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`(x-3)/5+1<2x-5`
`<=>x-3+5<5(2x-5)`
`<=>x+2<10x-25`
`<=>8x>27`
`<=>x>27/8`
Vậy `S={x|x>27/8}`
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left|1-x\right|+\left|2x-1\right|>5\)(*)
* Xét khoảng \(x< \frac{1}{2}\)thì \(\hept{\begin{cases}1-x>0\\2x-1< 0\end{cases}}\Rightarrow\hept{\begin{cases}\left|1-x\right|=1-x\\\left|2x-1\right|=1-2x\end{cases}}\)
(*)\(\Leftrightarrow\left(1-x\right)+\left(1-2x\right)>5\Leftrightarrow-3x>3\Leftrightarrow x< -1\)
Nghiệm của bất phương trình thuộc khoảng này là \(x< -1\)
* Xét khoảng \(\frac{1}{2}\le x\le1\)thì \(\hept{\begin{cases}1-x\ge0\\\left|2x-1\right|\ge0\end{cases}}\Rightarrow\hept{\begin{cases}\left|1-x\right|=1-x\\\left|2x-1\right|=2x-1\end{cases}}\)
(*)\(\Leftrightarrow\left(1-x\right)+\left(2x-1\right)>5\Leftrightarrow x>5\)(Nghiệm này không thuộc khoảng đang xét)
* Xét khoảng \(x>1\)thì \(\hept{\begin{cases}1-x< 0\\2x-1>0\end{cases}}\Rightarrow\hept{\begin{cases}\left|1-x\right|=x-1\\\left|2x-1\right|=2x-1\end{cases}}\)
(*)\(\Leftrightarrow\left(x-1\right)+\left(2x-1\right)>5\Leftrightarrow3x>7\Leftrightarrow x>\frac{7}{3}\)
Nghiệm của bất phương trình thuộc khoảng này là \(x>\frac{7}{3}\)
Vậy nghiệm của bất phương trình đã cho là \(x< -1\);\(x>\frac{7}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\dfrac{x-3}{x}=\dfrac{x-3}{x+3}\)\(\left(đk:x\ne0,-3\right)\)
\(\Leftrightarrow\dfrac{x-3}{x}-\dfrac{x-3}{x+3}=0\)
\(\Leftrightarrow\dfrac{\left(x-3\right)\left(x+3\right)-x\left(x-3\right)}{x\left(x+3\right)}=0\)
\(\Leftrightarrow x^2-9-x^2+3x=0\)
\(\Leftrightarrow3x-9=0\)
\(\Leftrightarrow3x=9\)
\(\Leftrightarrow x=3\left(n\right)\)
Vậy \(S=\left\{3\right\}\)
\(b,\dfrac{4x-3}{4}>\dfrac{3x-5}{3}-\dfrac{2x-7}{12}\)
\(\Leftrightarrow\dfrac{4x-3}{4}-\dfrac{3x-5}{3}+\dfrac{2x-7}{12}>0\)
\(\Leftrightarrow\dfrac{3\left(4x-3\right)-4\left(3x-5\right)+2x-7}{12}>0\)
\(\Leftrightarrow12x-9-12x+20+2x-7>0\)
\(\Leftrightarrow2x+4>0\)
\(\Leftrightarrow2x>-4\)
\(\Leftrightarrow x>-2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
a) Ta có: \(2\left(3-4x\right)=10-\left(2x-5\right)\)
\(\Leftrightarrow6-8x-10+2x-5=0\)
\(\Leftrightarrow-6x+11=0\)
\(\Leftrightarrow-6x=-11\)
hay \(x=\dfrac{11}{6}\)
b) Ta có: \(3\left(2-4x\right)=11-\left(3x-1\right)\)
\(\Leftrightarrow6-12x-11+3x-1=0\)
\(\Leftrightarrow-9x-6=0\)
\(\Leftrightarrow-9x=6\)
hay \(x=-\dfrac{2}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: Ta có: \(3x-5\ge2\left(x-6\right)-12\)
\(\Leftrightarrow3x-5\ge2x-24\)
hay \(x\ge-19\)
b: Ta có: \(2\left(5-2x\right)\ge3-x\)
\(\Leftrightarrow10-4x-3+x\ge0\)
\(\Leftrightarrow-3x\ge-7\)
hay \(x\le\dfrac{7}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{x}{2x-6}-\dfrac{x}{2x+2}=\dfrac{2x}{\left(x+1\right)\left(x-3\right)}\left(ĐKXĐ:x\ne-1,x\ne3\right)\)
\(\Leftrightarrow\dfrac{x}{2\left(x-3\right)}-\dfrac{x}{2\left(x+1\right)}=\dfrac{2x}{\left(x+1\right)\left(x-3\right)}\)
\(\Leftrightarrow\dfrac{x\left(x+1\right)}{2\left(x+1\right)\left(x-3\right)}-\dfrac{x\left(x-3\right)}{2\left(x+1\right)\left(x-3\right)}=\dfrac{2x\cdot2}{2\left(x+1\right)\left(x-3\right)}\)
\(\Rightarrow x\left(x+1\right)-x\left(x-3\right)=4x\)
\(\Leftrightarrow x^2+x-x^2+3x=4x\)
\(\Leftrightarrow x^2+x-x^2+3x-4x=0\)
\(\Leftrightarrow0x=0\)
Phương trình có vô số nghiệm , trừ x = -1,x = 3
Vậy ...
\(\dfrac{12x+1}{12}< \dfrac{9x+1}{3}-\dfrac{8x+1}{4}\)
\(\Leftrightarrow12\cdot\dfrac{12x+1}{12}< 12\cdot\dfrac{9x+1}{3}-12\cdot\dfrac{8x+1}{4}\)
\(\Leftrightarrow12x+1< 4\left(9x+1\right)-3\left(8x+1\right)\)
\(\Leftrightarrow12x+1< 36x+4-24x-3\)
\(\Leftrightarrow12x+1< 12x+1\)
\(\Leftrightarrow12x-12x< 1-1\)
\(\Leftrightarrow0x< 0\)
Vậy S = {x | x \(\in R\)}
![](https://rs.olm.vn/images/avt/0.png?1311)
a: Ta có: \(3x+5\le4x-9\)
\(\Leftrightarrow-x\le-14\)
\(\Leftrightarrow x\ge14\)
b: Ta có: \(6-2x< 6-x\)
\(\Leftrightarrow-x< 0\)
hay x>0
c: Ta có: \(7\left(x-1\right)+5>-3x\)
\(\Leftrightarrow7x-7+5+3x>0\)
\(\Leftrightarrow10x>2\)
hay \(x>\dfrac{1}{5}\)