a/ \(3.\left|9-2x\right|-17=16\)
\(\Leftrightarrow3.\left|9-2x\right|=33\)
\(\Leftrightarrow\left|9-2x\right|=11\)
\(\Leftrightarrow\left[{}\begin{matrix}9-2x=11\left(x\le\dfrac{9}{2}\right)\\2x-9=11\left(x>\dfrac{9}{2}\right)\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-1\left(tm\right)\\x=10\left(tm\right)\end{matrix}\right.\)
Vậy pt đã cho có \(S=\left\{-1;10\right\}\)
b/ \(3-4.\left|5-6x\right|=7\)
\(\Leftrightarrow4.\left|5-6x\right|=-4\)
\(\Leftrightarrow\left|5-6x\right|=-1\)
\(\Leftrightarrow\left[{}\begin{matrix}5-6x=-1\left(x\le\dfrac{5}{6}\right)\\6x-5=-1\left(x>\dfrac{5}{6}\right)\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=1\left(ktm\right)\\x=\dfrac{2}{3}\left(ktm\right)\end{matrix}\right.\)
Vậy pt đã cho vô nghiệm
c) \(\left|9-7x\right|=5x-3\)
\(\Rightarrow\left|9-7x\right|-5x=-3\)
\(\Leftrightarrow\left[{}\begin{matrix}9-7x-5x=-3\left(đk:9-7x\ge0\right)\\-\left(9-7x\right)-5x=-3\left(đk:9-7x< 0\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(đk:x\le\dfrac{9}{7}\right)\\x=3\left(đk:x>\dfrac{9}{7}\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)
Vậy \(x_1=1;x_2=3\)
d) \(8x-\left|4x+1\right|=x+2\)
\(\Rightarrow8x-\left|4x+1\right|-x=2\)
\(\Leftrightarrow7x-\left|4x+1\right|=2\)
\(\Leftrightarrow\left[{}\begin{matrix}7x-\left(4x+1\right)=2\left(đk:4x+1\ge0\right)\\7x-\left(-\left(4x+1\right)\right)=2\left(đk:4x+1< 0\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(đk:x\ge-\dfrac{1}{4}\right)\\x=\dfrac{1}{11}\left(đk:x< -\dfrac{1}{4}\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x\in\varnothing\end{matrix}\right.\)
Vậy \(x=1\)
a , 3 . | 9 - 2x | - 17 = 16
3 . | 9 - 2x | = 16 +17
3 . | 9 - 2x | = 33
| 9 - 2x | = 33 : 3
| 9 - 2x | = 11
* Trường hợp 1 :
9 - 2x = 11
2x = 9 - 11
2x = -2
x = - 2 : 2
x = -1
Trường hợp 2 :
9 - 2x = -11
2x = 9 - ( -11 )
2x = 9 + 11
2x = 20
x = 20 : 2
x = 10
Vậy x = -1 hoặc x = 10
b , 3 - 4 . | 5 - 6x | = 7
4 . | 5 - 6 x | = 3 - 7
4 . | 5 - 6x | = -4
| 5 - 6x | = ( -4 ) : 4
| 5 - 6x | = -1
* Trường hợp 1 :
5 - 6x = 1
6x = 5- 1
6x = 4
x = 6 : 4
x = \(\dfrac{3}{2}\)
* Trường hợp 2 :
5 - 6x = -1
6x = 5 - ( -1 )
6x = 5+1
6x = 6
x = 6 : 6
x =1
Kết luận : Vậy x = 1 hoặc x = \(\dfrac{3}{2}\)
e/ \(\left|2x-3\right|-\left(2x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3-2x+3=0\left(x\ge\dfrac{3}{2}\right)\\3-2x-2x+3=0\left(x< \dfrac{3}{2}\right)\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}0x=0\left(x\ge\dfrac{3}{2}\right)\\-4x+6=0\left(x< \dfrac{3}{2}\right)\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=\dfrac{3}{2}\left(ktm\right)\end{matrix}\right.\)
Vậy phương trình đã cho vô nghiệm
g/ \(\left|4-x\right|+\left(4-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}4-x+4-x=0\left(x\le4\right)\\x-4+4-x=0\left(x>4\right)\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}8-2x=0\left(x\le4\right)\\0x=0\left(x>4\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\left(tm\right)\\x=0\left(ktm\right)\end{matrix}\right.\)
Vậy phương trình đã cho có \(S=\left\{4\right\}\)
