Bài 1: 

c) |2x - 1| = x + 2

<=> 2x - 1 = +(x + 2) hoặc -(x + 2)

* 2x - 1 = x + 2      

<=> 2x - x = 2 + 1

<=> x = 3

* 2x - 1 = -(x + 2)

<=> 2x - 1 = x - 2

<=> 2x - x = -2 + 1

<=> x = -1

Vậy.....

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

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

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

<=> \(\dfrac{2x-3}{35}\) + \(\dfrac{5x\left(x-2\right)}{7.5}\) < \(\dfrac{5x^2}{7.5}\) - \(\dfrac{7\left(2x-3\right)}{7.5}\)

<=> 2x-3 + 5x²-10x < 5x² - 14x + 21

<=> 5x² - 5x² + 2x -10x + 14x < 21 + 3

<=> 6x < 24

<=> x < 4

vậy bpt có tập nghiệm S={ x < 4 }

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

<=> \(\dfrac{6\left(3x-2\right)}{6.4}\) < \(\dfrac{4\left(3x+3\right)}{6.4}\)

<=> 18x - 12 < 12x +12

<=> 18x - 12x < 12 + 12

<=>6x < 24

<=> x < 4

vậy bpt có tập nghiệm S={ x < 4 }

\(\left(x^2+5\right)\left(2x+3\right)\left(3x-1\right)< 0\)

Do \(\left(x^2+5\right)>0\)

\(\Rightarrow bpt\Leftrightarrow\left(2x+3\right)\left(3x-1\right)< 0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2x+3>0\\3x-1< 0\end{matrix}\right.\\\left\{{}\begin{matrix}2x+3< 0\\3x-1>0\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>\frac{-3}{2}\\x< \frac{1}{3}\end{matrix}\right.\\\left\{{}\begin{matrix}x< \frac{-3}{2}\\x>\frac{1}{3}\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\frac{-3}{2}< x< \frac{1}{3}\left(chon\right)\\\frac{1}{3}< x< \frac{-3}{2}\left(loai\right)\end{matrix}\right.\)

Vậy...

Gợi ý là : DK 2x+5>0 

Bình phương 2 vế không âm 

9x^2+6x+1<4x^2+20x+25

<=> 5x^2-14x-24<0 

<=> -6/5<X<4  NHỚ ĐỐI CHIẾU ĐIỀU KIỆN :)

bạn ns rõ hơn đi

\(\Leftrightarrow\left|2x-1\right|< 3x+5\Leftrightarrow\left[{}\begin{matrix}2x-1< 3x+5\\1-2x< 3x+5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x-3x< 5+1\\-2x-3x< 5-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-x< 6\\-5x< 4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>-6\\x>-\frac{4}{5}\end{matrix}\right.\)