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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
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 ...
\(\left(2017-x\right)^3+\left(2019-x\right)^3+\left(2x-4036\right)^3=0\)
\(\Leftrightarrow\left(2017-x\right)^3+\left(2019-x\right)^3+\left(2x-4036\right)^3=0^3\)
\(\Rightarrow\hept{\begin{cases}2017-x=0\\2019-x=0\\2x-4036=0\end{cases}\Rightarrow\hept{\begin{cases}x=2017\\x=2019\\x=2018\end{cases}}}\)
Vì x có 3 giá trị nên phương trình vô nghiệm
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....
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) x( x + 2018 ) - 2x - 4036 = 0
<=> x( x + 2018 ) - 2( x + 2018 ) = 0
<=> ( x + 2018 )( x - 2 ) = 0
<=> \(\orbr{\begin{cases}x+2018=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-2018\\x=2\end{cases}}\)
b) x + 5 = 2( x + 5 )2
<=> x + 5 = 2( x2 + 10x + 25 )
<=> x + 5 = 2x2 + 20x + 50
<=> 2x2 + 20x + 50 - x - 5 = 0
<=> 2x2 + 19x + 45 = 0
<=> 2x2 + 10x + 9x + 45 = 0
<=> 2x( x + 5 ) + 9( x + 5 ) = 0
<=> ( x + 5 )( 2x + 9 ) = 0
<=> \(\orbr{\begin{cases}x+5=0\\2x+9=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-5\\x=-\frac{9}{2}\end{cases}}\)
c) ( x2 + 1 )( 2x - 1 ) + 2x = 1
<=> 2x3 - x2 + 4x - 1 - 1 = 0
<=> 2x3 - x2 + 4x - 2 = 0
<=> x2( 2x - 1 ) + 2( 2x - 1 ) = 0
<=> ( 2x - 1 )( x2 + 2 ) = 0
<=> \(\orbr{\begin{cases}2x-1=0\\x^2+2=0\end{cases}\Leftrightarrow}x=\frac{1}{2}\)( vì x2 + 2 ≥ 2 > 0 ∀ x )
d) \(\frac{x}{3}-\frac{x^2}{4}=0\)
\(\Leftrightarrow\frac{4x}{12}-\frac{3x^2}{12}=0\)
\(\Leftrightarrow\frac{4x-3x^2}{12}=0\)
\(\Leftrightarrow4x-3x^2=0\)
\(\Leftrightarrow x\left(4-3x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\4-3x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{4}{3}\end{cases}}\)
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