a) x.(x-1)=0

\(\Rightarrow\)x=0 hoặc x-1=0

\(\Rightarrow\)x=0+1

\(\Rightarrow\)x=1

vậy x=1 hoặc x=0

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

\(\Rightarrow\)-x = 0 hoặc x+3 = 0

\(\Rightarrow\)x= 0-3

\(\Rightarrow\)x=-3

vậy x=0 hoặc x=-3

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

(2x-4)= 0

2x=0+4

2x=4

x=4:2

x=2

hoặc (x+2)=0

x= 0-2

x=-2

vậy x=2 hoặc x=-2

d) (3-x).|x+5|=0

3-x = 0

x= 3-0

x=3

hoặc |x+5|=0

x+ 5=0

x=0-5

x=-5

vậy x=3 hoặc x=-5

e) (|x|+1).( 4-2x) = 0

(|x|+1) =0

|x|= 0-1

|x|=-1

hoặc( 4-2x) = 0

2x=4-0

2x=4

x=4:2

x=2

g) x²+5x=0

x²=0

x=0

hoặc 5x=0

x= 0: 5

x=0

vậy x=0

2)

a) (x+3).(y-5)= 7

(x+3)và (y-5)\(\in\)Ư(7)=\(\left\{1;-1;7;-7\right\}\)

b) xy + 3x - 2y= 11

x( y+3) -2y=11

x(y-3)- 2( y+3) +6 = 11

( y+3) ( x-2) = 5

vì x,y thuộc Z \(\Leftrightarrow\)y+3 và x-2 \(\in\)Z

do đó y+3 và x-2 \(\in\)Ư ( 5)= \(\left\{1;5;-1;-5\right\}\)

\(\in\)\(\in\)

c) xy + 3x - 7y= 21

x( y+3) -7y= 21

x( y+3) - 7( y+3)+21= 21

(y+3)( x-7) =0

a) \(x.\left(x+2\right)=0\)

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

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

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

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

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

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

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

Vậy \(x=2\)

d) \(\left(x+1\right)\left(x^2-4\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+1=0\\x^2-4=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=\pm2\end{matrix}\right.\)

Vậy \(\left[{}\begin{matrix}x=-1\\x=\pm2\end{matrix}\right.\)

A) x={0;-2}

B) x={1;2}

C) x=2

D) x={-2;-1;2}

Bài 1:

a) \(\left(x-1\right).\left(x+2\right)< 0\)

\(\Rightarrow\) \(x-1\) và \(x+2\) khác dấu.

Mà \(x-1< x+2.\)

Ta có:

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

Vậy nếu \(1>x>-2\) thì \(\left(x-1\right).\left(x+2\right)< 0.\)

Chúc bạn học tốt!

a) \(\left|x-2\right|=x\)

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

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

\(\Leftrightarrow2x=2\)

\(\Leftrightarrow x=1\left(tm\right)\)

Vậy ......................

b) \(\left|x+2\right|=x\)

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

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

\(\Leftrightarrow2x=-2\)

\(\Leftrightarrow x=-1\left(tm\right)\)

Vậy ...............

c) Ta có ;

\(\left|x-3,4\right|+\left|2,6-x\right|=0\)

Mà :

\(\left\{{}\begin{matrix}\left|x-3,4\right|\ge0\\\left|2,6-x\right|\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left|x-3,4\right|+\left|2,6-x\right|\ge\left|x-3,4+2,6-x\right|=\left|-0,8\right|=0,8>0\)

\(\Leftrightarrow\) ko tồn tại \(x\)

Chọn C

Bài 2:

a: =>x+32=0

=>x=-32

b: =>x-1=0

=>x=1

c: =>45-x=0 hoặc x=0

=>x=0 hoặc x=45

d: =>x-12=0 hoặc x+27=0

=>x=12 hoặc x=-27

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

<=> \(\hept{\begin{cases}x^2-5=0\\x^2+1=0\end{cases}}\)

<=> \(\hept{\begin{cases}x^2=5\\x^2=-1\end{cases}}\)

<=> \(\hept{\begin{cases}x=\sqrt{5};x=-\sqrt{5}\\x\in\varnothing\end{cases}}\)

câu còn lại tương tự nha

\(A.x^2-16x=0\)

\(x^2-\left(4x\right)^2=0\)

\(\left(x-4x\right)\left(x+4x\right)=0\)

\(\left(-3x\right)\left(5x\right)=0\)

\(\Rightarrow\) \(-3x=0\) hoặc \(5x=0\)

\(x=\dfrac{0}{-3}\) hoặc \(x=\dfrac{0}{5}\)

Vậy \(x=0\) hoặc \(x=0.\)

B. 4x² - 4x + 1 = 0

(2x)² - (2x)² + 1² = 0

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

1 (4x + 1) =0

=> 1 (4x + 1) =0

4x + 1 = 0

4x = 0-1

Vậy x = \(\dfrac{-1}{4}.\)

C. (3x-1)² - (2x+3)² = 0

(3x -1 -2x +3) (3x -1 +2x +3) = 0

(x + 2)(5x + 2) = 0

=> x + 2 =0 hoặc 5x + 2 =0

x = 0 - 2 hoặc 5x = 0 - 2

Vậy x = -2 hoặc x = \(\dfrac{-2}{5}.\)

Còn về câu d thì mình hơi phân vân, tại mình dốt toán lắm

a/ \(x^2-16x=0\)

\(\Leftrightarrow x\left(x-16\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-16=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=16\end{matrix}\right.\)

Vậy...

b/ \(4x^2-4x+1=0\)

\(\Leftrightarrow\left(2x-1\right)^2=0\)

\(\Leftrightarrow2x-1=0\)

\(\Leftrightarrow x=\dfrac{1}{2}\)

Vậy..

c/ \(\left(3x-1\right)^2-\left(2x+3\right)^2=0\)

\(\Leftrightarrow\left(3x-1-2x-3\right)\left(3x-1+2x+3\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(5x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\5x+2=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{2}{5}\end{matrix}\right.\)

Vậy...

d/ \(2013x^2-2014x+1=0\)

\(\Leftrightarrow2013x^2-x-2013x+1=0\)

\(\Leftrightarrow x\left(2013x-1\right)-\left(2013x-1\right)=0\)

\(\Leftrightarrow\left(2013x-1\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2013x-1=0\\x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2013}\\x=1\end{matrix}\right.\)

Vậy..