Question

 giải các bất phương trình sau

1, 2 ( -2x+1) ≤ -x+3

2, 2( x+1) ≤ -x+3

3, 5-3(x-1) >2

4, \(x^2-12x+3-\left(x-3\right)^2>0\)

Answer 1

1.

$2(-2x+1)\leq -x+3$

$\Leftrightarrow -4x+2\leq -x+3$

$\Leftrightarrow -1\leq 3x$

$\Leftrightarrow x\geq \frac{-1}{3}$ 

2.

$2(x+1)\leq  -x+3$

$\Leftrightarrow 2x+2\leq -x+3$

$\Leftrightarrow 3x\leq 1$

$\Leftrightarrow x\leq \frac{1}{3}$

 

Answer 2

3.

$5-3(x-1)>2$

$\Leftrightarrow 5-(3x-3)>2$

$\Leftrightarrow 8-3x>2$

$\Leftrightarrow 8-3x-2>0$

$\Leftrightarrow 6-3x>0$

$\Leftrightarrow 6>3x$

$\Leftrightarrow x< 2$

4.

$x^2-12x+3-(x-3)^2>0$

$\Leftrightarrow x^2-12x+3-(x^2-6x+9)>0$

$\Leftrightarrow -6x-6>0$

$\Leftrightarrow -6>6x$

$\Leftrightarrow x< -1$

 

Answer 3

1: Ta có: \(2\left(-2x+1\right)\le-x+3\)

\(\Leftrightarrow-4x+x\le3-2=1\)

\(\Leftrightarrow x\ge-\dfrac{1}{3}\)

3: Ta có: \(5-3\left(x-1\right)>2\)

\(\Leftrightarrow3\left(x-1\right)< 3\)

\(\Leftrightarrow x-1< 1\)

hay x<2

Answer 4

2: Ta có: \(2\left(x+1\right)\le-x+3\)

\(\Leftrightarrow2x+x\le3-2=1\)

hay \(x\le\dfrac{1}{3}\)

4: Ta có: \(x^2-12x+3-\left(x-3\right)^2>0\)

\(\Leftrightarrow x^2-12x+3-x^2+6x-9>0\)

\(\Leftrightarrow-6x>6\)

hay x<-1