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}

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.

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)

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

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(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

1) Sửa lại đề là \(7.\left(x-1\right)+2x.\left(1-x\right)=0\)

⇒ \(7.\left(x-1\right)-2x.\left(x-1\right)=0\)

⇒ \(\left(x-1\right).\left(7-2x\right)=0\)

⇒ \(\left[{}\begin{matrix}x-1=0\\7-2x=0\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}x=0+1\\2x=7-0=7\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}x=1\\x=7:2\end{matrix}\right.\)

⇒ \(\left[{}\begin{matrix}x=1\\x=\frac{7}{2}\end{matrix}\right.\)

Vậy \(x\in\left\{1;\frac{7}{2}\right\}.\)

Mình chỉ làm câu 1) thôi nhé.

Chúc bạn học tốt!