\(x-\frac{2}{4}-\frac{2}{3}\ge5x-\frac{9}{12}\)

\(\Leftrightarrow x-\frac{7}{6}\ge5x-\frac{3}{4}\)

\(\Leftrightarrow-4x\ge\frac{5}{12}\)

\(\Leftrightarrow-\frac{5}{56}\ge x\)

a: 3x-5>15-x

=>4x>20

hay x>5

b: \(3\left(x-2\right)\left(x+2\right)< 3x^2+x\)

=>3x²+x>3x²-12

=>x>-12

1.a)|−7x|=3x+16

Vì |-7x| ≥ 0  nên 3x+16 ≥ 0 ⇔ x ≥ \(\dfrac{-16}{3}\)    (*)

Với đk (*), ta có: |-7x|=3x+16

\(\left[\begin{array}{} -7x=3x+16\\ -7x=-3x-16 \end{array} \right.\) ⇔  \(\left[\begin{array}{} -7x-3x=16\\ -7x+3x=-16 \end{array} \right.\)

⇔ \(\left[\begin{array}{} x=-1,6 (t/m)\\ x= 4 (t/m) \end{array} \right.\)

b) \(\dfrac{x-1}{x+2}\) - \(\dfrac{x}{x-2}\) = \(\dfrac{5x-8}{x^2-4}\)

⇔ \(\dfrac{(x-1)(x-2)}{x^2-4}\) - \(\dfrac{x(x+2)}{x^2-4}\) = \(\dfrac{5x-8}{x^2-4}\)

⇒ x² - 2x - x + 2 - x² - 2x = 5x - 8  

⇔ -5x - 5x = -8 - 2

⇔ -10x = -10

⇔ x=1

2.7x+5 < 3x−11

⇔ 7x - 3x < -11 - 5

⇔ 4x < -16

⇔ x < -4

bạn tự biểu diễn trên trục số nha !

c: =>2x+4>=2x+2-3

=>4>=-1(luôn đúng)

a: 5x+10>3x+3

=>2x>-7

=>x>-7/2

bạn coi lại đề nhé!

a: 3x-5>15-x

=>3x+x>15+5

=>4x>20

=>x>5

b: \(3\left(x-2\right)\left(x+2\right)<3x^2+x\)

=>\(3\left(x^2-4\right)<3x^2+x\)

=>\(3x^2-12-3x^2-x<0\)

=>-x-12<0

=>x+12>0

=>x>-12

c: \(\left(2x+1\right)^2+3x\left(1-x\right)\le\left(x+2\right)^2\)

=>\(4x^2+4x+1+3x-3x^2\le x^2+4x+4\)

=>\(x^2+7x+1\le x^2+4x+4\)

=>7x+1<=4x+4

=>7x-4x<=4-1

=>3x<=3

=>x<=1

d: \(\frac{5x-20}{3}-\frac{2x^2+x}{2}>\frac{x\left(1-3x\right)}{3}-\frac{5x}{4}\)

=>\(\frac{4\left(5x-20\right)-6\left(2x^2+x\right)}{12}>\frac{4x\left(1-3x\right)-15x}{12}\)

=>\(4\left(5x-20\right)-6\left(2x^2+x\right)>4x\left(1-3x\right)-15x\)

=>\(20x-80-12x^2-6x>4x-12x^2-15x\)

=>14x-80>-11x

=>25x>80

=>\(x>\frac{80}{25}=\frac{16}{5}\)

e: 4-2x<=3x-6

=>-2x-3x<=-6-4

=>-5x<=-10

=>x>=2

f: \(\left(x+4\right)\left(5x-1\right)>5x^2+16x+2\)

=>\(5x^2-x+20x-4>5x^2+16x+2\)

=>19x-4>16x+2

=>3x>6

=>x>2

g: \(x\left(2x-1\right)-8<5-2x\left(1-x\right)\)

=>\(2x^2-x-8<5-2x+2x^2\)

=>-x-8<-2x+5

=>-x+2x<5+8

=>x<13

h: \(\frac{3x-1}{4}-\frac{3\left(x-2\right)}{8}-1>\frac{5-3x}{2}\)

=>\(\frac{2\left(3x-1\right)}{8}-\frac{3\left(x-2\right)}{8}-\frac88>\frac{4\left(5-3x\right)}{8}\)

=>2(3x-1)-3(x-2)-8>4(5-3x)

=>6x-2-3x+6-8>20-12x

=>3x-4>20-12x

=>15x>24

=>\(x>\frac{24}{15}\)

=>x>1,6

2x + 4 > 5x - 11

<=> 2x - 5x > -11 - 4

<=> -3x > -15

<=> -3x : ( -3 ) < -15 : ( -3 )

<=> x < 5

Vậy tập nghiệm của bất phương trình là x < 5 

Câu 1a : tự kết luận nhé 

\(2\left(x+3\right)=5x-4\Leftrightarrow2x+6=5x-4\Leftrightarrow-3x=-10\Leftrightarrow x=\frac{10}{3}\)

Câu 1b : \(\frac{1}{x-3}-\frac{2}{x+3}=\frac{5-2x}{x^2-9}\)ĐK : \(x\ne\pm3\)

\(\Leftrightarrow x+3-2x+6=5-2x\Leftrightarrow-x+9=5-2x\Leftrightarrow x=-4\)

c, \(\frac{x+1}{2}\ge\frac{2x-2}{3}\Leftrightarrow\frac{x+1}{2}-\frac{2x-2}{3}\ge0\)

\(\Leftrightarrow\frac{3x+3-4x+8}{6}\ge0\Rightarrow-x+11\ge0\Leftrightarrow x\le11\)vì 6 >= 0 

1) 2(x + 3) = 5x - 4

<=> 2x + 6 = 5x - 4

<=> 3x = 10

<=> x = 10/3

Vậy x = 10/3 là nghiệm phương trình 

b) ĐKXĐ : \(x\ne\pm3\)

\(\frac{1}{x-3}-\frac{2}{x+3}=\frac{5-2x}{x^2-9}\)

=> \(\frac{x+3-2\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{5-2x}{\left(x-3\right)\left(x+3\right)}\)

=> x + 3 - 2(x - 3) = 5 - 2x

<=> -x + 9 = 5 - 2x

<=> x = -4 (tm) 

Vậy x = -4 là nghiệm phương trình 

c) \(\frac{x+1}{2}\ge\frac{2x-2}{3}\)

<=> \(6.\frac{x+1}{2}\ge6.\frac{2x-2}{3}\)

<=> 3(x + 1) \(\ge\)2(2x - 2)

<=> 3x + 3 \(\ge\)4x - 4

<=> 7 \(\ge\)x

<=> x \(\le7\)

Vậy x \(\le\)7 là nghiệm của bất phương trình 

Biểu diễn

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