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

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

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

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

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

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

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

Dễ thấy \(x=2017\)không là nghiệm của phương trình.

Ta có:

\(\frac{1+\frac{x-2018}{2017-x}+\left(\frac{x-2018}{2017-x}\right)^2}{1-\frac{x-2018}{2017-x}+\left(\frac{x-2018}{2017-x}\right)}=\frac{13}{37}\)

Đặt \(\frac{x-2018}{2017-x}=a\)

\(\Rightarrow\frac{1+a+a^2}{1-a+a^2}=\frac{13}{37}\)

\(\Leftrightarrow24a^2+50a+24=0\)

\(\Leftrightarrow\orbr{\begin{cases}a=-\frac{3}{4}\\a=-\frac{4}{3}\end{cases}}\)

Ta có: \(\frac{a}{2017}=\frac{b}{2018}=\frac{c}{2019}.\)

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

\(\frac{a}{2017}=\frac{b}{2018}=\frac{c}{2019}=\frac{a-b}{2017-2018}=\frac{b-c}{2018-2019}=\frac{a-c}{2017-2019}.\)

\(\Rightarrow\frac{a-b}{-1}=\frac{b-c}{-1}=\frac{a-c}{-2}\)

\(\Rightarrow\frac{a-b}{-1}.\frac{b-c}{-1}=\left(\frac{a-c}{-2}\right)^2\)

\(\Rightarrow\frac{\left(a-b\right).\left(b-c\right)}{1}=\frac{\left(a-c\right)^2}{\left(-2\right)^2}\)

\(\Rightarrow\frac{\left(a-b\right).\left(b-c\right)}{1}=\frac{\left(a-c\right)^2}{4}.\)

\(\Rightarrow4.\left(a-b\right).\left(b-c\right)=\left(a-c\right)^2.1\)

\(\Rightarrow4.\left(a-b\right).\left(b-c\right)=\left(a-c\right)^2\left(đpcm\right).\)

Chúc bạn học tốt!

nhanh lên nhé sáng mai mình ktra rồi

Lời giải:

\(A=2018^2-2017.2019=2018^2-(2018-1)(2018+1)\)

\(=2018^2-(2018^2-1^2)=1\)

\(B=9^8.2^8-(18^4-1)(18^4+1)\)

\(=(9.2)^8-[(18^4)^2-1^2]\)

\(=18^8-(18^8-1)=1\)

\(C=163^2+74.163+37^2=163^2+2.37.163+37^2\)

\(=(163+37)^2=200^2=40000\)

\(D=\frac{2018^3-1}{2018^2+2019}=\frac{(2018-1)(2018^2+2018+1)}{2018^2+2019}\)

\(=\frac{2017(2018^2+2019)}{2018^2+2019}=2017\)

Sử dụng công thức \((a-b)(a+b)=a^2-b^2\)

\(E=(2+1)(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)-2^{32}\)

\(=(2-1)(2+1)(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)-2^{32}\)

\(=(2^2-1)(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)-2^{32}\)

\(=(2^4-1)(2^4+1)(2^8+1)(2^{16}+1)-2^{32}\)

\(=(2^8-1)(2^8+1)(2^{16}+1)-2^{32}\)

\(=(2^{16}-1)(2^{16}+1)-2^{32}\)

\(=(2^{32}-1)-2^{32}=-1\)

khá dài đó đợi chút nha

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

Ta có: \(2017-x=0\Leftrightarrow x=2017\)

          \(2018-x=0\Leftrightarrow x=2018\)

          \(2019-x=0\Leftrightarrow x=2019\)

Lập bảng xét dấu : 

+) Với \(x\le2017\Rightarrow\hept{\begin{cases}2017-x\ge0\\2018-x>0\\2019-x>0\end{cases}\Rightarrow\hept{\begin{cases}|2017-x|=2017-x\\|2018-x|=2018-x\\|2019-x|=2019-x\end{cases}\left(2\right)}}\)

Thay (2) vào(1) ta được : 

\(2017-x+2018-x+2019-x=2\)

\(6054-3x=2\)

\(3x=6052\)

\(x=\frac{6052}{3}>2017\)( loại )

+) Với \(2017< x\le2018\Rightarrow\hept{\begin{cases}2017-x< 0\\2018-x>0\\2019-x>0\end{cases}\Rightarrow\hept{\begin{cases}|2017-x|=x-2017\\|2018-x|=2018-x\\|2019-x|=2019-x\end{cases}\left(3\right)}}\)

Thay (3) vào (1) ta được :

\(x-2017+2018-x+2019-x=2\)

\(2020-x=2\)

\(x=2018\)( chọn )

+) Với \(2018< x\le2019\Rightarrow\hept{\begin{cases}2017-x< 0\\2018-x< 0\\2019-x\ge0\end{cases}\Rightarrow\hept{\begin{cases}|2017-x|=x-2017\\|2018-x|=x-2018\\|2019-x|=2019-x\end{cases}\left(4\right)}}\)

Thay (4) vào (1) ta được :

\(x-2017+x-2018+2019-x=2\)

\(x-2016=2\)

\(x=2018\)( loại )

+) Với \(x>2019\Rightarrow\hept{\begin{cases}2017-x< 0\\2018-x< 0\\2019-x< 0\end{cases}\Rightarrow\hept{\begin{cases}|2017-x|=x-2017\\|2018-x|=x-2018\\|2019-x|=x-2019\end{cases}\left(5\right)}}\)

Thay (5) vào (1) ta được :

\(x-2017+x-2018+x-2019=2\)

\(3x-6054=2\)

\(3x=6056\)

\(x=\frac{6056}{3}< 2019\)( loại )

Vậy x=2018

Lê Tài Bảo ChâuE thấy a phân 4 TH sai\(x< 2017\)

\(2017\le x\le2018\); \(2018\le x\le2019\)và \(x>2019\)

KQ là 2018 vẫn đúng nhưng hơi dài anh ạ,e giải lại ngắn hơn

Giải: Có \(\left|2018-x\right|\ge0\)

và \(\left|2017-x\right|+\left|2019-x\right|=\left|x-2017\right|+\left|2019-x\right|\)

\(\ge\left|2\right|=2\)

Dấu "="\(\Leftrightarrow\left(x-2017\right)\left(2019-x\right)\ge0\)và 2018-x = 0

\(\Leftrightarrow2017\le x\le2019\)và x= 2018

Vậy x = 2018