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![](https://rs.olm.vn/images/avt/0.png?1311)
a, (x-5).(x-1) >0
<=> x-5>0 và x-1>0
<=> x-5>0
<=> x>5
x-1>0
<=> x>1
Vậy x>5
b, (2x-3).(x+1) <0
<=> 2x-3<0 và x+1<0
2x-3<0 <=> 2x<3 <=> x<2/3
x+1<0 <=> x<-1
Vậy x<2/3
c, 2x2 - 3x +1>0
<=> 2x2 - 2x- x +1>0
<=>(x-1). (2x-1) >0
<=> x-1>0 và 2x-1>0
x-1>0 <=> x>1
2x-1>0 <=> 2x>1 <=> x>1/2
Vậy x>1/2
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(x-1\right)\left(x+1\right)-2\left(2x+3\right)\le\left(x-2\right)^2+x\)
\(\Leftrightarrow x^2-1-4x-6\le x^2-4x+4+x\)
\(\Leftrightarrow x^2-4x-7\le x^2-3x+4\)
\(\Leftrightarrow x^2-4x-x^2+3x\le7+4\)
\(\Leftrightarrow-x\le11\)
\(\Leftrightarrow x\le-11\)
![](https://rs.olm.vn/images/avt/0.png?1311)
c) \(\frac{x+1}{3}>\frac{2x-1}{6}-2\)
⇔\(\frac{2\left(x+1\right)}{6}>\frac{2x-1-12}{6}\)
⇔2x + 2 > 2x - 13
⇔0x > -15
Vậy S=Φ
b) \(\frac{x-2}{6}-\frac{x-1}{3}\le\frac{x}{2}\)
⇔\(\frac{x-2-2\left(x-1\right)}{6}\le\frac{3x}{6}\)
⇔x - 2 - 2x +2 ≤ 3x
⇔-4x ≤ 0
⇔x ≥ 0
Vậy S={x | x ≥ 0}
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Leftrightarrow\frac{5\left(x+5\right)-3\left(x-3\right)}{15}=\frac{5\left(x+5\right)-3\left(x-3\right)}{\left(x-3\right)\left(x+5\right)}\)
\(\Leftrightarrow\frac{2x+34}{15}=\frac{2x+34}{x^2+2x-15}\Leftrightarrow\orbr{\begin{cases}2x+34=0\\x^2+2x-15=15\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-17\\x^2+2x-30=0\end{cases}}\)
Từ đó tìm được \(S=\left\{-17;\sqrt{31}-1;-\sqrt{31}-1\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a)\) \(3-2x>4x+5\)
\(\Leftrightarrow\)\(3-2x+2x>4x+2x+5\)
\(\Leftrightarrow\)\(6x+5< 3\)
\(\Leftrightarrow\)\(6x+5-5< 3-5\)
\(\Leftrightarrow\)\(6x< -2\)
\(\Leftrightarrow\)\(\frac{6x}{6}< \frac{-2}{6}\)
\(\Leftrightarrow\)\(x< \frac{-1}{3}\)
Vậy \(x< \frac{-1}{3}\)
Chúc bạn học tốt ~
\(\Leftrightarrow x+\frac{11x-1}{\frac{5}{3}}=1-\frac{3x-6x^2}{\frac{3}{5}}\)\(\Leftrightarrow x+\frac{11x-1}{15}=1-\frac{3x-6x^2}{15}\)\(\Leftrightarrow\frac{26x-1}{15}=\frac{15-3x+6x^2}{15}\)\(\Leftrightarrow26x-1=15-3x+6x^2\)\(\Leftrightarrow6x^2-29x+16=0\)\(\Leftrightarrow6x^2-2\cdot\sqrt{6}\cdot\frac{29}{2\sqrt{6}}+\left(\frac{29}{2\sqrt{6}}\right)^2-\frac{697}{24}=0\)\(\Leftrightarrow\left(\sqrt{6}x-\frac{29}{2\sqrt{6}}\right)^2=\frac{697}{24}\)\(\Leftrightarrow\orbr{\begin{cases}\sqrt{6}x-\frac{29}{2\sqrt{6}}=\sqrt{\frac{697}{24}}\\\sqrt{6}x-\frac{29}{2\sqrt{6}}=-\sqrt{\frac{697}{24}}\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=\frac{29+\sqrt{457}}{12}\\x=\frac{29-\sqrt{457}}{12}\end{cases}}\)