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\(\left|x-1\right|+2\left|x-2\right|+3\left|x-3\right|+4\left|x-4\right|+5\left|x-5\right|+20x=0\left(1\right)\)
TH1: x<1
(1) trở thành 1-x+2(2-x)+3(3-x)+4(4-x)+5(5-x)+20x=0
=>\(1-x+4-2x+9-3x+16-4x+25-5x+20x=0\)
=>\(5x+55=0\)
=>x=-11(nhận)
TH2: 1<=x<2
Phương trình (1) sẽ trở thành:
\(x-1+2\left(2-x\right)+3\left(3-x\right)+4\left(4-x\right)+5\left(5-x\right)+20x=0\)
=>\(x-1+4-2x+9-3x+16-4x+25-5x+20x=0\)
=>\(7x+53=0\)
=>\(x=-\dfrac{53}{7}\left(loại\right)\)
TH3: 2<=x<3
Phương trình (1) sẽ trở thành:
\(x-1+2\left(x-2\right)+3\left(3-x\right)+4\left(4-x\right)+5\left(5-x\right)+20x=0\)
=>\(x-1+2x-4+9-3x+16-4x+25-5x+20x=0\)
=>\(11x+45=0\)
=>\(x=-\dfrac{45}{11}\left(loại\right)\)
TH4: 3<=x<4
Phương trình (1) sẽ trở thành:
\(x-1+2\left(x-2\right)+3\left(x-3\right)+4\left(4-x\right)+5\left(5-x\right)+20x=0\)
=>\(x-1+2x-4+3x-9+16-4x+25-5x+20x=0\)
=>\(-3x+27=0\)
=>x=9(loại)
TH5: 4<=x<5
Phương trình (1) sẽ trở thành:
\(\left(x-1\right)+2\left(x-2\right)+3\left(x-3\right)+4\left(x-4\right)+5\left(5-x\right)+20x=0\)
=>\(x-1+2x-4+3x-9+4x-16+25-5x+20x=0\)
=>\(25x-5=0\)
=>x=1/5(loại)
TH6: x>=5
Phương trình (1) sẽ trở thành:
\(\left(x-1\right)+2\left(x-2\right)+3\left(x-3\right)+4\left(x-4\right)+5\left(x-5\right)+20x=0\)
=>\(x-1+2x-4+3x-9+4x-16+5x-25+20x=0\)
=>35x-55=0
=>x=55/35(loại)
`(x+3)^2014 = (x+3)^2012`
`(x+3)^2014 -(x+3)^2012 =0`
`(x+3)^2012 [(x+3)^2 -1]=0`
TH1 :`(x+3)^2012 =0 => x+3 =0 => x=-3`
TH2 :`(x+3)^2 -1 =0 => (x+3)^2 =1 => [(x+3=1),(x+3=-1):}`
`=> [(x=-2),(x=-4):}`
`(x+3)^2014 = (x+3)^2012`
`=> (x+3)^2014 - (x+3)^2012 = 0`
`=> (x+3)^2 * (x+3)^2012 - (x+3)^2012 = 0`
`=> (x+3)^2012 * [ (x+3)^2 - 1] =0`
`=>`\(\left[{}\begin{matrix}\left(x+3\right)^{2012}=0\\\left(x+3\right)^2-1=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x+3=0\\\left(x+3\right)^2=1\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=-3\\x+3=1\\x+3=-1\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=-3\\x=-2\\x=-4\end{matrix}\right.\)
Vậy, `x = {-3; -2; -4}.`
Ta có: \(\left(x+1\right)+\left(x+3\right)+\left(x+5\right)+...+\left(x+99\right)=0\)
\(\dfrac{\left[\left(x+1\right)+\left(x+99\right)\right].50}{2}=0\)
\(\left(x+50\right).50=0\)
\(x+50=0\)
\(x=-50\)
(4x - 3) - (x + 5) = 3(10 - x)
<=> 4x - 3 - x - 5 = 30 - 3x
<=> 4x -x + 3x = 3 + 5 + 30
<=> 6x = 38
<=> x = 19/3
\(\left(4x-3\right)-\left(x+5\right)=3\left(10-x\right)\)
\(4x-3-x-5=30-3x\)
\(3x-8=30-3x\)
\(3x+3x=30+8\)
\(6x=38\)
\(x=\frac{19}{3}\)
a)\(\Leftrightarrow\left[{}\begin{matrix}x^3-x-1=x^3+x+1\\x^3-x-1=-x^3-x-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2\left(x+1\right)=0\\2x^3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=0\end{matrix}\right.\)
câu b) tương tự
b) Ta có: \(\left|x^4+x^2+1\right|=x^2+x+1\)
\(\Leftrightarrow x^4+x^2+1=x^2+x+1\)
\(\Leftrightarrow x^4-x=0\)
\(\Leftrightarrow x\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
Ta có: \(3\left(x-\dfrac{1}{2}\right)-5\left(x+\dfrac{3}{5}\right)=-x+\dfrac{1}{15}\)
\(\Leftrightarrow3x-5x+x=\dfrac{1}{15}+\dfrac{3}{2}+3\)
\(\Leftrightarrow x=-\dfrac{137}{30}\)
\(\left(x-3\right)^3=x-3\)
\(\Leftrightarrow\left(x-3\right)^3-\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left[\left(x-3\right)^2-1\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\\left(x-3\right)^2-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=3\\\left(x-3\right)^2=1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=3\\x\in\left\{4;2\right\}\end{cases}}\)
Vậy \(x\in\left\{2;3;4\right\}\)