Giải phương trình:
(2017 - x)3 + (2019 - x)3 + (2x - 4036)3 = 0
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W
2
10 tháng 3 2019
\(\left(2017-x\right)^3+\left(2019-x\right)^3+\left(2x-4036\right)^3=0\)
\(\Leftrightarrow\left(2017-x\right)^3+\left(2019-x\right)^3+\left(2x-4036\right)^3=0^3\)
\(\Rightarrow\hept{\begin{cases}2017-x=0\\2019-x=0\\2x-4036=0\end{cases}\Rightarrow\hept{\begin{cases}x=2017\\x=2019\\x=2018\end{cases}}}\)
Vì x có 3 giá trị nên phương trình vô nghiệm
WM
1
KB
10 tháng 3 2019
Đặt \(2017-x=a;2019-x=b;2x-4036=c\)
\(\Rightarrow a+b+c=0\)
Do \(a+b+c=0\Rightarrow a+b=-c\Leftrightarrow\left(a+b\right)^3=-c^3\)
Có : \(a^3+b^3+c^3=\left(a+b\right)^3-3ab\left(a+b\right)+c^3=-c^3-3ab.\left(-c\right)+c^3=3abc\)
Do \(\left(2017-x\right)^3+\left(2019-x\right)^3+\left(2x-4036\right)^3=0\)
\(\Rightarrow3\left(2017-x\right)\left(2019-x\right)\left(2x-4036\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2017-x=0\\2019-x=0\\2x-4036=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2017\\x=2019\\x=2018\end{matrix}\right.\)
Vậy ...
A
3
18 tháng 3 2021
Đặt 2017-x=a; 2019-x=b
\(\Leftrightarrow a+b=4036-2x\)
\(\Leftrightarrow-\left(a+b\right)=2x-4036\)
Phương trình trở thành: \(a^3+b^3-\left(a+b\right)^3=0\)
\(\Leftrightarrow\left(a+b\right)^3-3ab\left(a+b\right)-\left(a+b\right)^3=0\)
\(\Leftrightarrow-3ab\left(a+b\right)=0\)
mà -3<0
nên \(ab\left(a+b\right)=0\)
\(\Leftrightarrow\left(2017-x\right)\left(2019-x\right)\left(4036-2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2017-x=0\\2019-x=0\\4036-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2017\\x=2019\\x=2018\end{matrix}\right.\)
Vậy: S={2017;2018;2019}
TM
18 tháng 3 2021
Cho \(\left(2017-x\right)^3=x;\left(2019-x\right)^3=y;\left(2x-4036\right)^3=z\)
Ta có: \(x+y+z=0\)
\(=>x+y=-z\) \(=>\left(x+y\right)^3=-z^3\)
Ta có: \(x^3+y^3+z^3=\left(x+y\right)^3-3xy\left(x+y\right)+z^3=-z^3-3xy\left(-z\right)+z^3=3xyz\)
Vì (2017-x)3 + (2019-x)3 + (2x-4036)3 =0
=>\(3\left(2017-x\right)\left(2019-x\right)\left(2x-4036\right)=0\)
Gải phương trình được x=2017; x=2019; x=2018
H9
HT.Phong (9A5)
CTVHS
6 tháng 3 2023
Đặt \(2017-x=m,2019-x=n\)
\(\rightarrow m+n=2x-4036\)
Phương trình ban đầu trở thành :
\(m^3+n^3=\left(m+n\right)^3\)
\(\rightarrow3mn.\left(m+n\right)^3=0\)
\(\rightarrow\left(2017-x\right)\left(2019-x\right)\left(2x-4036\right)=0\)
\(\rightarrow\left[{}\begin{matrix}x=2017\\x=2018\\x=2019\end{matrix}\right.\)
Vậy \(S=\left\{2017;2018;2019\right\}\)
NB
6 tháng 3 2023
(2017-X)3+(2019-X)3+(2X-4036)3=0
<=>(2017-x).(2018-x).(2019-x)=0
<=>x=2017
x=2018
x=2019
#YQ
17 tháng 3 2019
Bài 2:
Đặt \(2017-x=a;2019-x=b;2x-4036=c\)
\(\Rightarrow a+b+c=0\)
Do \(a+b+c=0\Rightarrow a+b=-c\Leftrightarrow\left(a+b\right)^3=-c^3\)
Có : \(a^3+b^3+c^3=\left(a+b\right)^3-3ab\left(a+b\right)+c^3=-c^3-3ab.\left(-c\right)+c^3=3abc\)
Do \(\left(2017-x\right)^3+\left(2019-x\right)^3+\left(2x-4036\right)^3=0\)
\(\Rightarrow3\left(2017-x\right)\left(2019-x\right)\left(2x-4036\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2017-x=0\\2019-x=0\\2x-4036=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2017\\x=2019\\x=2018\end{matrix}\right.\)
HN
19 tháng 2 2019
a) \(22-x\left(1-4x\right)=\left(2x+3\right)^3\)
\(\Leftrightarrow22-x+4x^2=8x^3+36x^2+54x+27\)
\(\Leftrightarrow-x-54x+4x^2-36x^2-8x^3=-22+27\)
\(\Leftrightarrow-8x^3-32x^2-55x=5\Leftrightarrow-8x^3-32x^2-55x-5=0\)
Bn tự làm tiếp nhé
b) \(\frac{2x}{3}+\frac{2x-1}{6}=\frac{4-x}{3}\Leftrightarrow\frac{2.2x}{6}+\frac{2x-1}{6}=\frac{2\left(4-x\right)}{6}\)
\(\Leftrightarrow2.2x+2x-1=2\left(4-x\right)\Leftrightarrow4x+2x-1=8-2x\)
\(\Leftrightarrow6x-1=8-2x\Leftrightarrow8x=9\Leftrightarrow x=\frac{9}{8}\)
Vậy phương trình có tập nghiệm S ={9/8}
T
19 tháng 2 2019
c) \(\frac{x-1}{2019}+\frac{x-2}{2018}=\frac{x-3}{2017}+\frac{x-4}{2016}\)
\(\Leftrightarrow\left(\frac{x-1}{2019}-1\right)+\left(\frac{x-2}{2018}-1\right)=\left(\frac{x-3}{2017}-1\right)+\left(\frac{x-4}{2016}-1\right)\)
\(\Leftrightarrow\frac{x-2020}{2019}+\frac{x-2020}{2018}-\frac{x-2020}{2017}-\frac{x-2020}{2016}=0\)
\(\Leftrightarrow\left(x-2020\right)\left(\frac{1}{2019}+\frac{1}{2018}+\frac{1}{2017}+\frac{1}{2016}\right)=0\)
Do \(\frac{1}{2019}+\frac{1}{2018}+\frac{1}{2017}+\frac{1}{2016}>0\)
Nên \(x-2020=0\Leftrightarrow x=2020\)
H9
HT.Phong (9A5)
CTVHS
12 tháng 3 2023
a) \(3\left(2x-x\right)=5x+1\)
\(\Leftrightarrow6x-3x=5x+1\)
\(\Leftrightarrow6x-3x-5x=1\)
\(\Leftrightarrow-2x=1\)
\(\Leftrightarrow x=\dfrac{1}{-2}=-\dfrac{1}{2}\)
b) \(\dfrac{x+1}{2021}+\dfrac{x+2}{2020}+\dfrac{x+3}{2019}+\dfrac{x+4}{2018}=0\)
\(\Leftrightarrow\dfrac{x+1}{2021}+1+\dfrac{x+2}{2020}+1=\dfrac{x+3}{2019}+1+\dfrac{x+4}{2018}+1\)
\(\Leftrightarrow\dfrac{x+2022}{2021}+\dfrac{x+2022}{2020}=\dfrac{x+2022}{2019}+\dfrac{x+2022}{2018}\)
\(\Leftrightarrow\left(x+2022\right)\left(\dfrac{1}{2021}+\dfrac{1}{2020}+\dfrac{1}{2019}+\dfrac{1}{2018}\right)\)
\(\Leftrightarrow x+2022=0\)
\(\Leftrightarrow x=-2022\)
20 tháng 4 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.
20 tháng 4 2020
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 \(\left(\frac{1}{2017}+\frac{1}{2018}-\frac{1}{2019}-\frac{1}{2020}\right)>0\)
\(\Rightarrow x+2015=0\)
\(\Rightarrow x=-2015\)
nhận thấy (2017 - x) + (2019 -x) + (2x-4036) = 0
gọi 2017 - x = a ; 2019-x = b và 2x-4036 = c
có a+b+c=0 (=) a+b=-c (=) a3+b3+3ab.(a+b) = -c3 (=) a3+b3+c3 = 3abc (vì a+b=-c)
hay (2017 - x)3 + (2019 -x)3 + (2x-4036)3 = 3.(2017 - x).(2019 -x).(2x-4036) (1)
mà theo đề bài (2017 - x)3 + (2019 -x)3 + (2x-4036)3 =0 (2)
từ (1) và (2) =) 3.(2017 - x).(2019 -x).(2x-4036) =0
=) 2017 - x=0 hoặc 2019 -x=0 hoặc 2x-4036=0
(=) x=2017 hoặc x=2019 hoặc x=2018
vậy....