1) Do x ∈ Z và 0 < x < 3

⇒ x ∈ {1; 2}

2) Do x ∈ Z và 0 < x ≤ 3

⇒ x ∈ {1; 2; 3}

3) Do x ∈ Z và -1 < x ≤ 4

⇒ x ∈ {0; 1; 2; 3; 4}

3) \(x\left(x-4\right)+\left(x-4\right)^2=0\Leftrightarrow\left(x-4\right)\left(x+x-4\right)=0\Leftrightarrow2\left(x-4\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)

4x.(x+1)-8(x+1)=0

(4x-8)(x+1)=0

suy ra x=2 hoặc x=-1

1) \(4x\left(x+1\right)=8\left(x+1\right)\Leftrightarrow4x^2+4x=8x+8\Leftrightarrow4x^2-4x-8=0\Leftrightarrow x^2-x-2=0\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\)

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

\(\left(x-2\right)\left(x-3\right)>0\)

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

\(\Rightarrow x>2;x< 3\)

\(\dfrac{x+1}{x+2}< 0\)

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

\(\Rightarrow-2< x< -1\)

\(\left(x-1\right)\left(x+3\right)>0\)

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

\(\Rightarrow-3< x< 1\)

\(\dfrac{x+3}{x-1}< 0\)

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

\(\dfrac{x+5}{x+8}>1\)

\(\Rightarrow x+5>x+8\)

(đến đây chịu)

\(\Rightarrow-3< x< 1\)

x²( x + 1 ) + 2x( x + 1 ) = 0 <=> x( x + 1 )( x + 2 ) = 0 <=> x = 0 hoặc x = -1 hoặc x = -2

x( 3x - 1 ) - 5( 1 - 3x ) = 0 <=> x( 3x - 1 ) + 5( 3x - 1 ) = 0 <=> ( 3x - 1 )( x + 5 ) = 0 <=> x = 1/3 hoặc x = -5

Trả lời:

1, \(x^2\left(x+1\right)+2x\left(x+1\right)=0\)

\(\Leftrightarrow x\left(x+1\right)\left(x+2\right)=0\)

\(\Leftrightarrow x=0;x=-1;x=-2\)

Vậy x = 0; x = - 1; x = - 2 là nghiệm của pt.

2, \(x\left(3x-1\right)-5\left(1-3x\right)=0\)

\(\Leftrightarrow x\left(3x-1\right)+5\left(3x-1\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(x+5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}3x-1=0\\x+5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=-5\end{cases}}}\)

Vậy x = 1/3; x = - 5 là nghiệm của pt.

\(\text{1 , ( x - 3 ) . ( 4 - x ) = 0}\)

\(\Rightarrow\orbr{\begin{cases}x-3=0\\4-x=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=3\in Z\\x=4\in Z\end{cases}}\)

vậy______

\(2,\left(x-5\right)\left(x^2+1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-5=0\\x^2+1=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=5\in Z\\x\in\varnothing\end{cases}}\)

vậy x = 5

3, ( x + 1 ) + ( x + 2 ) + (x + 3 ) + ... +( x + 99 ) = 0

(x+x+x+....+x)+(1+2+3+.....+99) = 0

(x.99) + 5050 = 0

x.99 = 0-5050

x.99 = -5050

x = -5050 : 99

x = \(\frac{-5050}{99}\notin Z\Rightarrow x\in\varnothing\)

vậy_____

1, suy ra x-3=0 suy ra x=3

  suy ra  4-x=0 suy ra x=4

a) \(\Rightarrow x\left(x-3\right)+\left(x-3\right)=0\)

\(\Rightarrow\left(x+1\right)\left(x-3\right)=0\)

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

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

b) \(\Rightarrow7x\left(x-1\right)-\left(x-1\right)=0\)

\(\Rightarrow\left(7x-1\right)\left(x-1\right)=0\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{7}\\x=1\end{matrix}\right.\).

a) \(x\left(x-3\right)+x-3=0\) (1)

\(\Leftrightarrow x\left(x-3\right)+\left(x-3\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x-3\right)=0\)

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

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

Vậy tập nghiệm phương trình (1) là \(S=\left\{-1;3\right\}\)

b) \(7x\left(x-1\right)-x+1=0\) (2)

\(\Leftrightarrow7x\left(x-1\right)-\left(x-1\right)=0\)

\(\Leftrightarrow\left(7x-1\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}7x-1=0\\x-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{7}\\x=1\end{matrix}\right.\)

Vậy tập nghiệm phương trình (2) là \(S=\left\{\dfrac{1}{7};1\right\}\)

Câu a : \(x\left(x-3\right)+x-3=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+1\right)=0\)

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

Câu b : \(7x\left(x-1\right)-x+1=0\)

\(\Leftrightarrow7x\left(x-1\right)-\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(7x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\Rightarrow x=1\\7x-1=0\Rightarrow x=\dfrac{1}{7}\end{matrix}\right.\)

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

<=> x+5=0 hoặc x-4=0

<=> x=-5 hoặc x=4

b) (x-1)(x-3)=0

<=> x-1=0 hoặc x-3=0

<=> x=1 hoặc x=3

a) (x+5).(9x-4)=0

=> x+5=0 hoặc 9x-4=0

Nếu x+5=0: x=0-5=-5

Nếu 9x-4=0: 9x=0+4=4

                     x=4/9

b) (x-1).(x-3)=0

=> x-1=0 hoặc x-3=0

Nếu x-1=0: x=0+1=1

Nếu x-3=0: x=0+3=3

c) (3-x).(x-3)=0

=> 3-x=0 hoặc x-3=0

Nếu 3-x=0: x=3-0=0

Nếu x-3=0: x=0+3=3

d) x.(x+1)=0

=> x=0 hoặc x+1=0

Nếu x+1=0: x=0-1=-1

\(a,\left(x+5\right)\left(9x-4\right)=0\)

<=>\(\orbr{\begin{cases}x+5=0\\9x-4=0\end{cases}}\)

<=>\(\orbr{\begin{cases}x=-5\\x=\frac{4}{9}\end{cases}}\)

\(b,\left(x-1\right)\left(x-3\right)=0\)

<=>\(\orbr{\begin{cases}x-1=0\\x-3=0\end{cases}}\)

<=>\(\orbr{\begin{cases}x=1\\x=3\end{cases}}\)

\(c,\left(3-x\right)\left(x-3\right)=0\)

<=>\(\orbr{\begin{cases}3-x=0\\x-3=0\end{cases}}\)

<=>\(\orbr{\begin{cases}x=3\\x=3\end{cases}}\)

\(d,x\left(x+1\right)=0\)

<=>\(\orbr{\begin{cases}x=0\\x+1=0\end{cases}}\)

<=>\(\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)

a) x(x-1) - (x+1)(x+2) = 0

    x\(^2\)- x -x\(^{^2}\)-2x +x+2=0

     -2x+2=0

      -2x=0+2

       -2x=2

         x=-1

Vậy x bằng -1