a) Ta có: \(3\left(x-2\right)-\left(x-5\right)>21\)

\(\Leftrightarrow3x-6-x+5>21\)

\(\Leftrightarrow2x-1>21\)

\(\Leftrightarrow2x>22\)

hay x>11

Vậy: S={x|x>11}

b) Ta có: \(5\left(x+1\right)-7\left(x-3\right)< 10\)

\(\Leftrightarrow5x+5-7x+21-10< 0\)

\(\Leftrightarrow-2x+16< 0\)

\(\Leftrightarrow-2x< -16\)

hay x>8

Vậy: S={x|x>8}

Cậu tách ra `2->3` câu thôi nhe

a: =>17x-5x-15-2x-5=0

=>10x-20=0

=>x=2

b: =>\(\dfrac{3x-6-5x-10}{\left(x+2\right)\left(x-2\right)}=\dfrac{11x+23}{\left(x+2\right)\left(x-2\right)}\)

=>11x+23=-2x-16

=>13x=-39

=>x=-3(nhận)

c: =>5x+7>=3x-3

=>2x>=-10

=>x>=-5

d: =>5(3x-1)=-2(x+1)

=>15x-5=-2x-2

=>17x=3

=>x=3/17

e: =>4x^2-1-4x^2-3x-2=0

=>-3x-3=0

=>x=-1

g: =>7x-5-8x+2-7<0

=>-x-10<0

=>x+10>0

=>x>-10

a: =>17x-5x-15-2x-5=0

=>10x-20=0

=>x=2

b: =>\(\dfrac{3x-6-5x-10}{\left(x+2\right)\left(x-2\right)}=\dfrac{11x+23}{\left(x+2\right)\left(x-2\right)}\)

=>11x+23=-2x-16

=>13x=-39

=>x=-3(nhận)

c: =>5x+7>=3x-3

=>2x>=-10

=>x>=-5

d: =>5(3x-1)=-2(x+1)

=>15x-5=-2x-2

=>17x=3

=>x=3/17

e: =>4x^2-1-4x^2-3x-2=0

=>-3x-3=0

=>x=-1

g: =>7x-5-8x+2-7<0

=>-x-10<0

=>x+10>0

=>x>-10

Lúc mới đọc đề tớ tưởng x+\(\frac{3}{2}\)chứ. Nhìn lại thì...

Câu B đây;vừa bị lag

B, \(\frac{x+1}{35}\)+\(\frac{x+3}{33}\)=\(\frac{x+5}{31}\)+\(\frac{x+7}{29}\)

⇔ \(\frac{x+1}{35}\)+1+\(\frac{x+3}{33}\)+1=\(\frac{x+5}{31}\)+1+\(\frac{x+7}{29}\)+1

⇔ \(\frac{x+36}{35}\)+\(\frac{x+36}{33}\)-\(\frac{x+36}{31}\)-\(\frac{x+36}{29}\)=0

⇔ (x+36)(\(\frac{1}{35}\)+\(\frac{1}{33}\)-\(\frac{1}{31}\)-\(\frac{1}{29}\))=0

Mà \(\frac{1}{35}\)+\(\frac{1}{33}\)-\(\frac{1}{31}\)-\(\frac{1}{29}\)<0

⇔ x+36=0

⇔ x=-36

Vậy tập nghiệm của phương trình đã cho là:S={-36}

câu C tương tự nhé

Câu 1 : 

a, \(\frac{3\left(2x+1\right)}{4}-\frac{5x+3}{6}=\frac{2x-1}{3}-\frac{3-x}{4}\)

\(\Leftrightarrow\frac{6x+3}{4}+\frac{3-x}{4}=\frac{2x-1}{3}+\frac{5x+3}{6}\)

\(\Leftrightarrow\frac{5x+6}{4}=\frac{9x+1}{6}\Leftrightarrow\frac{30x+36}{24}=\frac{36x+4}{24}\)

Khử mẫu : \(30x+36=36x+4\Leftrightarrow-6x=-32\Leftrightarrow x=\frac{32}{6}=\frac{16}{3}\)

tương tự 

\(\frac{19}{4}-\frac{2\left(3x-5\right)}{5}=\frac{3-2x}{10}-\frac{3x-1}{4}\)

\(< =>\frac{19.5}{20}-\frac{8\left(3x-5\right)}{20}=\frac{2\left(3-2x\right)}{20}-\frac{5\left(3x-1\right)}{20}\)

\(< =>95-24x+40=6-4x-15x+5\)

\(< =>-24x+135=-19x+11\)

\(< =>5x=135-11=124\)

\(< =>x=\frac{124}{5}\)

a: \(\dfrac{2x-3}{35}+\dfrac{x\left(x-2\right)}{7}< \dfrac{x^2}{7}-\dfrac{2x-3}{5}\)

\(\Leftrightarrow2x-3+5x\left(x-2\right)< 5x^2-7\left(2x-3\right)\)

\(\Leftrightarrow2x-3+5x^2-10x< 5x^2-14x+21\)

=>-8x-3<-14x+21

=>6x<24

hay x<4

3: \(\dfrac{3x-2}{4}< \dfrac{3x+3}{6}\)

\(\Leftrightarrow3\left(3x-2\right)< 2\left(3x+3\right)\)

=>9x-6<6x+6

=>3x<12

hay x<4