a) 2x - 3 > 3(x - 2)

⇔ 2x - 3 > 3x - 6

⇔ 2x - 3x > -6 + 3

⇔ -x > -3

⇔ x < 3

Vậy S = {x | x < 3}

b) (12x + 1)/12 ≤ (9x + 1)/3 - (8x + 1)/4

⇔ 12x + 1 ≤ 4(9x + 1) - 3(8x + 1)

⇔ 12x + 1 ≤ 36x + 4 - 24x - 3

⇔ 12x - 36x + 24x ≤ 4 - 3 - 1

⇔ 0x ≤ 0 (luôn đúng với mọi x)

Vậy S = R

a: =>2x-3>3x-6

=>-x>-3

=>x<3

b: =>12x+1<=36x+4-24x-3

=>12x+1<=12x+1

=>0x<=0(luôn đúng)

a) \(2x-3>3\left(x-2\right)\)

\(\Leftrightarrow2x-3>3x-6\)

\(\Leftrightarrow2x-3x>-6+3\)

\(\Leftrightarrow-x>-3\)

\(\Leftrightarrow x< 3\)

Vậy bất phương trình có nghiệm là \(x< 3\)

b) \(\dfrac{12x+1}{12}\le\dfrac{9x+1}{3}-\dfrac{8x+1}{4}\)

\(\Leftrightarrow\dfrac{12x+1}{12}\le\dfrac{\left(9x+1\right).4}{3.4}-\dfrac{\left(8x+1\right)3}{4.3}\)

\(\Leftrightarrow12x+1\le36x+4-24x-3\)

\(\Leftrightarrow12x-36x+24x\le4-3-1\)

\(\Leftrightarrow0x\le0\)

Vậy bất phương trình vô nghiệm

\(\left(x-3\right)\sqrt{x^2-4}\le x^2-9\)

\(\Leftrightarrow\left(x-3\right)\sqrt{x^2-4}\le\left(x-3\right)\left(x+3\right)\)

\(\Leftrightarrow\sqrt{x^2-4}\le x+3\)

\(\Leftrightarrow x^2-4\le x^2+6x+9\)

\(\Leftrightarrow6x+9\le-4\)

\(\Leftrightarrow6x\le-13\)

\(\Leftrightarrow x\le\dfrac{-13}{6}\)

Vậy...

Nguyễn Lưu Ngọc Tường Vy xem lại đề giùm cái

\(\left(x-3\right)\sqrt{x^2-4}\le x^2-9\) (1)

-3 ;-2;2-3

đk xđ: |x| >=2

xét x<=-3 VP >=0VT <=0 => đúng mọi x<=-3

xét -3<=x<=-2 hoặc 2<=x<=3 <=> x-3 <0

(1) <=>\(\sqrt{x^2-4}\ge x+3\) <=> x^2 -4 >=x^2 +6x+9

<=> 6x<=-13 => x<= -13/6 <-2

vậy nghiệm là -3 <=x <=-13/2

xét khi x>3 => x-3>0

<=> \(\sqrt{x^2-4}\le\left(x+3\right)\) <=> x>= -13/6

vậy nghiệm x>= 3

kết luận

x<=-13/6 hoặc x>= 3

a: \(\Leftrightarrow x-5-2\left(2x-1\right)< 12\)

=>x-5-4x+2<12

=>-3x-3<12

=>-3x<15

hay x>-5

b: \(\Leftrightarrow4\left(5x^2-3\right)+5\left(3x-1\right)>10x\left(2x+3\right)-100\)

\(\Leftrightarrow20x^2-12+15x-5-20x^2-30x+100>0\)

=>-15x+83>0

hay x<83/15

=>2x^3+2-2x^2-1=0

=>2x^3-2x^2+1=0

=>\(x\simeq-0,57\)

Bạn xem đã viết đúng phương trình chưa vậy?

\(\dfrac{3x-2}{4}-x< \dfrac{x+2}{5}\\ \Leftrightarrow\dfrac{5\left(3x-2\right)}{20}-\dfrac{20x}{20}-\dfrac{4\left(x+2\right)}{20}< 0\\ \Leftrightarrow15x-10-20x-4x-8< 0\\ \Leftrightarrow-9x-18< 0\\ \Leftrightarrow-9x< 18\\ \Leftrightarrow x>-2\)

  `[ 7x - 1 ] / 6 + 2x >= [ 16 - x ] / 5`

`<=> [ 5 ( 7x - 1 ) + 60x ] / 30 >= [ 6 ( 16 - x ) ] / 30`

`<=> 35x - 5 + 60x >= 96 - 6x`

`<=>101x >= 101`

`<=> x >= 1`

Vậy `S = { x | x >= 1 }`

a, \(x^2-4x+4x-5\ge x^2+6\Leftrightarrow-5\ge6\)

vô lí bpt vô nghiệm 

b, \(9x^2-6x+1-9x^2+9< 5x-2\Leftrightarrow-6x+10< 5x-2\)

\(\Leftrightarrow-11x< -12\Leftrightarrow x>\dfrac{12}{11}\)

bài 1: x.(x+7) = 0

Th1:x=0              Th2:x+7=0

                          =>x=-7

bài 2 (x+12).(x-3)= 0

Th1:x+12=0                                         Th2:x-3=0

=>x=-12                                                =>x=3

bài 3 (-x+5).(3-x)=0

Th1 (-x)+5=0                                          Th2:3-x=0

=>-x=-5                                                  =>x=3

bài 4 x.(2+x).(7-x)=0

Th1:x=0                                               Th3:7-x=0

Th2:2+x=0                                             =>x=7

=>x=-2

bài 5 (x-1).(x+2).(-x-3)=0

Th1:x-1=0                                               Th2:x+2=0

=>x=1                                                   =>x=-2

Th3:-x-3=0

=>-x=-3