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4 tháng 10 2021
\(x=2021\Leftrightarrow x+1=2022\\ \Leftrightarrow P=x^5-\left(x+1\right)x^4+\left(x+1\right)x^3-\left(x+1\right)x^2+\left(x+1\right)x-x\\ P=x^5-x^5-x^4+x^4+x^3-x^3-x^2+x^2+x-x\\ P=0\)
4 tháng 10 2021
\(P=x^5-2022x^4+2022x^3-2022x^2+2022x-2021=x^4\left(x-2021\right)-x^3\left(x-2021\right)+x^2\left(x-2021\right)-x\left(x-2021\right)+\left(x-2021\right)\)
\(=\left(x-2021\right)\left(x^4-x^3+x^2-x+1\right)\)
\(=\left(2021-2021\right)\left(x^4-x^3+x^2-x+1\right)=0\)
18 tháng 2 2021
Ta có: \(x^3-5x^2+6x=0\)
\(\Leftrightarrow x\left(x^2-5x+6\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=3\end{matrix}\right.\)
Vậy: S={0;2;3}
MN
18 tháng 2 2021
\(x^3-5x^2+6x=0\)
\(\Leftrightarrow x^3-2x^2-3x^2+6x=0\)
\(\Leftrightarrow x^2\left(x-2\right)-3x\left(x-2\right)=0\)
\(\Leftrightarrow\left(x^2-3x\right)\left(x-2\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-2=0\\x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=3\end{matrix}\right.\)
\(S=\left\{0,2,3\right\}\)
25 tháng 2 2022
a,\(\left(x-4-5\right)\left(x-4+5\right)=0\Leftrightarrow\left(x-9\right)\left(x+1\right)=0\Leftrightarrow x=9;x=-1\)
b, \(\left(x-3-x-1\right)\left(x-3+x+1\right)=0\Leftrightarrow2x-2=0\Leftrightarrow x=1\)
c, \(\left(x^2-4\right)\left(2x-3\right)-\left(x^2-4\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(x^2-4\right)\left(2x-3-x+1\right)=0\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x-2\right)=0\Leftrightarrow x=-2;x=2\)
d, \(\left(3x-7\right)^2-\left(2x+2\right)^2=0\Leftrightarrow\left(3x-7-2x-2\right)\left(3x-7+2x+2\right)=0\)
\(\Leftrightarrow\left(x-9\right)\left(5x-5\right)=0\Leftrightarrow x=1;x=9\)
DT
2
7 tháng 3 2021
Ta có: \(\dfrac{x-25}{75}+\dfrac{x-15}{85}+\dfrac{x-5}{95}+\dfrac{x-145}{15}=0\)
\(\Leftrightarrow\dfrac{x-25}{75}-1+\dfrac{x-15}{85}-1+\dfrac{x-5}{95}-1+\dfrac{x-145}{15}+3=0\)
\(\Leftrightarrow\dfrac{x-100}{75}+\dfrac{x-100}{85}+\dfrac{x-100}{95}+\dfrac{x-100}{15}=0\)
\(\Leftrightarrow\left(x-100\right)\left(\dfrac{1}{75}+\dfrac{1}{85}+\dfrac{1}{95}+\dfrac{1}{15}\right)=0\)
mà \(\dfrac{1}{75}+\dfrac{1}{85}+\dfrac{1}{95}+\dfrac{1}{15}>0\)
nên x-100=0
hay x=100
Vậy: S={100}
TG
13 tháng 3 2022
gọi 2021-x = a
2023-x=b
2x-4044=c
ta có a + b + c=2021-x+2023-x+2x-4044=0
suy ra a + b = -c
suy ra (a+b)^3 =-c^3
ta có a^3 + b^3 + c^3=(a+b)^3 -3ab(a+b) + c^3 = -c^3 +3abc +c^3 = 3abc
ta có (2021-x)^3 + (2023-x)^3 + (2x-4044)^3 = 0
=> 3(2021-x)(2023-x)(2x-4044)=0
=> th 1 x = 2021, th 2 x = 2023; th3 x = 2022
NV
1
27 tháng 1 2022
sửa đề :
\(x^3-3x^2+3x-1=0\)
\(\Leftrightarrow\left(x-1\right)^3=0\Leftrightarrow x=1\)
28 tháng 7 2021
\(5x^2-3=0\Leftrightarrow x^2=\dfrac{3}{5}\Leftrightarrow x=\pm\sqrt{\dfrac{3}{5}}=\pm\dfrac{\sqrt{15}}{5}\)
\(4x^3+x=0\Leftrightarrow x\left(4x^2+1\right)=0\Leftrightarrow x=0;4x^2+1>0\)
LN
28 tháng 7 2021
\(5x^2-3=0\\ \Leftrightarrow5x^2=3\\ \Leftrightarrow x^2=\dfrac{3}{5}\\\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{\dfrac{3}{5}}\\x=-\sqrt{\dfrac{3}{5}}\end{matrix}\right. \)
vậy \(x=\sqrt{\dfrac{3}{5}}\) ;\(x=-\sqrt{\dfrac{3}{5}}\)
\(4x^3+x=0\\ \Leftrightarrow x\left(4x^2+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\4x^2+1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\4x^2=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2=\dfrac{-1}{4}\left(vl\right)\end{matrix}\right.\)
vậy x=0
\(PT\Leftrightarrow2022x^2+2022x-2021x-2021=0\)
\(\Leftrightarrow2022x\left(x+1\right)-2021\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(2022x-2021\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\2022x-2021=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{2021}{2022}\end{matrix}\right.\)
Vậy: \(S=\left\{-1;\dfrac{2021}{2022}\right\}\)
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