Bài 2. Trường hợp 1. Với x = - 2 thay vào ta có:     - 2y - 4 + 2y = - 9 suy ra - 4 = - 9 không xẩy ra.

Trường hợp 2.  Với     x \(\ne\) - 2

Ta có: \(xy+2x+2y=-9\Leftrightarrow\left(x+2\right)y=-2x-9\Rightarrow y=-\frac{2x+9}{x+2}\)

Do \(y\in Z\Rightarrow\frac{2x+9}{x+2}\in Z\Rightarrow\frac{2x+4+5}{x+2}\in Z\Rightarrow\frac{2\left(x+2\right)+5}{x+2}\in Z\)

\(\Rightarrow2+\frac{5}{x+2}\in Z\Rightarrow\frac{5}{x+2}\in Z\Rightarrow x+2\inƯ\left(5\right)=\left\{-5;-1;1;5\right\}\)

\(\)Với x + 2 = - 5    =>   x = - 7    => y = 1

   Với x + 2 = -1     =>   x = -3     => y = -3

   Với x + 2 = 1      =>   x = -1     => y = 7

   Với x + 2 = 5      =>   x = 3      => y = 3

a, ( x-3) (2x+5)=0                           b, (2x-1)(5-x)=0                      c, (x-1)(x^2-1)=0

=> x-3=0 hoặc 2x+5=0              => 2x-1=0 hoặc 5-x=0             => x-1=0 hoặc x^2-1=0

=> x=3 ; x=-5/2                            => x=1/2; x=5                        => x=1 ; x= -1

\(\left|2x\right|+2x=0\)

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

\(\Rightarrow2x\le0\)

\(\Rightarrow x\le0\)

Vậy \(x\le0\)

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

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

Vậy \(\orbr{\begin{cases}x=1\\x=-2\end{cases}}\)

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

\(\left|x-3\right|=-x+3\)

\(\left|x-3\right|=-\left(x-3\right)\)

\(\Rightarrow x-3\le0\)

\(\Rightarrow x\le3\)

Vậy \(x\le3\)

\(\left(x+1\right)^3=\left(x+1\right)^5\)

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

\(\left(x+1\right)^3.\left[\left(x+1\right)^2-1\right]=0\)

\(\orbr{\begin{cases}\left(x+1\right)^3=0\\\left(x+1\right)^2-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=0\end{cases}}}\)hoặc \(x=-2\)

Vậy \(x\in\left\{-1;0;-2\right\}\)

\(\left(x-2\right)^3=2^9\)

\(\left(x-2\right)^3=\left(2^3\right)^3\)

\(\Rightarrow x-2=2^3\)

\(x=8+2\)

\(x=10\)

Vậy \(x=10\)

Câu 6 tương tự câu 4

Tham khảo nhé~

P/S: nên chia nhỏ đăng thành nhiều bài khác nhau

Bài 1: Tìm x , biết :

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

<x-1=0

|

<x=0+1=1

-<x-2=0

-<x=0+2=2

Vậy x E {1;2}

b) (x-2).(x^2+1)=0

[<x-2=0

[<x=0+2=2

[>x²+1=0

   x²=0-1

   x²=1.(-1)

c) (x+`1).(x^2-4)=0

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

1-3x-3=0

1-3 .x=0+3

1-3.x=3

3.x=3+1

3.x=4

x=4:3

x=\(\frac{4}{3}\)

/2.x-1/+2.x=1

 /2.x-1/+x=0,5

=>2x-1+x=0,5

     2x+x=1+0,5

        3x=1,5

          x=0,5

a, \(\frac{1}{3}x+\frac{2}{5}\left(x-1\right)=0\)

\(\frac{1}{3}x+\frac{2}{5}x-\frac{2}{5}=0\)

\(\left(\frac{1}{3}+\frac{2}{5}\right)x-\frac{2}{5}=0\)

\(\frac{11}{15}x-\frac{2}{5}=0\)

\(\frac{11}{15}x=\frac{2}{5}\)

x = \(\frac{2}{5}:\frac{11}{15}\)

x = \(\frac{6}{11}\)

( 2x - 3 )( 6-2x ) = 0 

=> \(\orbr{\begin{cases}2x-3=0\\6-2x=0\end{cases}}\)

=> \(\orbr{\begin{cases}2x=3\\2x=6\end{cases}}\)

=>\(\orbr{\orbr{\begin{cases}x=\frac{3}{2}\\x=3\end{cases}}}\)

                                                 Xong rùi :3 Chúc bạn hok thật tốt nhé ^_^ 

a) 3x - 2 = 0    =>   3x = 2    => x = 2/3

b) 2x - 1 = 0     =>  2x = 1      =>  x = 1/2

c) 5 ( 4+2x) = 8+5x

<=> 20 + 10x = 8 + 5x

<=> 10x - 5x = 8 - 20

<=>  5x  =  -12

x = -12/5

d) \(\frac{1}{2}+\frac{3}{4}x=6-\frac{4}{5}x\)

\(\frac{3}{4}x+\frac{4}{5}x=6-\frac{1}{2}\)

\(\frac{31}{20}x=\frac{11}{2}\)

\(x=\frac{11}{2}:\frac{31}{20}=\frac{110}{31}\)

e) 3 + 2x = 4 - 8x

<=> 2x + 8x = 4 - 3

10 x = 1

x = 1/10

f \(5+\frac{1}{2}\left(x+5\right)=3\)

\(\frac{1}{2}\left(x+5\right)=3-5=-2\)

\(x+5=-2:\frac{1}{2}=-4\)

\(x=-4-5=1\)

Vậy ......

a, 3x - 2 = 0

=> 3x = 2

=> x = 2/3

vậy_