\(\dfrac{x+2}{x-2}-\dfrac{2}{x^2-2x}=\dfrac{1}{x}\)

\(\Leftrightarrow\dfrac{x+2}{x-2}-\dfrac{2}{x\left(x-2\right)}=\dfrac{1}{x}\)

ĐKXĐ:\(\left\{{}\begin{matrix}x\ne0\\x\ne2\end{matrix}\right.\)

\(\Leftrightarrow\dfrac{x\left(x+2\right)}{x\left(x-2\right)}-\dfrac{2}{x\left(x-2\right)}=\dfrac{x-2}{x\left(x-2\right)}\)

\(\Leftrightarrow x^2+2x-2=x-2\)

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

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

Vậy phương trình đã cho có nghiệm \(x=-1\)

Ta có: \(\dfrac{x+2}{x-2}-\dfrac{2}{x^2-2x}=\dfrac{1}{x}\)

\(\Leftrightarrow x\left(x+2\right)-2=x-2\)

\(\Leftrightarrow x^2+2x-x=0\)

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

hay x=-1

Bạn tham khảo nhé : 

Ta có : 

\(\frac{x-3}{2012}+\frac{x-2}{2013}=\frac{x-2013}{2}+\frac{x-2012}{3}\)

\(\Leftrightarrow\)\(\left(\frac{x-3}{2012}-1\right)+\left(\frac{x-2}{2013}-1\right)=\left(\frac{x-2013}{2}-1\right)+\left(\frac{x-2012}{3}-1\right)\)

\(\Leftrightarrow\)\(\frac{x-2015}{2012}+\frac{x-2015}{2013}=\frac{x-2015}{2}+\frac{x-2015}{3}\)

\(\Leftrightarrow\)\(\frac{x-2015}{2012}+\frac{x-2015}{2013}-\frac{x-2015}{2}-\frac{x-2015}{3}=0\)

\(\Leftrightarrow\)\(\left(x-2015\right)\left(\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2}-\frac{1}{3}\right)=0\)

Mà \(\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2}-\frac{1}{3}\ne0\)

\(\Rightarrow\)\(x-2015=0\)

\(\Rightarrow\)\(x=2015\)

Vậy \(x=2015\)

Chsuc bạn học tốt ~

chuyển vế t cs 

x+x-x-x=....

0x=.....

Suy ra vô nghiệm

(=) [(x-3/2012)-1]+[(x-2/2013)-1]= [(x-2013/2)-1]+[(x-2012/3)-1]

(=) x-2015/2012 .   +.   x-2015/2013. = x-2015/2.     +.     x-2015/3

(=)( x-2015 ) × (1/2012    + . 1/2013.    -. 1/2 .   -   1/3 .) =0

Mà 1/2012+1/2013-1/2-1/3≠0

=) x-2015 =0

(=) x=2015

=(4h30-2h35) x 3

= 1h55 x 3

=5h45

a) |2+x|.3=3²

|2+x|=3²:3

|2+x|=3

+) TH1: 2+x=3                                          +) TH2: 2+x=-3

x=3-2                                                                    x= -3 - 2

x=1                                                                       x= -5

Vậy x= 1 hoặc -5

b) 2.|x-5⁰|=2²

|x-1|=2²:2

|x-1|=2

+) TH1: x-1=2

x=2+1

x=3

+) TH2: x-1= -2

x= -2 + 1

x= -1

Vậy x=3 hoặc -1

x-2=y.(x-2)

=> y=(x-2):(x-2)

=> y=1

=> x-2=x-2

=> x là mọi số tự nhiên.

Vậy x \(\in\)N ; y=1.

\(\left(x-2\right)-y\left(x-2\right)=0\)

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

+ x-2 =0 => x =2

+ hoặc 1-y =0 => y =1

Vậy x =2 hoặc y =1

a) \(4590:\left(x-24+14\right)=30\)
\(x-24+14=4590:30\)
\(x-24+14=153\)
\(x-24=153-14\)
\(x-24=139\)
\(x=139+24\)
\(x=163\)
b) \(61\times\left(126:x\right)=1281\)
\(126:x=1281:61\)
\(126:x=21\)
\(x=126:21\)
\(x=6\)

b. 61 × ( 126 : x ) = 1281

x = 126 : x = 1281 ÷ 61 = 21

 x = 126÷21

x = 6

a) 4590 ÷ ( x− 24 + 14 ) = 30

4590 ( x- 10 ) = 30

x - 10 = 4590 : 30

x-10 = 153

x = 153+10

x = 163

\(3x+6=15\\ 3x=15-6\\ 3x=9\\ x=9:3\\ x=3\)

Chưa tim được giải pháp cho lớp 6

a, Ix-2I*(x-5)<0
lx-2l lớn hơn or bằng 0 nên (x-5)<0 hay x<5
b, Ix-2I*(x-5)>0
lx-2l lớn hơn or bằng 0 nên (x-5)>0 hay x>5

 thế x= 2 thì sao nhỉ

\(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)=20\)

\(3x+6=20\)

\(\Leftrightarrow3x=14\)

\(\Leftrightarrow x=\frac{14}{3}\)

Vậy \(x=\frac{14}{3}\)

   (x+1) + (x+2) + (x+3) = 3x + 6 = 20

=> 3x + 6 =20

      3x       = 20 - 6

      3x       = 14

       x        = 14/3

(x + 1) + (x+2) + (x+3) = 20 

x + 1 + x + 2 + x + 3 = 20 

=> (x + x + x) + (1 + 2 + 3) = 20

=> 3x + 6 = 20 

=> 3x = 20 - 6 

=> 3x = 14 

=> x = 14 : 3 = 14/3 

Vậy x = 14/3