Ta có: \(\dfrac{x+2}{x-3}< 0\)

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

Vậy: S={x|-2<x<3}

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

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< -2\\x>3\end{matrix}\right.\\\left\{{}\begin{matrix}x>-2\\x< 3\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow-2< x< 3\)

Vậy ...

a) |x-7|=2x+3  (1)

Ta có:|x-7|=x-7<=>x-7 \(\ge\) 0<=>x\(\ge\)7

         |x-7|=-(x-7)<=>x-7<0<=>x<7

Nếu x\(\ge\)  7thì (1) <=>x-7=2x+3

                         <=>x-2x=7+3

                         <=>-x    =  10

                        <=>x=-10 (ko thỏa mãn đk)

Nếu x<7 thì (1) <=>-(x-7)=2x+3

                          <=>-x+7=2x+3

                        <=>-x-2x=-7+3

                        <=>-3x=-4

                       <=>x=4/3 (thỏa mãn đk)

=>(4x-1)(-x+4)>0

=>(4x-1)(x-4)<0

=>1/4<x<4

A, 3X+6>0

(=)3X>-6

(=)X>-2

VẬY ...

B,10-2X≥-4

(=)-2X≥-4-10

(=)-2X≥-14

(=)X≤7

VẬY....

C,

(=)

(=) -15X+10>-3+3X

(=)-15X-3X>-3-10

(=)-18X>-13

(=)X<

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: =>x+2<=0

=>x<=-2

b: =>-x>-1

=>x<1

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

=>4>=-1(luôn đúng)

d: =>2x+2<5x-10-3x

=>2x+2<2x-10

=>2<-10(loại)

1/  

\(x^2-2x+1< \left(x-1\right)\left(x-4\right)\)

\(\Rightarrow x^2-2x+1< x^2-4x-x+4\)

\(\Rightarrow x^2-2x+1< x^2-5x+4\)

\(\Rightarrow x^2-x^2-2x+5x< 4-1\)

\(\Rightarrow3x< 3\)

\(\Rightarrow x< 1\)

\(\Rightarrow S=\left\{x\in R;x< 1\right\}\)

câu 1 

a) 5x(x-2)=0 =>\(\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

b)(x+5)(2x-7)=0 =>\(\left[{}\begin{matrix}x+5=0\\2x-7=0\end{matrix}\right.\)=>\(\left[{}\begin{matrix}x=-5\\x=\dfrac{7}{2}\end{matrix}\right.\)