![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
1/ x(x+17)=0
⇒ \(\left[{}\begin{matrix}x=0\\x+17=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-17\end{matrix}\right.\)
2/ (x+1112)(x-3)=0
⇒\(\left[{}\begin{matrix}x+1112=0\\x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1112\\x=3\end{matrix}\right.\)
3/ (-x+25)(3-x)=0
⇒\(\left[{}\begin{matrix}-x+25=0\\3-x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=25\\x=3\end{matrix}\right.\)
4/ x(12+x)(7-x)=0
⇒ \(\left[{}\begin{matrix}x=0\\12+x=0\\7-x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-12\\x=7\end{matrix}\right.\)
5/ (x-15)(x+2)(-x-3)=0
⇒\(\left[{}\begin{matrix}x-15=0\\x+2=0\\-x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=15\\x=-2\\x=-3\end{matrix}\right.\)
\(x\left(x+17\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x+17=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-17\end{matrix}\right.\)
Vậy \(x\in\left\{0;-17\right\}\)
\(\left(x+1112\right)\left(x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+1112=0\\x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1112\\x=3\end{matrix}\right.\)
Vậy \(x\in\left\{-1112;3\right\}\)
\(\left(-x+25\right)\left(3-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}-x+25=0\\3-x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=25\\x=3\end{matrix}\right.\)
Vậy \(x\in\left\{25;3\right\}\)
\(x\left(12+x\right)\left(7-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\12+x=0\\7-x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-12\\x=7\end{matrix}\right.\)
Vậy \(x\in\left\{0;-12;7\right\}\)
\(\left(x-15\right)\left(x+2\right)\left(-x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-15=0\\x+2=0\\-x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=15\\x=-2\\x=-3\end{matrix}\right.\)
Vậy \(x\in\left\{15;-2;-3\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
x(x+2)=0
suy ra x=0 hoặc x+2=0
5-2x=-7
2x=-7+5
2x=-(7-5)
2x=-2
x=-2:2
x=-1
Vậy x=-1
NHỚ TÍCH MK NHA
Tự học giúp bạn có được một gia tài
Jim Rohn – Triết lý cuộc đời
![](https://rs.olm.vn/images/avt/0.png?1311)
(2x - 7) + 17 = 6
=> 2x - 7 = 6 - 17
=> 2x - 7 = -11
=> 2x = -11 + 7
=> 2x = -4
=> x = -4 : 2
=> x = -2
+) 12 -2(3 - 3x)= -2
=> 2(3 - 3x) = 12 + 2
=> 2(3 - 3x) = 14
=> 3 - 3x = 14 : 2
=> 3 - 3x = 7
=> 3x = 3 - 7
=> 3x = -4
=> x = -4/3
\(\left(x+1\right)\left(x-3\right)=0\)
=> \(\orbr{\begin{cases}x+1=0\\x-3=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=-1\\x=3\end{cases}}\)
Vậy...
![](https://rs.olm.vn/images/avt/0.png?1311)
14) (x - 2) . (x + 4) = 0
\(\Rightarrow\) x - 2 = 0 hoặc x + 4 = 0
Nếu x - 2 = 0
x = 0 + 2
x = 2
Nếu x + 4 = 0
x = 0 - 4
x = -4
Vậy x \(\in\) {2 ; -4)
15) (x - 2) . (x + 15) = 0
\(\Rightarrow\) x - 2 = 0 hoặc x + 15 = 0
Nếu x - 2 = 0
x = 0 + 2
x = 2
Nếu x + 15 = 0
x = 0 - 15
x = -15
Vậy x \(\in\) {-15 ; 2}
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1.\left(x-2\right).\left(x+14\right)=0\)
<=>\(\orbr{\begin{cases}x-2=0\\x+14=0\end{cases}}\)
<=>\(\orbr{\begin{cases}x=2\\x=-14\end{cases}}\)
\(2.\left(x-2\right).\left(x+4\right)=0\)
<=>\(\orbr{\begin{cases}x-2=0\\x+4=0\end{cases}}\)
<=>\(\orbr{\begin{cases}x=2\\x=-4\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Mình trả lời câu b cho bạn nhé
2[x+7]-3[x-2]=-2
2x-14-3*x+6=-2
2x-3x-14+6=-2
-x-14+6=-2
-x-14=-2-6
-x-14=-8
-x=-8+14
-x=6
<=>x=-6
![](https://rs.olm.vn/images/avt/0.png?1311)
a) x-14=3x + 18
x - 3x = 18 + 14
-2x = 32
=> x = -16
b) (x+7)(x-9)=0
=> TH1: x+7=0 => x = -7
=> TH2: x-9=0 => x = 9
c) x(x+3) =0
=> TH1: x=0
=> TH2: x+3 =0 => x = -3
d) (x-2)(5-x)=0
=> TH1: x-2=0 => x=2
=> Th2: 5-x=0 => x=5
![](https://rs.olm.vn/images/avt/0.png?1311)
a, => |15-x| = 2+3
=> |15-x| = 5
=> 15-x = -5 hoặc 15-x = 5
=> x = 10 hoặc x = 10
Vậy ......
Tk mk nha
a/|15-x|=|-2|+|-3|
=>|15-x|=2+3=5
=>\(15-x=\hept{\begin{cases}5\\-5\end{cases}}\)
=>\(x=\hept{\begin{cases}10\\15\end{cases}}\)
.........
k nha , thanks
a. \(|x|=3\)
\(\Rightarrow\hept{\begin{cases}x=3\\x=-3\end{cases}}\)
b. \(x(x+2)\)
\(\Rightarrow\hept{\begin{cases}x=0\\x+2=0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=0\\x=-2\end{cases}}\)
c.\(-3.|x|+5=14\)
\(-3.|x|\) \(=9\)
\(|x|=-3\)[ vô lí ]
Học tốt
Làm
a) | x | = 3
=> x = 3 hoặc x = -3
b) x(x + 2) = 0
x = 0
=>
x + 2 = 0
x = 0
=>
x = -2
c) -3 . |x| + 4 = 14
-3 . |x| = 14 - 5
-3 . |x| = 9
|x| = -3
Ta có : | x | luôn lớn hơn hoặc bằng 0
mà -3 < 0
=> |x| = -3 ( vô lý )
HỌC TỐT