Tìm x : | 3.x -15 | =0
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![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(\Leftrightarrow\left[{}\begin{matrix}x>3\\x< -4\end{matrix}\right.\)
b: -2<x<5
![](https://rs.olm.vn/images/avt/0.png?1311)
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)
a: \(\Leftrightarrow x\in\left\{0;-3\right\}\)
b: =>(x-3)(x-5)=0
hay \(x\in\left\{3;5\right\}\)
c: =>x(x-4)=0
hay \(x\in\left\{0;4\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(\Leftrightarrow\left[{}\begin{matrix}x>3\\x< -4\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a
\(x^2\left(2x+15\right)+4\left(2x+15\right)=0\\ \Leftrightarrow\left(2x+15\right)\left(x^2+4\right)=0\\ \Leftrightarrow2x+15=0\left(x^2+4>0\forall x\right)\\ \Leftrightarrow2x=-15\\ \Leftrightarrow x=-\dfrac{15}{2}\)
b
\(5x\left(x-2\right)-3\left(x-2\right)=0\\ \Leftrightarrow\left(x-2\right)\left(5x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-2=0\\5x-3=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0+2=2\\x=\dfrac{0+3}{5}=\dfrac{3}{5}\end{matrix}\right.\)
c
\(2\left(x+3\right)-x^2-3x=0\\ \Leftrightarrow2\left(x+3\right)-\left(x^2+3x\right)=0\\ \Leftrightarrow2\left(x+3\right)-x\left(x+3\right)=0\\ \Leftrightarrow\left(x+3\right)\left(2-x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+3=0\\2-x=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0-3=-3\\x=2-0=2\end{matrix}\right.\)
a: =>(2x+15)(x^2+4)=0
=>2x+15=0
=>2x=-15
=>x=-15/2
b; =>(x-2)(5x-3)=0
=>x=2 hoặc x=3/5
c: =>(x+3)(2-x)=0
=>x=2 hoặc x=-3
![](https://rs.olm.vn/images/avt/0.png?1311)
Dễ mak
nhưng mik nhìn đề thấy dài quá nên ko muốn làm
hihi^_$
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(x-3\right)\left(x-1\right)\left(x+1\right)\left(x+3\right)+15=0\)
\(\Leftrightarrow\left(x^2-1\right)\left(x^2-9\right)+15=0\)
Đặt \(x^2-1=t\)nên \(pt\Leftrightarrow t\left(t-8\right)+15=0\)
\(\Leftrightarrow t^2-8t+15=0\)
\(\Leftrightarrow t^2-3t-5t+15=0\)
\(\Leftrightarrow t\left(t-3\right)-5\left(t-3\right)=0\)
\(\left(t-5\right)\left(t-3\right)=0\Rightarrow\orbr{\begin{cases}t-5=0\\t-3=0\end{cases}\Rightarrow\orbr{\begin{cases}t=5\\t=3\end{cases}}}\)
Với \(t=5\) thì \(x^2-1=5\Leftrightarrow x^2=6\Rightarrow\orbr{\begin{cases}x=\sqrt{6}\\x=-\sqrt{6}\end{cases}}\)
Với \(t=3\) thì \(x^2-1=3\Leftrightarrow x^2=4\Rightarrow\orbr{\begin{cases}x=-2\\x=2\end{cases}}\)
Vậy \(x\in\left\{-\sqrt{6};-2;2\sqrt{6}\right\}\)
(x-3)x-1)(x+1)(x+3)+15=0
<=>(x^2-9)(x^2-1)=-15
<=>x^4-10x^2-9=-15
<=>-(x^4+6x^2+9+4X^2)=-15
x:|3.x-15|=0
0 chia số nào cũng bằng 0
=> x=0
Bài toán kiểu gì mà logic quá vậy???
Còn cái gạch thẳng kia là gì???