Ta có : \(\frac{2}{2x-6}+\frac{1}{x+2}+\frac{2.x}{\left(x+1\right).\left(3-x\right)}=0\)

ĐKXĐ : x \(\ne\)-1 ; x \(\ne\)-2 ; x \(\ne\)3

MTC : ( x + 1 ) . ( x+ 2 ) . ( x - 3 ) 

<=> ( x + 1 ) . ( x + 2 ) + ( x + 1 ) . ( x + 3 ) - 2.x. ( x + 2 ) = 0

<=> x² + x + 2.x + 2 + x² -3.x + x -3 - 2.x² -4.x               = 0

<=> -3.x                                                                             = 1

<=> x = \(\frac{-1}{3}\)

Vậy S = { ​​​\(\frac{-1}{3}\)​}

ĐKXĐ: x khác 3, x khác -1

\(\frac{2}{2x-6}+\frac{2}{2x+2}+\frac{2}{\left(x+1\right)\left(3-x\right)}=0\)

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

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

<=> \(\frac{-2x+4}{\left(3-x\right)\left(x+1\right)}=0\)

<=> -2x+4=0

<=>x=-2

vậy ....

<=> x^2+2x=x^2+3x

<=>x=0

vậy ....

hok tốt

...

\(A=\frac{12x^2}{x+3}\)

\(12x^2\ge0\forall x\Rightarrow A< 0\Leftrightarrow x+3< 0\)

\(\Leftrightarrow x< -3\)

~~

ĐKXĐ: \(x\ne\pm1;x\ne0\)

a, \(A=\left(\frac{x^2+2x+1}{\left(x-1\right)\left(x+1\right)}-\frac{x^2-2x+1}{\left(x-1\right)\left(x+1\right)}\right):\frac{2x}{5\left(x-1\right)}\)

= \(\frac{4x}{\left(x-1\right)\left(x+1\right)}\cdot\frac{5\left(x-1\right)}{2x}\)

= \(\frac{20x\left(x-1\right)}{2x\left(x-1\right)\left(x+1\right)}\)

= \(\frac{10}{x+1}\)

Vậy ......

b, Thay x=3 vào A

A= \(\frac{10}{4}=\frac{5}{2}\)

Vì x khác -1 nên ko cần tính TH này

c, Cho A = 2

=> \(\frac{10}{x+1}=2\)

=> \(2x+2=10\)

=> x= 4

vậy ......

hok tốt

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

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

=2

vậy ....

hok tốt

.....

=2 bạn nhé

\(\left(2x-1\right)\left(x-5\right)-2x^2+10x-25=0\)

\(\Leftrightarrow2x^2-11x+5-2x^2+10x-25=0\)

\(\Leftrightarrow2x^2-2x^2-11x+10x=25-5\)

\(\Leftrightarrow-x=20\)

\(\Leftrightarrow x=-20\)

<=> 2x^2 - 11x + 5 - 2x^2 + 10x -25 =0

em mới lớp 7

oh, mk cũng lp 7, bài trên mk bó tay. com

\(5\left(x+4\right)\left(x-4\right)+96=\left(2x-1\right)\left(x-4\right)+\left(3x-1\right)\left(x+4\right)\)

\(\Leftrightarrow5x^2-80+96=2x^2-9x+4+3x^2+11x-4\)

\(\Leftrightarrow5x^2-2x^2-3x^2+9x-11x=4-4-96+80\)

\(\Leftrightarrow-2x=-16\)

\(\Leftrightarrow x=8\)

x = 8 đó bạn