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Ta có : \(\left|5x-4\right|=\left|x+2\right|\)
\(\Leftrightarrow\orbr{\begin{cases}5x-4=x+2\\5x-4=-x-2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}5x-x=2+4\\5x+x=-2+4\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}4x=6\\6x=2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=\frac{1}{3}\end{cases}}\)
b) \(\left|2x-3\right|-\left|3x+2\right|=0\)
\(\Rightarrow\orbr{\begin{cases}2x-3=3x+2\\2x-3=-3x-2\end{cases}\Rightarrow\orbr{\begin{cases}2x-3x=2+3\\2x+3x=-2+3\end{cases}\Rightarrow}\orbr{\begin{cases}-x=5\\5x=1\end{cases}\Rightarrow}\orbr{\begin{cases}x=-5\\x=\frac{1}{5}\end{cases}}}\)
c)/2+3x/=/4x-3/
\(\Rightarrow\orbr{\begin{cases}2+3x=4x-3\\2+3x=-\left(4x-3\right)\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}3x-4x=-3-2\\3x+4x=3-2\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}-x=-5\\7x=1\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=\frac{1}{7}\end{cases}}}\)
d)/7x+1/-/5x+6|=0
\(\Rightarrow\left|7x+1\right|=\left|5x+6\right|\)
\(\Rightarrow\orbr{\begin{cases}7x+1=5x+6\\7x+1=-\left(5x+6\right)\end{cases}\Rightarrow\orbr{\begin{cases}7x-5x=6-1\\7x+1=-5x-6\end{cases}\Rightarrow}\orbr{\begin{cases}2x=5\\7x+5x=-6-1\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{5}{2}\\x=-\frac{7}{12}\end{cases}}}\)
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a) \(\left(x-9\right)^4=\left(x-9\right)^7\)
\(\Rightarrow\left[{}\begin{matrix}x-9=1\\x-9=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=10\\x=9\end{matrix}\right.\)
b) \(\left(3x-15\right)^{10}=\left(3x-15\right)^{15}\)
\(\Rightarrow\left[{}\begin{matrix}3x-15=0\\3x-15=1\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{15}{3}\\x=\dfrac{16}{3}\end{matrix}\right.\)
c) \(\left(x-8\right)^3=\left(x-8\right)^6\)
\(\Rightarrow\left[{}\begin{matrix}x-8=0\\x-8=1\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=8\\x=9\end{matrix}\right.\)
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a) x - 96 = (433 - x) - 15
x - 96 = 433 - x - 15
2x = 433 - 15 + 96
2x = 514
x = 257
b) 7x (-x-10) = 0
\(\Leftrightarrow\orbr{\begin{cases}7x=0\\-x-10=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-10\end{cases}}\)
c) 17 (3x - 6) (2x - 8) = 0
(3x - 6) (2x - 8) = 0
\(\Leftrightarrow\orbr{\begin{cases}3x-6=0\\2x-8=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}3x=6\\2x=8\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=4\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(-3x+2\right)-\left(5-3x\right)=-3\)
\(\Rightarrow-3x+2-5+3x=-3\)
\(\Rightarrow-3x+3x=-3+5-2\)
\(\Rightarrow0x=0\Rightarrow x\in Z\)
\(3+x-\left(3x-1\right)=6-2x\)
\(\Rightarrow3+x-3x+1=6-2x\)
\(\Rightarrow x-3x+2x=6-1-3\)
\(\Rightarrow0x=2\left(loại\right)\)
\(\left(x-5\right)\left(3x+4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-5=0\\3x+4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=5\\x=-\frac{4}{3}\end{cases}}}\)
\(7x\left(2x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}7x=0\\2x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{2}\end{cases}}}\)
\(\left(3x-1\right)2x=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x-1=0\\2x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=0\end{cases}}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, 7x . (x-10 ) = 0
=> \(\hept{\begin{cases}7x=0\\x-10=0\end{cases}}\) => \(\hept{\begin{cases}x=0\\x=10\end{cases}}\)
b,17 . (3x-6).(2x-8) = 0
(3x-6).(2x-8) = 0 : 17
(3x-6).(2x-8) = 0
=>\(\hept{\begin{cases}3x-6=0\\2x-8=0\end{cases}}\)=>\(\hept{\begin{cases}x=2\\x=4\end{cases}}\)
Nhớ nha
![](https://rs.olm.vn/images/avt/0.png?1311)
a, `(x-9)^4=(x-9)^7`
`(x-9)^4-(x-9)^7=0`
`(x-9)^4 . [(1-(x-9)^3]=0`
TH1: `(x-9)^4=0`
`x-9=0`
`x=9`
TH2: `1-(x-9)^3=0`
`(x-9)^3=1^3`
`x-9=1`
`x=10`
b, `(3x-15)^10=(3x-15)^15`
`(3x-15)^10 . [1-(3x-15)^5]=0`
TH1: `(3x-15)^10=0`
`3x-15=0`
`x=5`
TH2: `1-(3x-15)^5=0`
`(3x-15)^5=1^5`
`3x-15=1`
`x=16/3` (Loại)
c, `(x-8)^3=(x-8)^6`
`(x-8)^3 .[1-(x-8)^3]=0`
TH1: `(x-8)^3=0`
`x=8`
TH2: `1-(x-8)^3=0`
`x-8=1`
`x=9`
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\text{a, 7x. (x-10)= 0 }\)
\(\Rightarrow\orbr{\begin{cases}7x=0\\x-10=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=10\end{cases}}}\)
\(\text{ b, 17.(3x-6) . (2x-8) =0}\)
\(\left(51x-102\right).\left(2x-8\right)=0\)
\(\Rightarrow\orbr{\begin{cases}51x-102=0\\2x-8=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=4\end{cases}}}\)
a) \(7x\left(x-10\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}7x=0\\x-10=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=10\end{cases}}}\)
Vậy \(x\in\left\{0;10\right\}\)
b) \(17\left(3x-6\right)\left(2x-8\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x-6=0\\2x-8=0\end{cases}\Leftrightarrow\orbr{\begin{cases}3x=6\\2x=8\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=2\\x=4\end{cases}}}\)
Vậy \(x\in\left\{2;4\right\}\)
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a, 7\(x\).(\(x\) - 10) = 0
\(\left[{}\begin{matrix}7x=0\\x-10=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\x=10\end{matrix}\right.\)
Vậy \(x\in\) {0; 10}
b, 17.(3\(x\) - 6).(2\(x\) - 18) = 0
\(\left[{}\begin{matrix}3x-6=0\\2x-18=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}3x=6\\2x-18=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=6:3\\x=18:2\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=2\\x=9\end{matrix}\right.\)
a) \(\orbr{\begin{cases}x=0\\x=10\end{cases}}\)
b) \(\orbr{\begin{cases}x=2\\x=4\end{cases}}\)
a) 7x(x-10)=0
<=> 7x=0 hoặc x-10=0
<=>x=0 hoặc x=10
Vậy \(x\in\left\{0;10\right\}\)
b) 17(3x-6)(2x-8)=0
<=>3x-6=0 hoặc 2x-8=0
<=>3x=6 hoặc 2x=8
<=>x=2 hoặc x=4
Vậy \(x\in\left\{2;4\right\}\)