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19 tháng 2 2020
|x+19|+|x+5|+|x+2020|=5x(*)
+)Ta có:|x+19|\(\ge\)0;|x+5|\(\ge\)0;|x+2020|\(\ge\)0
=>VT(*)=|x+19|+|x+5|+|x+2020|\(\ge\)0
Mà |x+19|+|x+5|+|x+2020|=5x
=>5x\(\ge\)0
=>x\(\ge\)0
+)Ta lại có:x\(\ge\)0=>x+19\(\ge\)19=>|x+19|=x+19
x\(\ge\)0=>x+5\(\ge\)5=>|x+5|=x+5
x\(\ge\)0=>x+2020\(\ge\)2020=>|x+2020|=x+2020
=>VT(*)=x+19+x+5+x+2020=5x
x+x+x+19+5+2020=5x
3x+2044 =5x
2044 =5x-3x
2044 =2x
=> 2x =2044
x =\(\frac{2044}{2}=1022\)\(\in\)Z
Vậy x=1022
Chúc bn học tốt
MQ
14 tháng 5 2019
neu 2020-x\(\ge\)0 thi |2020-x|=2020-x
\(\Leftrightarrow\)x\(\le\)2020
ta co phuong trinh :
2020-x+9x=2019
\(\Leftrightarrow\)8x=2020-2019
\(\Leftrightarrow\)8x=1
\(\Leftrightarrow\)x=1/8(thoa man dk x\(\le\)2020)
neu 2020-x<0 thi |2020-x|=x-2020
\(\Leftrightarrow\)x>2020
ta co phuong trinh :
x-2020+9x=2019
\(\Leftrightarrow\)10x=2019+2020
\(\Leftrightarrow\)10x=4039
\(\Leftrightarrow\)x=403,9(thoa man dk x>2020)
vay s={1/8,403,9}
chuc bn hoc tot
MQ
14 tháng 5 2019
mk xin loi nha mk sai cho dau mk chuyen dau nham .mk lam lai nha:
ta co phuong trinh : 2020-x+9x=2019
\(\Leftrightarrow\)8x=-2020+2019
\(\Leftrightarrow\)8x=-1
\(\Leftrightarrow\)x=-1/8(thoa man dieu kien x\(\le\)2020)
vay s={-1/8;403,9}
NL
Nguyễn Lê Phước Thịnh
CTVHS
21 tháng 12 2021
=>|2x+2020|=2
\(\Leftrightarrow\left[{}\begin{matrix}2x=-2022\\2x=-2018\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1011\\x=-1009\end{matrix}\right.\)
DV
18 tháng 1 2022
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AF
1
10 tháng 3 2020
a) Ta có: x-7=x-5-2
Để x-7 chia hết cho x-5 thì x-5-2 chia hết cho x-5
=> 2 chia hết cho x-5
Mà x nguyên => x-5 nguyên
=> x-5 thuôc Ư (2)={-2;-1;1;2}
Ta có bảng
| x-5 | -2 | -1 | 1 | 2 |
| x | -3 | 4 | 6 | 7 |
b) x2-5x=0
<=> x(x-5)=0
<=> x=0 hoặc x-5=0
<=> x=0 hoặc x=5
Vậy x=0; x=5
TN
2
30 tháng 8 2021
\(2021x\left(x-2020\right)-x+2020=0\)
\(\Rightarrow2021x\left(x-2020\right)-\left(x-2020\right)=0\)
\(\Rightarrow\left(x-2020\right)\left(2021x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-2020=0\\2021x-1=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=2020\\x=\dfrac{1}{2021}\end{matrix}\right.\)
NL
Nguyễn Lê Phước Thịnh
CTVHS
30 tháng 8 2021
Ta có: \(2021x\left(x-2020\right)-x+2020=0\)
\(\Leftrightarrow\left(x-2020\right)\left(2021x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2020\\x=\dfrac{1}{2021}\end{matrix}\right.\)
CI
1
10 tháng 10 2021
\(a,\Rightarrow x^2+4x+4+x^2-2x+1+x^2-9-3x^2=-8\\ \Rightarrow2x=-4\\ \Rightarrow x=-2\\ b,\Rightarrow2021x\left(x-2020\right)-\left(x-2020\right)=0\\ \Rightarrow\left(2021x-1\right)\left(x-2020\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-2020=0\\2021x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2020\\x=\dfrac{1}{2021}\end{matrix}\right.\)
CI
1
12 tháng 10 2021
a) \(\Rightarrow x^2+4x+4+x^2-2x+1+x^2-9-3x^2=-8\)
\(\Rightarrow2x=-4\Rightarrow x=-2\)
b) \(\Rightarrow2021x\left(x-2020\right)-\left(x-2020\right)=0\)
\(\Rightarrow\left(x-2020\right)\left(2021x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=2020\\x=\dfrac{1}{2021}\end{matrix}\right.\)
Câu a:
(x - 2020)^(x+ 1) - (x - 2020)^(x + 11) = 0
(x - 2020)^(x+1).[1 - (x -2020)^10 ] = 0
x - 2020 = 0 hoặc (x - 2020)^10 = 1
x - 2020 = 0
x = 2020
(x - 2020)^10 = 1
x - 2020 = - 1 hoặc x - 2020 = 1
x - 2020 = - 1
x = - 1 + 2020
x = 2019
x - 2020 = 1
x = 2020 + 1
x = 2021
Vậy x ∈ {2019; 2020; 2021}
Bài 8
a)
\(\left(\right. x - 2020 \left.\right)^{x + 1} - \left(\right. x - 2020 \left.\right)^{x + 11} = 0\) \(\left(\right. x - 2020 \left.\right)^{x + 1} \left[\right. 1 - \left(\right. x - 2020 \left.\right)^{10} \left]\right. = 0\) \(\Rightarrow x-2020=0\text{ ho}ặ\text{c }\left(\right.x-2020\left.\right)^{10}=1\) \(\Rightarrow x=2020\text{ ho}ặ\text{c }x=2019,\textrm{ }2021\) \(\boxed{x = 2019 , \textrm{ } 2020 , \textrm{ } 2021}\)
b)
\(\left(\right. x - 2015 \left.\right)^{x + 2} - \left(\right. x - 2015 \left.\right)^{x + 12} = 0\) \(\left(\right. x - 2015 \left.\right)^{x + 2} \left[\right. 1 - \left(\right. x - 2015 \left.\right)^{10} \left]\right. = 0\) \(\Rightarrow x-2015=0\text{ ho}ặ\text{c}\left(\right.x-2015\left.\right)^{10}=1\) \(\Rightarrow x=2015\text{ ho}ặcx=2014,\textrm{ }2016\) \(\boxed{x = 2014 , \textrm{ } 2015 , \textrm{ } 2016}\)
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