \(c=2020k\)

Ta có: \(4\left(a-b\right)\left(b-c\right)=4\left(2018k-2019k\right)\left(2019k-2020k\right)\)

                                                 \(=4.\left(-k\right).\left(-k\right)=4k^2=\left(2k\right)^2\)

Ta lại có: \(\left(a-c\right)^2=\left(2018k-2020k\right)^2=\left(-2k\right)^2=\left(2k\right)^2\)

Vậy \(4\left(a-b\right)\left(b-c\right)=\left(a-c\right)^2\)

Đặt \(\frac{a}{2018}=\frac{b}{2019}=\frac{c}{2020}=k\Rightarrow\hept{\begin{cases}a=2018k\\b=2019k\\c=2020k\end{cases}}\)

Thế vị trí tương ứng ta được :

VT = 4( a - b )( b - c )

       = 4( 2018k - 2019k )( 2019k - 2020k )

       = 4(-k)(-k)

       = 4k²

VP = ( a - c )² 

       = ( 2018k - 2020k )²

       = ( -2k )²

       = 4k²

=> VT = VP

=> đpcm

Ta có: \(\frac{a}{2018}=\frac{b}{2019}=\frac{c}{2020}.\)

Áp dụng tính chất dãy tỉ số bằng nhau ta được:

\(\frac{a}{2018}=\frac{b}{2019}=\frac{c}{2020}=\frac{a-b}{2018-2019}=\frac{b-c}{2019-2020}=\frac{a-c}{2018-2020}.\)

Đặt \(\frac{a}{2018}=\frac{b}{2019}=\frac{c}{2020}=k\Rightarrow\hept{\begin{cases}a=2018k\\b=2019k\\c=2020k\end{cases}}\)

Khi đó 4(a - b)(b - c) = 4(2018k - 2019k)(2019k - 2020k)

= 4(-k).(-k) 

= 4k² (1)

Lại có (c - a)² = (2020k - 2018k)² = (2k)² = 4k² (2)

Từ (1)(2) => 4(a - b)(b - c) = (c - a)²

Đề sai sai gì đó nhá xem lại dùm

Đặt \(\frac{a}{2018}=\frac{b}{2019}=\frac{c}{2020}=k\)

\(\Rightarrow\left\{{}\begin{matrix}a=2018k\\b=2019k\\c=2020k\end{matrix}\right.\)

\(\Rightarrow\left(a-c\right)^3=\left(2018k-2020k\right)^3=\left(-2k\right)^3=-8k^3\) (1)

\(8\left(a-b\right)^2.\left(b-c\right)=8\left(2018k-2019k\right)^2.\left(2019k-2020k\right)=8k^2\left(-k\right)=8\left(-k\right)^3=-8k^3\left(2\right)\)

Từ (1) và (2) ⇒ \(\left(a-c\right)^3=8\left(a-b\right)^2.\left(b-c\right)\left(đpcm\right)\)

mn giúp mk vs

chiều mk nộp rùi

\(\frac{2\left|2018x-2019\right|+2019}{\left|2018x-2019\right|+1}\)

\(=\frac{\left(2\left(\left|2018x-2019\right|+1\right)\right)+2017}{\left|2018x-2019\right|+1}\)

\(=2+\frac{2017}{\left|2018x-2019\right|+1}\)có giá trị lớn nhất

\(\Rightarrow\frac{2017}{\left|2018x-2019\right|+1}\)có giá trị lớn nhất

\(\Rightarrow\left|2018x-2019\right|+1\)có giá trị nhỏ nhất

Mà \(\left|2018x-2019\right|\ge0\)

\(\Rightarrow\left|2018x-2019\right|+1\ge1\)

Dấu "=" xảy ra khi và chỉ khi:

\(\left|2018x-2019\right|=0\)

\(\Leftrightarrow x=\frac{2019}{2018}\)

Vậy \(M_{MAX}=2019\)tại \(x=\frac{2019}{2018}\)

\(\frac{5^x+5^{x+1}+5^{x+2}}{31}=\frac{3^{2x}+3^{2x+1}+3^{2x+2}}{13}\)

\(\Rightarrow\frac{5^x\left(1+5+5^2\right)}{31}=\frac{3^{2x}\left(1+3+3^2\right)}{13}\)

\(\Rightarrow\frac{5^x\cdot31}{31}=\frac{3^{2x}\cdot13}{13}\)

\(\Rightarrow5^x=3^{2x}\)

Mà \(\left(5;3\right)=1\)

\(\Rightarrow x=2x=0\)