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\(\left(2017-x\right)^3+\left(2019-x\right)^3+\left(2x-4036\right)^3=0\)

\(\Leftrightarrow\left(2017-x\right)^3+\left(2019-x\right)^3+\left(2x-4036\right)^3=0^3\)

\(\Rightarrow\hept{\begin{cases}2017-x=0\\2019-x=0\\2x-4036=0\end{cases}\Rightarrow\hept{\begin{cases}x=2017\\x=2019\\x=2018\end{cases}}}\)

Vì x có 3 giá trị nên phương trình vô nghiệm

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 ...

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}

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

⇒ 2017-x=0                                                 ⇒ x= 2017

 ⇒ 2019-x=0                                               ⇒ x= 2019

 ⇒ 2x-4036=0                                             ⇒x= 2018

Vì x có 3 giá trị nên phương trình vô nghiệm.

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 \(S=\left\{2017;2018;2019\right\}\)

(2017-X)³+(2019-X)³+(2X-4036)³=0

<=>(2017-x).(2018-x).(2019-x)=0

<=>x=2017

x=2018

x=2019

#YQ

Câu 3b

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 \(\left(2017-x\right)^3+\left(2019-x\right)^3+\left(2x-4036\right)^3=0\)

\(\Rightarrow3\left(2017-x\right)\left(2019-x\right)\left(2x-4036\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2017-x=0\\2019-x=0\\2x-4036=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2017\\x=2019\\x=2018\end{matrix}\right.\)

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}

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 \(x-2020=0\Leftrightarrow x=2020\)

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) \(\dfrac{x+1}{2021}+\dfrac{x+2}{2020}+\dfrac{x+3}{2019}+\dfrac{x+4}{2018}=0\)

\(\Leftrightarrow\dfrac{x+1}{2021}+1+\dfrac{x+2}{2020}+1=\dfrac{x+3}{2019}+1+\dfrac{x+4}{2018}+1\)

\(\Leftrightarrow\dfrac{x+2022}{2021}+\dfrac{x+2022}{2020}=\dfrac{x+2022}{2019}+\dfrac{x+2022}{2018}\)

\(\Leftrightarrow\left(x+2022\right)\left(\dfrac{1}{2021}+\dfrac{1}{2020}+\dfrac{1}{2019}+\dfrac{1}{2018}\right)\)

\(\Leftrightarrow x+2022=0\)

\(\Leftrightarrow x=-2022\)

a)3(2x-3)=5x+1

⇔6x-9=5x+1

⇔6x-5x=1+9

⇔x=10

vậy phương trình có nghiệm là S={10}

b)\(\dfrac{x+1}{2021}\)+\(\dfrac{x+2}{2020}\)+\(\dfrac{x+3}{2019}\)+\(\dfrac{x+2028}{2}\)=0

⇔2020(x+1)+2021(x+2)+2041210(x+2028)=0

⇔2045251x+4139579942=0

⇔2045251x=-4139579942=0

⇔x=-\(\dfrac{4139579942}{2045251}\)

vậy phương trình có tập nghiệm là S={\(-\dfrac{4139579942}{2045251}\)}

\(\frac{x-2}{2017}+\frac{x-3}{2018}=\frac{x-4}{2019}+\frac{x-5}{2020}\)

<=> \(\frac{x-2}{2017}+1+\frac{x-3}{2018}+1=\frac{x-4}{2019}+1+\frac{x-5}{2020}+1\)

<=> \(\frac{x+2015}{2017}+\frac{x+2015}{2018}-\frac{x+2015}{2019}-\frac{x+2015}{2020}=0\)

<=> \(\left(x+2015\right)\left(\frac{1}{2017}+\frac{1}{2018}-\frac{1}{2019}-\frac{1}{2020}\right)=0\)

<=> x + 2015 = 0  ( vì \(\frac{1}{2017}+\frac{1}{2018}-\frac{1}{2019}-\frac{1}{2020}\ne0\)) 

<=> x = - 2015 

Vậy x = -2015.

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 \(\left(\frac{1}{2017}+\frac{1}{2018}-\frac{1}{2019}-\frac{1}{2020}\right)>0\)

\(\Rightarrow x+2015=0\)

\(\Rightarrow x=-2015\)

\(\frac{x-2}{2017}+\frac{x-3}{2018}=\frac{x-4}{2019}+\frac{x-5}{2020}\)

\(\Rightarrow\left(\frac{x-2}{2017}+1\right)+\left(\frac{x-3}{2018}+1\right)=\left(\frac{x-4}{2019}+1\right)+\left(\frac{x-5}{2020}+1\right)\)

\(\Rightarrow\frac{x-2+2017}{2017}+\frac{x-3+2018}{2018}=\frac{x-4+2019}{2019}+\frac{x-5+2020}{2020}\)

\(\Rightarrow\frac{x+2015}{2017}+\frac{x+2015}{2018}=\frac{x+2015}{2019}+\frac{x+2015}{2020}\)

\(\Rightarrow\frac{x+2015}{2017}+\frac{x+2015}{2018}-\frac{x+2015}{2019}-\frac{x+2015}{2020}=0\)

\(\Rightarrow\left(x+15\right)\left(\frac{1}{2017}+\frac{1}{2018}-\frac{1}{2019}-\frac{1}{2020}\right)=0\)

\(\Rightarrow x+2015=0\)

\(\Rightarrow x=-2015\)

Vậy x = - 2015