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)
\(\Leftrightarrow\left|\dfrac{5}{2}x-\dfrac{2}{5}\right|=0\) (1)
+ Nếu \(\dfrac{5}{2}x-\dfrac{2}{5}\ge0\Leftrightarrow x\ge\dfrac{4}{25}\)
\(\left(1\right)\Leftrightarrow\dfrac{5}{2}x-\dfrac{2}{5}=0\Leftrightarrow x=\dfrac{4}{25}\) Thảo mãn diều kiện
+ Nếu \(\dfrac{5}{2}x-\dfrac{2}{5}< 0\Leftrightarrow x< \dfrac{4}{25}\)
\(\left(1\right)\Leftrightarrow-\dfrac{5}{2}x+\dfrac{2}{5}=0\Leftrightarrow x=\dfrac{4}{25}\) (loại)
![](https://rs.olm.vn/images/avt/0.png?1311)
1) \(|5x-3|=|7-x|\)
\(\Leftrightarrow\orbr{\begin{cases}5x-3=7-x\\5x-3=x-7\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}6x=10\\4x=-4\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{3}\\x=-1\end{cases}}\)
Vậy...
2) \(2.|3x-1|-3x=7\)
\(\Leftrightarrow2.|3x-1|=7+3x\)
\(\Leftrightarrow\orbr{\begin{cases}2.\left(3x-1\right)=7+3x\\2.\left(3x-1\right)=-7-3x\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}6x-2=7+3x\\6x-2=-7-3x\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}3x=9\\9x=-5\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=\frac{-5}{9}\end{cases}}\)
Vậy...
![](https://rs.olm.vn/images/avt/0.png?1311)
1) 1/4 + 1/3 : 3x = -5
1/3 : 3x = -21/4
3x = -4/63
x= -4/189
2) 1- ( 5/3/8 + x - 7/5/24 ) : ( -16/2/3 ) = 0
(5/3/8 + x - 7/5/24) : (-16/2/3) =1
(43/8 + x -173/24) : (-50/3) =1
43/8 +x - 173/24 = -50/3
43/8 +x = -227/24
x= -89/6
3) ( 1/1/3 - 25% - 5/12 ) + 2x = 1,6 : 3/5
Tìm x biết:
| x - 1/3 | . 2 + 4/5 = 0,24
24/7x-3 = -4/25
![](https://rs.olm.vn/images/avt/0.png?1311)
a)25/x+1 - 7/6 = 1/3 - 0,5
25/x+1 - 7/6 = 1/3 -1/2
25/x+1 - 7/6 = -1/6
25/x+1 = -1/6 + 7/6
25/x+1 = 1
25/x+1 = 25/25
=>25 = x + 1
x = 25 - 1
x = 24
![](https://rs.olm.vn/images/avt/0.png?1311)
a: (2x-3)(3x+6)>0
=>(2x-3)(x+2)>0
=>x<-2 hoặc x>3/2
b: (3x+4)(2x-6)<0
=>(3x+4)(x-3)<0
=>-4/3<x<3
c: (3x+5)(2x+4)>4
\(\Leftrightarrow6x^2+12x+10x+20-4>0\)
\(\Leftrightarrow6x^2+22x+16>0\)
=>\(6x^2+6x+16x+16>0\)
=>(x+1)(3x+8)>0
=>x>-1 hoặc x<-8/3
f: (4x-8)(2x+5)<0
=>(x-2)(2x+5)<0
=>-5/2<x<2
h: (3x-7)(x+1)<=0
=>x+1>=0 và 3x-7<=0
=>-1<=x<=7/3