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a: Ta có: \(2x+3>1-x\)

\(\Leftrightarrow3x>-2\)

hay \(x>-\dfrac{2}{3}\)

b: Ta có: \(15-2\left(x-3\right)< -2x+5\)

\(\Leftrightarrow15-2x+6+2x-5< 0\)

\(\Leftrightarrow16< 0\left(vôlý\right)\)

c: Ta có: \(\left(x+1\right)\left(x-3\right)\le\left(x+4\right)\left(x-1\right)\)

\(\Leftrightarrow x^2-3x+x-3-x^2+x-4x+4\le0\)

\(\Leftrightarrow-5x\le-1\)

hay \(x\ge\dfrac{1}{5}\)

d: Ta có: \(\dfrac{2x+1}{3}-\dfrac{1-x}{2}\ge1-\dfrac{x}{4}\)

\(\Leftrightarrow8x+4-6+6x\ge12-3x\)

\(\Leftrightarrow14x+3x\ge12+2=14\)

\(\Leftrightarrow x\ge\dfrac{14}{17}\)

e: Ta có: \(\dfrac{x+1}{2}-\dfrac{2-x}{3}< \dfrac{2x-3}{4}\)

\(\Leftrightarrow6x+12+4x-8< 6x-9\)

\(\Leftrightarrow4x< -9+8-12=-13\)

hay \(x< -\dfrac{13}{4}\)

f: Ta có: \(\left(x+1\right)\left(x-2\right)-\left(2-x\right)\left(3-x\right)>0\)

\(\Leftrightarrow x^2-2x+x-2-\left(x-2\right)\left(x-3\right)>0\)

\(\Leftrightarrow x^2-x-2-x^2+5x-6>0\)

\(\Leftrightarrow4x>8\)

hay x>2

g: Ta có: \(\left(2x-1\right)^2\le2\left(x-1\right)^2\)

\(\Leftrightarrow4x^2-4x+1-2x^2+4x-2\le0\)

\(\Leftrightarrow2x^2\le1\)

\(\Leftrightarrow x^2\le\dfrac{1}{2}\)

\(\Leftrightarrow-\dfrac{\sqrt{2}}{2}\le x\le\dfrac{\sqrt{2}}{2}\)

12 tháng 10 2017

a)

1: \(\Leftrightarrow x^2+6x+9-6x+3>x^2-4x\)

=>-4x<12

hay x>-3

2: \(\Leftrightarrow6+2x+2>2x-1-12\)

=>8>-13(đúng)

4: \(\dfrac{2x+1}{x-3}\le2\)

\(\Leftrightarrow\dfrac{2x+1-2x+6}{x-3}< =0\)

=>x-3<0

hay x<3

6: =>(x+4)(x-1)<=0

=>-4<=x<=1

3 tháng 12 2019

a: Ta có: x+17<10

nên x<-7

b: Ta có: 9-2x<0

\(\Leftrightarrow2x>9\)

hay \(x>\dfrac{9}{2}\)

c: Ta có: \(-3x-11\ge0\)

\(\Leftrightarrow-3x\ge11\)

hay \(x\le-\dfrac{11}{3}\)

1:

a: 2x-3=5

=>2x=8

=>x=4

b: (x+2)(3x-15)=0

=>(x-5)(x+2)=0

=>x=5 hoặc x=-2

2:

b: 3x-4<5x-6

=>-2x<-2

=>x>1

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