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a \ |x-2|+|x-5|=5x
==> x - 2 + x - 5=5x
x + x - 2 - 5 =5x
2x - 7 =5x
2x : x -7=5
x-7=5
x=5+7
x=12[22222222222222222222222222222222222222222 =]]]
hoac -(x-2)-(x-5)=5
-x+2-x+5=5
-x-x+2+5=5x
-2x+7=5x
-2x:x+7=5
-x+7=5
-x=5-7
-x=-2
==> x=2
vay x=12 hoac x=2
hinh nhu t sai cho nao do ;{
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\(\frac{x-4}{2021}+\frac{x-3}{2020}=\frac{x-2}{2019}+\frac{x-1}{2018}\)
\(\Leftrightarrow\left(\frac{x-4}{2021}+1\right)+\left(\frac{x-3}{2020}+1\right)=\left(\frac{x-2}{2019}+1\right)+\left(\frac{x-1}{2018}+1\right)\)
\(\Leftrightarrow\frac{x+2017}{2021}+\frac{x+2017}{2020}=\frac{x+2017}{2019}+\frac{x+2017}{2018}\)
\(\Leftrightarrow\frac{x+2017}{2021}+\frac{x+2017}{2020}-\frac{x+2017}{2019}-\frac{x+2017}{2018}=0\)
\(\Leftrightarrow\left(x+2017\right)\left(\frac{1}{2021}+\frac{1}{2020}-\frac{1}{2019}-\frac{1}{2018}\right)=0\)
Mà \(\left(\frac{1}{2021}+\frac{1}{2020}-\frac{1}{2019}-\frac{1}{2018}\right)\ne0\)
\(\Leftrightarrow x+2017=0\)
\(\Leftrightarrow x=-2017\)
Vậy ..
=> (x-4/2021 +1) + (x-3/2020 +1) = (x-2/2019 +1)+ (x-1/2018 +1)
=> x+2017/2021 + x+2017/2020 = x+2017/2019 + x+2017/2018
=> x+2017/2018 + x+2017/2018 - x+2017/2020 - x+2017/2021 = 0
=> (x+2017).(1/2018+1/2019+1/2020+1/2021) = 0
=> x+2017 = 0 ( vì 1/2018+1/2019+1/2020+1/2021 > 0 )
=> x=-2017
Vậy x=-2017
k mk nha
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\(a,Taco:\)
\(\left(x-1\right)^2,\left(y-3\right)^8\ge0\)
\(\Rightarrow\left(x-1\right)^2+\left(y-3\right)^8=0\Leftrightarrow\hept{\begin{cases}x-1=0\Leftrightarrow x=1\\y-3=0\Leftrightarrow y=3\end{cases}}\)
\(b,Taco:\)
\(|x-2018|+\left(y-2019\right)^{2018}\ge0\)
\(\Rightarrow|x-2018|+\left(y-2019\right)^{2018}=0\Leftrightarrow\hept{\begin{cases}x-2018=0\Leftrightarrow x=2018\\y-2019=0\Leftrightarrow y=2019\end{cases}}\)
\(a,\left(x-1\right)^2+\left(y-3\right)^8=0\)
Vì \(\left(x-1\right)^2\ge0vs\forall x;\left(y-3\right)^8\ge0vs\forall y\)
\(\Rightarrow\hept{\begin{cases}\left(x-1\right)^2=0\\\left(y-3\right)^8=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x-1=0\\y-3=0\end{cases}}\) \(\Rightarrow\hept{\begin{cases}x=1\\y=3\end{cases}}\)
Vậy x = 1, y = 3
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b, tìm x,y biết |x-2018|+|y+2019|=0
\(\Rightarrow\hept{\begin{cases}|x-2018|=0\\|y+2019|=0\end{cases}}\Rightarrow\hept{\begin{cases}x-2018=0\\y+2019=0\end{cases}}\Rightarrow\hept{\begin{cases}x=2018\\y=-2019\end{cases}}\)
vậy x=2018 ; y=-2019
a)
ta có \(\hept{\begin{cases}\left|x\right|\ge0\\\left|y+1\right|\ge0\end{cases}}\Rightarrow\left|x\right|+\left|y+1\right|\ge0\Rightarrow A_{min}=0\Leftrightarrow\hept{\begin{cases}x=0\\y+1=0\end{cases}}\Rightarrow\hept{\begin{cases}x=0\\y=-1\end{cases}}\)
b)
ta có \(\hept{\begin{cases}\left|x-2018\right|\ge0\\\left|y+2019\right|\ge0\end{cases}}\)
mà \(\left|x-2018\right|+\left|y+2019\right|=0\Rightarrow\hept{\begin{cases}x-2018=0\\y+2019=0\end{cases}\Rightarrow\hept{\begin{cases}x=2018\\y=-2019\end{cases}}}\)
Lời giải:
\(|x-2019|-|x-1|=0\)
\(\Leftrightarrow |x-2019|=|x-1|\)
\(\Rightarrow \left[\begin{matrix} x-2019=x-1\\ x-2019=-(x-1)=1-x\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} 2019=1(\text{vô lý})\\ x=1010\end{matrix}\right.\)
Vậy $x=1010$