e)

A = \(\frac{x+5}{x-2}\) = \(\frac{\left(x-2\right)+7}{x-2}=1+\frac{7}{x-2}\)

Muốn A nguyên thì:

=> \(\frac{7}{x-2}\) ∈ Z

=> 7 ⋮ x - 2

=> x - 2 ∈ Ư (7)

=> x - 2 ∈ { 1; 7; -1; -7 }

=> x ∈ { 3; 9; -5; 1 }

a) (5x - 1)(2x - 1/3) = 0

\(\Rightarrow5x-1=0\) hoặc \(2x-\frac{1}{3}=0\)

\(\Rightarrow\left[{}\begin{matrix}5x=0+1\\2x=0+\frac{1}{3}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}5x=1\\2x=\frac{1}{3}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=1:5=\frac{1}{5}\\x=\frac{1}{3}:2=\frac{1}{3}.\frac{1}{2}=\frac{1}{6}\end{matrix}\right.\)

Vậy x = 1/5 hoặc x = 1/6

a: =>x+3>0

hay x>-3

b: =>4-x<0

hay x>4

c: =>x²-1=0 hoặc x+5=0

hay \(x\in\left\{1;-1;-5\right\}\)

j) (x + 2).(3 - x) = 0

TH1:  x + 2 = 0

          x = 0 - 2

          x = (-2)

TH2: 3 - x = 0

              x = 3 - 0

              x = 3

⇒ Vậy x = (-2) hoặc x = 3.

m) (x + 5).(x.2 - 4) = 0

TH1: x + 5 = 0

         x = 0 - 5

         x = (-5)

TH2: x.2 - 4 = 0

         x.2 = 0 + 4

         x.2 = 4

         x = 4 : 2

         x = 2

⇒ Vậy x = (-5) hoặc x = 2.

g) 0,3x + 0,6x = 9

x.(0,3 + 0,6) = 9

x. 0,9 = 9

x = 9 : 0,9

x = 10

j) (x + 2)(3 - x) = 0

x + 2 = 0 hoặc 3 - x = 0

*) x + 2 = 0

x = 0 - 2

x = -2

*) 3 - x = 0

x = 3

Vậy x = -2; x = 3

m) (x + 5)(x.2 - 4) = 0

x + 5 = 0 hoặc x.2 - 4 = 0

*) x + 5 = 0

x = 0 - 5

x = -5

*) x.2 - 4 = 0

2x = 0 + 4

2x = 4

x = 4 : 2

x = 2

Vậy x = -5; x = 2

g) 0,3x + 0,6x = 9

0,9x = 9

x = 9 : 0,9

x = 10

Bài 3: 

a: x(x-1)=0

=>x=0 hoặc x-1=0

=>x=0 hoặc x=1

b: (x-3)(x+4)=0

=>x-3=0 hoặc x+4=0

=>x=3 hoặc x=-4

c: (2x-4)(x+2)=0

=>2x-4=0 hoặc x+2=0

=>x=2 hoặc x=-2

d: (x+1)²(x-2)²=0

=>x+1=0 hoặc x-2=0

=>x=-1 hoặc x=2

Bài 2: - Xét dấu :

P1 : (-).(+).(-).(-) -> Kết quả cuối cùng là số âm.

P2 : (-).(-).(-).(-).(+) -> Kết quả cuối cùng là số dương.

===> P1 < P2.

Bài 3 :

a) \(x\cdot\left(x-1\right)=0\)

\(\Rightarrow\left[\begin{matrix}x=0\\x-1=0\end{matrix}\right.\rightarrow\left[\begin{matrix}x=0\\x=1\end{matrix}\right.\)

b) \(\left(x-3\right)\cdot\left(x+4\right)=0\)

\(\left[\begin{matrix}x-3=0\\x+4=0\end{matrix}\right.\rightarrow\left[\begin{matrix}x=3\\x=-4\end{matrix}\right.\)

c) \(\left(2x-4\right)\cdot\left(x+2\right)=0\rightarrow\left[\begin{matrix}2x-4=0\\x+2=0\end{matrix}\right.\rightarrow\left[\begin{matrix}x=2\\x=-2\end{matrix}\right.\)

d) \(\left(x+1\right)^2\cdot\left(x-2\right)^2=0\rightarrow\left[\begin{matrix}x+1=0\\x-2=0\end{matrix}\right.\rightarrow\left[\begin{matrix}x=-1\\x=2\end{matrix}\right.\)

e) \(x\cdot\left(x+1\right)\cdot\left(x+2\right)^2\cdot\left(x+3\right)^3=0\)

\(\Rightarrow\left[\begin{matrix}x=0\\x+1=0\\x+2=0\\x+3=0\end{matrix}\right.\rightarrow\left[\begin{matrix}x=0\\x=-1\\x=-2\\x=-3\end{matrix}\right.\)

f) \(\left(x-9^5\right)\cdot\left(x-5\right)^8=0\)

\(\Rightarrow\left[\begin{matrix}x-9=0\\x-5=0\end{matrix}\right.\rightarrow\left[\begin{matrix}x=9\\x=5\end{matrix}\right.\)

g) \(x\cdot\left(x+100\right)^{10}\cdot\left(x+2000\right)^{20}\cdot\left(x+300\right)^{3000}=0\)

\(\Rightarrow\left[\begin{matrix}x=0\\x+100=0\\x+2000=0\\x+300=0\end{matrix}\right.\rightarrow\left[\begin{matrix}x=0\\x=-100\\x=-2000\\x=-300\end{matrix}\right.\)

h) \(\left(x-2\right)^2=0\rightarrow x=2\)

a: (x+5)(3x-12)>0

=>(x-4)(x+5)>0

=>x>4 hoặc x<-5

b: (x+4)(x-6)<0

=>x+4>0 và x-6<0

=>-4<x<6

c: (5x+5)(x+2)<0

=>(x+1)(x+2)<0

=>-2<x<-1

d: =>(x-2)(x+2)<=0

=>-2<=x<=2

a/ \(\left(x+5\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\x-4=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=4\end{matrix}\right.\)

Vậy ........

b/ \(x\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)

c/ \(\left(x+2\right)\left(x+5\right)>0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+2< 0\\x+5< 0\end{matrix}\right.\\\left\{{}\begin{matrix}x+5>0\\x+2>0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< -2\\x< -5\end{matrix}\right.\\\left\{{}\begin{matrix}x>-5\\x>-2\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x< -5\\x>-2\end{matrix}\right.\)

Vậy ..

Bài mình làm nha. Tại thấy giống quá, lười nên chụp!

1/\(x.\left(x+7\right)=0\)

\(\Rightarrow\left[\begin{matrix}x=0\\x+7=0\end{matrix}\right.\)

\(\Rightarrow\left[\begin{matrix}x=0\\x=0-7\end{matrix}\right.\)

\(\Rightarrow\left[\begin{matrix}x=0\\x=-7\end{matrix}\right.\)

2/\(\left(x+12\right).\left(x-3\right)=0\)

\(\Rightarrow\left[\begin{matrix}x+12=0\\x-3=0\end{matrix}\right.\)

\(\Rightarrow\left[\begin{matrix}x=0-12\\x=0+3\end{matrix}\right.\)

\(\Rightarrow\left[\begin{matrix}x=-12\\x=3\end{matrix}\right.\)

3/\(\left(-x+5\right).\left(3-x\right)\)

\(\Rightarrow\left[\begin{matrix}-x+5=0\\3-x=0\end{matrix}\right.\)

\(\Rightarrow\left[\begin{matrix}-x=0-5\\x=3-0\end{matrix}\right.\)

\(\Rightarrow\left[\begin{matrix}-x=-5\\x=3\end{matrix}\right.\)

4/\(x.\left(2+x\right).\left(7-x\right)\)

\(\Rightarrow\left[\begin{matrix}x=0\\2+x=0\\7-x=0\end{matrix}\right.\)

\(\Rightarrow\left[\begin{matrix}x=0\\x=0-2\\x=7-0\end{matrix}\right.\)

\(\Rightarrow\left[\begin{matrix}x=0\\x=-2\\x=7\end{matrix}\right.\)

5/\(\left(x-1\right).\left(x+2\right).\left(-x-3\right)=0\)

\(\Rightarrow\left[\begin{matrix}x-1=0\\x+2=0\\-x-3=0\end{matrix}\right.\)

\(\Rightarrow\left[\begin{matrix}x=0+1\\x=0-2\\-x=0+3\end{matrix}\right.\)

\(\Rightarrow\left[\begin{matrix}x=1\\x=-2\\-x=3\end{matrix}\right.\)