1: Ta có: \(\dfrac{x+2}{x-2}+\dfrac{2}{x+2}=\dfrac{x^2}{x^2-4}\)

Suy ra: \(x^2+4x+4+2x-4=x^2\)

\(\Leftrightarrow6x=0\)

hay \(x=0\left(nhận\right)\)

2: Ta có: \(\dfrac{1}{x-6}-\dfrac{2}{x+6}=\dfrac{3x+6}{x^2-36}\)

Suy ra: \(x+6-2x+12=3x+6\)

\(\Leftrightarrow-x-3x=6-18=-12\)

hay \(x=3\left(nhận\right)\)

Lời giải:
1. ĐKXĐ: $x\neq \pm 2$

PT \(\Leftrightarrow \frac{(x+2)^2+2(x-2)}{(x-2)(x+2)}=\frac{x^2}{x^2-4}\)

\(\Leftrightarrow \frac{x^2+6x}{x^2-4}=\frac{x^2}{x^2-4}\)

\(\Rightarrow x^2+6x=x^2\Leftrightarrow x=0\) (tm)

2. ĐKXĐ: $x\neq \pm 6$

PT \(\Leftrightarrow \frac{6+x-2(x-6)}{(x-6)(6+x)}=\frac{3x+6}{x^2-36}\)

\(\Leftrightarrow \frac{18-x}{x^2-36}=\frac{3x+6}{x^2-36}\)

\(\Rightarrow 18-x=3x+6\Leftrightarrow 12=4x\Leftrightarrow x=3\) (tm)

1) \(\dfrac{x+2}{x-2}+\dfrac{2}{x+2}=\dfrac{x^2}{x^2-4}\)

\(\Leftrightarrow\dfrac{x+2}{x-2}+\dfrac{2}{x+2}-\dfrac{x^2}{\left(x-2\right)\left(x+2\right)}\)=0

\(\Leftrightarrow\dfrac{\left(x+2\right)^2+2\left(x-2\right)-x^2}{\left(x-2\right)\left(x+2\right)}\)=0

\(\Leftrightarrow\dfrac{x^2+2x2+2^2+2x-4-x^2}{\left(x-2\right)\left(x+2\right)}\)=0

\(\Leftrightarrow\dfrac{x^2-x^2+4x+2x+4-4}{\left(x-2\right)\left(x+2\right)}\)=0

\(\Leftrightarrow\dfrac{6x}{\left(x-2\right)\left(x+2\right)}\)=0

\(\Leftrightarrow6x=0\)

\(\Rightarrow x=0\)

2) \(\dfrac{1}{x-6}-\dfrac{2}{6+x}=\dfrac{3x+6}{x^2-36}\)

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

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

\(\Leftrightarrow\dfrac{x+6-2x+12-3x-6}{\left(x-6\right)\left(x+6\right)}\)=0

\(\Leftrightarrow\dfrac{x-2x-3x+6-6+12}{\left(x-6\right)\left(x+6\right)}\)=0

\(\Leftrightarrow\dfrac{-4x+12}{\left(x-6\right)\left(x+6\right)}\)=0

\(\Leftrightarrow-4x+12=0\)

\(\Leftrightarrow-4x=12\)

\(\Rightarrow x=3\)

\(\dfrac{9x-2}{x^2-x-6}+\dfrac{2x}{x+2}-\dfrac{x-1}{x-3}=1\left(x\ne-2;x\ne3\right)\\ < =>\dfrac{9x-2}{x^2-3x+2x-6}+\dfrac{2x}{x+2}-\dfrac{x-1}{x-3}=1\\ < =>\dfrac{9x-2}{x\left(x-3\right)+2\left(x-3\right)}+\dfrac{2x}{x+2}-\dfrac{x-1}{x-3}=1\\ < =>\dfrac{9x-2}{\left(x-3\right)\left(x+2\right)}+\dfrac{2x}{x+2}-\dfrac{x-1}{x-3}=1\)

suy ra: \(9x-2+2x\left(x-3\right)-\left(x-1\right)\left(x+2\right)=\left(x-3\right)\left(x+2\right)\)

\(< =>9x-2+2x^2-6x-\left(x^2+2x-x-2\right)=x^2+2x-3x-6\)

\(< =>9x-2+2x^2-6x-x^2-2x+x+2=x^2-x-6\)

\(< =>2x^2-x^2-x^2+9x-6x-2x+x+x=6+2-2\)

\(< =>3x=6\\ < =>x=2\left(tm\right)\)

ĐKXĐ: \(x\ne\left\{-2;3\right\}\)

\(\dfrac{9x-2}{x^2-x-6}+\dfrac{2x}{x+2}-\dfrac{x-1}{x-3}=1\)

\(\Leftrightarrow\dfrac{9x-2}{\left(x+2\right)\left(x-3\right)}+\dfrac{2x\left(x-3\right)}{\left(x+2\right)\left(x-3\right)}-\dfrac{\left(x-1\right)\left(x+2\right)}{\left(x+2\right)\left(x-3\right)}=\dfrac{\left(x+2\right)\left(x-3\right)}{\left(x+2\right)\left(x-3\right)}\)

\(\Leftrightarrow9x-2+2x\left(x-3\right)-\left(x-1\right)\left(x+2\right)=\left(x+2\right)\left(x-3\right)\)

\(\Leftrightarrow9x-2+2x^2-6x-x^2-x+2=x^2-x-6\)

\(\Leftrightarrow3x=-6\)

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

Vậy: PT vô nghiệm.

`(x-1)/2013+(x-2)/2012+(x-3)/2011=(x-4)/2010+(x-5)/2009 +(x-6)/2008`

`<=> ((x-1)/2013-1)+((x-2)/2012-1)+((x-3)/2011-1)=( (x-4)/2010-1)+((x-5)/2009-1)+((x-6)/2008-1)`

`<=> (x-2014)/2013 +(x-2014)/2012+(x-2014)/2011=(x-2014)/2010+(x-2014)/2009+(x-2014)/2008`

`<=> x-2014=0` (Vì `1/2013+1/2012+1/2011-1/2010-1/2009-1/2008 \ne 0`)

`<=>x=2014`

Vậy `S={2014}`.

=>x-2014=0

hay x=2014

\(\dfrac{x-1}{2013}-1+\dfrac{x-2}{2012}-1+\dfrac{x-3}{2011}-1=\dfrac{x-4}{2019}-1+\dfrac{x-5}{2010}-1+\dfrac{x-6}{2008}-1\)

\(\Leftrightarrow\dfrac{x-2014}{2013}+\dfrac{x-2014}{2012}+\dfrac{x-2014}{2011}-\dfrac{x-2014}{2010}-\dfrac{x-2014}{2009}-\dfrac{x-2014}{2008}=0\)

\(\Leftrightarrow\left(x-2014\right)\left(\dfrac{1}{2013}+\dfrac{1}{2012}+\dfrac{1}{2011}-\dfrac{1}{2010}-\dfrac{1}{2009}-\dfrac{1}{2008}\ne0\right)=0\Leftrightarrow x=2014\)

\(\dfrac{x-1}{2013}+\dfrac{x-2}{2012}+\dfrac{x-3}{2011}=\dfrac{x-4}{2010}+\dfrac{x-5}{2009}+\dfrac{x-6}{2008}\)

\(\Leftrightarrow\left(\dfrac{x-1}{2013}-1\right)+\left(\dfrac{x-2}{2012}-1\right)+\left(\dfrac{x-3}{2011}-1\right)=\left(\dfrac{x-4}{2010}-1\right)+\left(\dfrac{x-5}{2009}-1\right)+\left(\dfrac{x-6}{2008}-1\right)\)

\(\Leftrightarrow\dfrac{x-2014}{2013}+\dfrac{x-2014}{2012}+\dfrac{x-2014}{2011}=\dfrac{x-2014}{2010}+\dfrac{x-2014}{2009}+\dfrac{x-2014}{2008}\)

\(\Leftrightarrow\dfrac{x-2014}{2013}+\dfrac{x-2014}{2012}+\dfrac{x-2014}{2011}-\dfrac{x-2014}{2010}-\dfrac{x-2014}{2009}-\dfrac{x-2014}{2008}=0\)

\(\Leftrightarrow\left(x-2014\right)\left(\dfrac{1}{2013}+\dfrac{1}{2012}+\dfrac{1}{2011}-\dfrac{1}{2010}-\dfrac{1}{2009}-\dfrac{1}{2008}\right)=0\)

\(\Leftrightarrow\left(x-2014\right).A=0\)

\(\text{Vì A }\ne0\)

\(\Rightarrow x-2014=0\)

\(\Leftrightarrow x=2014\)

\(\text{Vậy phương trình có tập nghiệm là }S=\left\{2014\right\}\)

a:=>3x=15

=>x=5

b: =>8-11x<52

=>-11x<44

=>x>-4

c: \(VT=\left(\dfrac{x^2-\left(x-6\right)^2}{x\left(x+6\right)\left(x-6\right)}\right)\cdot\dfrac{x\left(x+6\right)}{2x-6}+\dfrac{x}{6-x}\)

\(=\dfrac{12x-36}{2x-6}\cdot\dfrac{1}{x-6}-\dfrac{x}{x-6}=\dfrac{6}{x-6}-\dfrac{x}{x-6}=-1\)

Ta có: \(\sqrt{25x-125}-3\cdot\sqrt{\dfrac{x-5}{9}}-\dfrac{1}{3}\sqrt{9x-45}=6\)

\(\Leftrightarrow5\sqrt{x-5}-3\cdot\dfrac{\sqrt{x-5}}{3}-\dfrac{1}{3}\cdot3\sqrt{x-5}=6\)

\(\Leftrightarrow3\sqrt{x-5}=6\)

\(\Leftrightarrow x-5=4\)

hay x=9

\(\dfrac{1}{x^2-3x+2}+\dfrac{1}{x^2-5x+6}-\dfrac{2}{x^2-4x+3}\) = 0  (ĐKXĐ: x \(\ne\) 1; x \(\ne\) 2; x \(\ne\) 3)

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

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

\(\Leftrightarrow\) \(\dfrac{0}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}=0\) (luôn đúng)

Vậy pt trên có vô số nghiệm và x \(\ne\) 1; x \(\ne\) 2; x \(\ne\) 3

Chúc bn học tốt!

3x+5

=0

\(pt\text{ tương đương:}\dfrac{1}{\left(x-1\right)\left(x-2\right)}+\dfrac{1}{\left(x-2\right)\left(x-3\right)}-\dfrac{2}{\left(x-1\right)\left(x-3\right)}=0\text{ tương đương với: }\dfrac{1}{x-1}-\dfrac{1}{x-2}+\dfrac{1}{x-2}-\dfrac{1}{x-3}-\dfrac{2}{\left(x-1\right)\left(x-3\right)}=0\text{ hay:}\dfrac{1}{x-1}-\dfrac{1}{x-3}-\dfrac{1}{x-1}+\dfrac{1}{x-3}=0\text{ đúng}\)

a, đk : x khác 5;-6 

\(x^2+12x+36+x^2-10x+25=2x^2+23x+61\)

\(\Leftrightarrow2x+61=23x+61\Leftrightarrow21x=0\Leftrightarrow x=0\)(tm) 

b, đk : x khác 1;3 

\(x^2+2x-15=x^2-1-8\Leftrightarrow2x-15=-9\Leftrightarrow x=3\left(ktmđk\right)\)

pt vô nghiệm 

a, đk : x khác 5;-6 

(tm) 

b, đk : x khác 1;3 

x2+2x−15=x2−1−8⇔2x−15=−9⇔x=3(ktmđk)x2+2x−15=x2−1−8⇔2x−15=−9⇔x=3(ktmđk)

pt vô nghiệm 

a: \(\Leftrightarrow\left(x+6\right)^2+\left(x-5\right)^2=2x^2+23x+61\)

\(\Leftrightarrow x^2+12x+36+x^2-10x+25=2x^2+23x+61\)

=>x=0(nhận)

b: \(\Leftrightarrow\left(x+5\right)\left(x-3\right)=\left(x+1\right)\left(x-1\right)-8\)

\(\Leftrightarrow x^2+2x-15=x^2-1-8\)

=>2x-15=-9

=>2x=-6

hay x=-3(nhận)