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Tìm x biết:
5. ( x-1 ) - 7.( x-2 ) = 2x -39
Tìm x thuộc Z biết:
x - 3 - 14.( x-2 )= -3x -3
\(3x+7⋮x-2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
5 ( x - 1 ) - 7 ( x - 2 ) = 2x - 39
<=> 5x - 5 - 7x + 14 = 2x - 39
<=> 5x - 7x - 2x = -39 + 5 - 14
<=> -4x = -48
<=> x = 12
x - 3 - 14.( x-2 )= -3x -3\(\Rightarrow\chi-3-28-14\chi-28=-3\chi-3\)
\(\Rightarrow\chi-3-28+3=-3\chi-3\)
\(\Rightarrow\chi-28=11\chi\)
\(\Rightarrow\chi-11\chi=28\)
\(\Rightarrow10\chi=28\Rightarrow\chi=2,8\left(kot.m\chi\inℤ\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1,\\ \left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\\ \Leftrightarrow\left(x-7\right)^{x+1}\left[1-\left(x-7\right)^{10}\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}\left(x-7\right)^{x+1}=0\\\left(x-7\right)^{10}=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-7=0\\x-7=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=8\end{matrix}\right.\)
\(2,\\ a,\left|2x-3\right|>5\Leftrightarrow\left[{}\begin{matrix}2x-3< -5\\2x-3>5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x< -1\\x>4\end{matrix}\right.\\ b,\left|3x-1\right|\le7\Leftrightarrow\left[{}\begin{matrix}3x-1\le7\\1-3x\le7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\le\dfrac{8}{3}\\x\ge-2\end{matrix}\right.\\ c,\cdot x< -\dfrac{3}{2}\\ \Leftrightarrow5-3x+\left(-2x-3\right)=7\Leftrightarrow2-5x=7\Leftrightarrow x=-1\left(ktm\right)\\ \cdot-\dfrac{3}{2}\le x\le\dfrac{5}{3}\\ \Leftrightarrow\left(5-3x\right)+\left(2x+3\right)=7\Leftrightarrow8-x=7\Leftrightarrow x=1\left(tm\right)\\ \cdot x>\dfrac{5}{3}\\ \Leftrightarrow\left(3x-5\right)+\left(2x+3\right)=7\Leftrightarrow5x-2=7\Leftrightarrow x=\dfrac{9}{5}\left(tm\right)\\ \Leftrightarrow S=\left\{1;\dfrac{9}{5}\right\}\)
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1) x (x-2016) + 2015 (2016-x) = 0
x (x-2016) - 2015 (x- 2016) = 0
(x-2015)(x-2016) =0
\(\Rightarrow\orbr{\begin{cases}x-2015=0\\x-2016=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2015\\x=2016\end{cases}}}\)
Vậy x= 2015; 2016
2) -5x (x-15) + (15-x) = 0
-5x (x-15) - (x-15) =0
(-5x -1) (x-15) =0
\(\Rightarrow\orbr{\begin{cases}-5x-1=0\\x-15=0\end{cases}\Rightarrow\orbr{\begin{cases}-5x=1\\x=15\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{1}{5}\\x=15\end{cases}}}\)
Vậy x= -1/5; 15
3) 3x (3x-7) - (7-3x) =0
3x(3x-7) + (3x -7) =0
(3x+1) (3x-7) =0
\(\Rightarrow\orbr{\begin{cases}3x+1=0\\3x-7=0\end{cases}\Rightarrow\orbr{\begin{cases}3x=-1\\3x=7\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{1}{3}\\x=\frac{7}{3}\end{cases}}}\)
Vậy x= -1/3 ; 7/3
![](https://rs.olm.vn/images/avt/0.png?1311)
3x + 7 = 3 - x
<=> 3x + x = 3 - 7
<=> 4x = -4
<=> x = -4/4
<=> x= -1
Vâỵ S{ -1}
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a) |3x - 5| = 3x - 5 => 3x - 5 > 0 => 3x > 5 => x > 5/3
b) |7 - x| = x - 7 => 7 - x < 0 => - x < - 7 => x > 7
c) |2x - 3| = 3 - 2x => 2x - 3 < 0 => 2x < 3 => x < 3/2
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{2}{7}\)x - \(\frac{1}{3}\)=\(\frac{3}{5}\)x-1
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
TH1:x < 0 . PT có dạng
\(-4\left(3x-1\right)-x+2\left(x-5\right)-7\left(x-3\right)=12\)
\(4-12x-x+2x-10-7x+21=12\)
\(15-18x=12\)
\(x=\frac{1}{6}\left(koTM\right)\)
TH2:\(0\le x\le\frac{1}{3}\) PT có dạng:
\(x-4\left(3x-1\right)-7\left(x-3\right)+2\left(x-5\right)=12\)
\(x=\frac{3}{16}\left(TM\right)\)
TH3:\(\frac{1}{3}\le x< 3\) PT có dạng:
\(x+4\left(3x-1\right)-7\left(x-3\right)+2\left(x-5\right)=12\)
\(x=\frac{5}{8}\left(TM\right)\)
TH4:\(3\le x< 5\) PT có dạng:
\(4\left(3x-1\right)+x+2\left(x-5\right)+7\left(x-3\right)=12\)
\(x=\frac{47}{22}\left(koTM\right)\)
\(TH5:x\ge5\)PT có dạng:
\(4\left(3x-1\right)+x-2\left(x-5\right)+7\left(x-3\right)=12\)
\(x=1,5\left(koTM\right)\)
Vậy nghiệm PT là \(\frac{3}{16};\frac{5}{8}\)
ta có :
\(\left|x+3\right|+\left|x+7\right|=3x\ge0\Rightarrow x\ge0\)
\(PT\Leftrightarrow x+3+x+7=3x\Leftrightarrow x=10\)