Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(|x^{2018}+|x-1||=x^{2018}+2404\)
\(\Leftrightarrow\orbr{\begin{cases}x^{2018}+|x-1|=-x^{2018}-2404\\x^{2018}+|x-1|=x^{2018}+2404\end{cases}\Leftrightarrow\orbr{\begin{cases}|x-1|=-\left(2x^{2018}+2404\right)\left(l\right)\\|x-1|=2404\left(n\right)\end{cases}}}\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=-2404\\x-1=2404\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-2403\\x=2405\end{cases}}}\)
V...
\(\left|x^{2018}+\left|x-1\right|\right|=x^{2018}+2404\)
Ta thấy: \(x^{2018}\ge0\);\(\left|x-1\right|\ge0\)\(\Rightarrow x^{2018}+\left|x-1\right|\ge0\)
\(\Rightarrow\left|x^{2018}+\left|x-1\right|\right|=x^{2018}+2404\)
\(\Leftrightarrow x^{2018}+\left|x-1\right|=x^{2018}+2404\)
\(\left|x-1\right|=2404\)
\(\Rightarrow\orbr{\begin{cases}x-1=2404\\x-1=-2404\end{cases}}\Rightarrow\orbr{\begin{cases}x=2405\\x=-2403\end{cases}}\)
Vậy \(x\in\left\{2405;-2403\right\}\)
\(|x^{2018}+|x+1||=x^{2018}+2404\)
\(\Rightarrow x^{2018}+\left|x+1\right|=x^{2018}+2404\)
\(\Rightarrow\left|x+1\right|=2404\)
\(\Rightarrow\orbr{\begin{cases}x+1=2404\\x+1=-2404\end{cases}\Rightarrow}\orbr{\begin{cases}x=2403\\x=-2405\end{cases}}\)
\(\frac{x+2}{2018}+\frac{x+3}{2017}+\frac{x+4}{2016}=-3\)
\(\frac{x+2}{2018}+1+\frac{x+3}{2017}+1+\frac{x+4}{2016}+1=0\)
\(\frac{x+2+2018}{2018}+\frac{x+3+2017}{2017}+\frac{x+4+2016}{2016}=0\)
\(\frac{x+2020}{2018}+\frac{x+2020}{2017}+\frac{x+2020}{2016}=0\)
\(\left(x+2020\right)\left(\frac{1}{2018}+\frac{1}{2017}+\frac{1}{2016}\right)=0\)
\(\Rightarrow x+2020=0\)
\(\Leftrightarrow x=-2020\)
#Sakura
\(\frac{x+2}{2018}+\frac{x+3}{2017}+\frac{x+4}{2016}=-\overrightarrow{3}\)
=>\(\frac{x+2}{2018}+1+\frac{x+3}{2017}+1+\frac{x+4}{2016}+1=0\)
=>\(\frac{x+2020}{2018}+\frac{x+2020}{2017}+\frac{x+2020}{2016}=0\)
=>\(\left(x+2020\right):\left(\frac{1}{2018}+\frac{1}{2017}+\frac{1}{2016}\right)=0\)
=>\(\left(x+2020\right)=0\)
=>\(x=0-2020\)
=>\(x=-2020\)
vậy ....
chúc bạn học tốt!
3. Tìm x biết: |15-|4.x||=2019
\(\Rightarrow\orbr{\begin{cases}15-\left|4x\right|=2019\\15-\left|4x\right|=-2019\end{cases}\Rightarrow\orbr{\begin{cases}\left|4x\right|=-2004\\\left|4x\right|=2034\end{cases}}}\)
vì \(4x\ge0\)\(\Rightarrow\)|4x|=2043\(\Rightarrow4x=2034\Rightarrow x=508,5\)
KL: x=508,5
b) để \(\left(x-7\right)^{x+2015}-\left(x-7\right)^{x+2016}=0\)
thì \(\left(x-7\right)^{x+2015}=\left(x-7\right)^{x+2016}\)
mà \(x+2015
nên \(x-7=x-7\Rightarrow x=7\)
bài a)
|2x+3|=x+2
2x+3=x+2 hoặc -(2x+3)=x+2
2x-x=2-3 -2x-3=x+2
1x=-1 -2x-x=2+3
x=-1 -3x =5
x=\(\frac{-5}{3}\)
1) ADTCDTSBN
có: \(\frac{x}{3}=\frac{y}{5}=\frac{z}{-7}=\frac{x-y-z}{3-5+7}=\frac{20}{5}=4.\)
=> ...
\(\left|x^{2018}+|x-1|\right|=x^{2018}+2404\)
\(\Leftrightarrow x^{2018}+\left|x-1\right|=x^{2018}+2404\)
\(\Leftrightarrow\left|x-1\right|=2404\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=2404\\x-1=-2404\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2405\\x=-2403\end{cases}}}\)