Giải phương trình sau: x^2 - x - 2018*2019.
Những câu hỏi liên quan
Giải phương trình : 4-x/2018-2=3-x/2019-x/1011
Giải phương trình
2-x/2017+1=x-1/2018-x/2019
Giải phương trình .x-2/2017+x-3/2018=x-4/2019+x-5/2020
\(\frac{x-2}{2017}+\frac{x-3}{2018}=\frac{x-4}{2019}+\frac{x-5}{2020}\)
<=> \(\frac{x-2}{2017}+1+\frac{x-3}{2018}+1=\frac{x-4}{2019}+1+\frac{x-5}{2020}+1\)
<=> \(\frac{x+2015}{2017}+\frac{x+2015}{2018}-\frac{x+2015}{2019}-\frac{x+2015}{2020}=0\)
<=> \(\left(x+2015\right)\left(\frac{1}{2017}+\frac{1}{2018}-\frac{1}{2019}-\frac{1}{2020}\right)=0\)
<=> x + 2015 = 0 ( vì \(\frac{1}{2017}+\frac{1}{2018}-\frac{1}{2019}-\frac{1}{2020}\ne0\))
<=> x = - 2015
Vậy x = -2015.
Giải phương trình :
\(\frac{x-2}{2017}+\frac{x-3}{2018}=\frac{x-4}{2019}+\frac{x-5}{2020}\)
\(\Rightarrow\frac{x-2}{2017}+1+\frac{x-3}{2018}+1=\frac{x-4}{2019}+1+\frac{x-5}{2020}+1\)
\(\Rightarrow\frac{x+2015}{2017}+\frac{x+2015}{2018}-\frac{x+2015}{2019}-\frac{x+2015}{2020}=0\)
\(\Rightarrow\left(x+2015\right)\left(\frac{1}{2017}+\frac{1}{2018}-\frac{1}{2019}-\frac{1}{2020}\right)=0\)
Mà \(\left(\frac{1}{2017}+\frac{1}{2018}-\frac{1}{2019}-\frac{1}{2020}\right)>0\)
\(\Rightarrow x+2015=0\)
\(\Rightarrow x=-2015\)
\(\frac{x-2}{2017}+\frac{x-3}{2018}=\frac{x-4}{2019}+\frac{x-5}{2020}\)
\(\Rightarrow\left(\frac{x-2}{2017}+1\right)+\left(\frac{x-3}{2018}+1\right)=\left(\frac{x-4}{2019}+1\right)+\left(\frac{x-5}{2020}+1\right)\)
\(\Rightarrow\frac{x-2+2017}{2017}+\frac{x-3+2018}{2018}=\frac{x-4+2019}{2019}+\frac{x-5+2020}{2020}\)
\(\Rightarrow\frac{x+2015}{2017}+\frac{x+2015}{2018}=\frac{x+2015}{2019}+\frac{x+2015}{2020}\)
\(\Rightarrow\frac{x+2015}{2017}+\frac{x+2015}{2018}-\frac{x+2015}{2019}-\frac{x+2015}{2020}=0\)
\(\Rightarrow\left(x+15\right)\left(\frac{1}{2017}+\frac{1}{2018}-\frac{1}{2019}-\frac{1}{2020}\right)=0\)
\(\Rightarrow x+2015=0\)
\(\Rightarrow x=-2015\)
Vậy x = - 2015
Xem thêm câu trả lời
Giải phương trình: \(|x-2017|+|2x-2018|+|3x-2019|=x-2020\)
Nhận thấy vế trái luôn dương nên \(x-2020\ge0\Leftrightarrow x\ge2020\)
Với \(x\ge2020\Rightarrow\left\{{}\begin{matrix}x-2017\ge0\\2x-2018\ge0\\3x-2019\ge0\end{matrix}\right.\)
PT trở thành: \(x-2017+2x-2018+3x-2019=x-2020\)
Hay kết hợp với điều kiện \(x=\dfrac{4034}{5}\) suy ra PT đã cho vô nghiệm
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Bình luận (2)
\(\left|x-2017\right|+\left|2x-2018\right|+\left|3x-2019\right|=x-2020\)
\(ĐK:x\ge2020\)
\(\Leftrightarrow x-2017+2x-2018+3x-2019=x-2020\)
\(\Leftrightarrow5x=4034\)
\(\Leftrightarrow x=806,8\left(tm\right)\)
Vậy \(S=\left\{806,8\right\}\)
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Bình luận (2)
giải phương trình \(|x-2017|^{2018}+|x-2018|^{2019}=1.\)
1. Giải phương trình: |2x-3|+|x-2|=7
2. Tìm x: |x-1|^2018+|x+2|^2019=1
Giải phương trình: |x-2017|+|2x-2018|+|3x-2019|=x-2020
Bài: giải các phương trình sau:
a/2x(27x^2-8)+4(2x-6)(2x+6)-(3x-4)(5x+2)=2(3x-4)(9x^2+12x+16).
b/ 4-x/2018-2=3-x/2019-x/1011
Giải các phương trình:
\(\dfrac{x+1}{2019}+\dfrac{x+2}{2018}=\dfrac{x+2017}{3}+\dfrac{x+2016}{4}\)
\(\frac{x+1}{2019}+\frac{x+2}{2018}=\frac{x+2017}{3}+\frac{x+2016}{4}\)
\(\Leftrightarrow\frac{x+1}{2019}+1+\frac{x+2}{2018}+1=\frac{x+2017}{3}+1+\frac{x+2016}{4}+1\)
\(\Leftrightarrow\frac{x+2020}{2019}+\frac{x+2020}{2018}-\frac{x+2020}{3}-\frac{x+2020}{4}=0\)
\(\Leftrightarrow\left(x+2020\right).\left(\frac{1}{2019}+\frac{1}{2018}-\frac{1}{3}-\frac{1}{4}\right)=0\)
Mà \(\left(\frac{1}{2019}+\frac{1}{2018}-\frac{1}{3}-\frac{1}{4}\right)\ne0\)
\(\Rightarrow x+2020=0\Leftrightarrow x=-2020\)
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