a) 3x+2>2b-3

\(\Leftrightarrow\)?

b) 5x-1>4x+3

\(\Leftrightarrow\)5x-4x>3+1

\(\Leftrightarrow\)x>4

Vậy phương trình có tập nghiệm S={x|x>4}

c)2-x/3>3-2x/5

\(\Leftrightarrow\)2-3>(-2x/5)+(x/3)

\(\Leftrightarrow\)-1>-x/15

\(\Leftrightarrow\)1<x/15

\(\Leftrightarrow\)x>1/15

Vậy phương trình có tập nghiệm S={x|x>1/15}

Câu a đề nó hơi lạ nhỉ?

a: \(\frac{2x+2}{5}+\frac{3}{10}<\frac{3x-2}{4}\)

=>\(\frac{4\left(2x+2\right)}{20}+\frac{6}{20}<\frac{5\left(3x-2\right)}{20}\)

=>4(2x+2)+6<5(3x-2)

=>8x+8+6<15x-10

=>8x+14<15x-10

=>-7x<-24

=>\(x>\frac{24}{7}\)

b: \(\frac{2+x}{3}<\frac{3+2x}{5}\)

=>\(\frac{x+2}{3}-\frac{2x+3}{5}<0\)

=>\(\frac{5\left(x+2\right)-3\left(2x+3\right)}{15}<0\)

=>5(x+2)-3(2x+3)<0

=>5x+10-6x-9<0

=>-x+1<0

=>-x<-1

=>x>1

d: \(1+\frac{3\left(x+1\right)}{10}>\frac{x-2}{5}\)

=>\(\frac{10+3\left(x+1\right)}{10}>\frac{2\left(x-2\right)}{10}\)

=>3(x+1)+10>2(x-2)

=>3x+3+10>2x-4

=>3x+13>2x-4

=>3x-2x>-4-13

=>x>-17

e: \(\frac{2x-7}{6}\ge\frac{3x-7}{2}\)

=>\(\frac{2x-7}{6}\ge\frac{3\left(3x-7\right)}{6}\)

=>2x-7>=3(3x-7)

=>2x-7>=9x-21

=>-7x>=-14

=>x<=2

f: \(\frac{2x-1}{3}>\frac{3x+1}{2}\)

=>\(\frac{2x-1}{3}-\frac{3x+1}{2}>0\)

=>\(\frac{2\left(2x-1\right)-3\left(3x+1\right)}{6}>0\)

=>2(2x-1)-3(3x+1)>0

=>4x-2-9x-3>0

=>-5x-5>0

=>5x+5<0

=>5x<-5

=>x<-1

\(\frac{x+4}{5}+\frac{3x+2}{10}< \frac{x-1}{3}\)

\(\Leftrightarrow\frac{6\left(x+4\right)}{30}+\frac{3\left(3x+2\right)}{30}< \frac{10\left(x-1\right)}{30}\)

\(\Leftrightarrow6x+24+9x+6< 10x-10\)

\(\Leftrightarrow5x+40< 0\)

\(\Leftrightarrow x< -8\)

Tự biểu diễn nha bạn

\(\frac{x+4}{5}+\frac{3x+2}{10}< \frac{x-1}{3}\)

\(\Rightarrow\frac{6\left(x+4\right)}{30}+\frac{3\left(3x+2\right)}{30}< \frac{10\left(x-1\right)}{30}\)

\(\Rightarrow6x+24+9x+6< 10x-10\)

\(5x< -40\)

\(\Rightarrow x< -8\)

=>4x>16

=>x>4

\(a,3x-11>5-x\\ \Leftrightarrow3x+x>5+11\\ \Leftrightarrow4x< 16\\ \Leftrightarrow x>4\)

Vậy bất phương trình có nghiệm là: \(S=\left\{x|x>4\right\}\)

biểu diễn

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