Pt <=>  \(\dfrac{1}{x}-\dfrac{1}{x+5}=\dfrac{1}{550}\)

<=>  \(\dfrac{\left(x+5\right)-x}{x\left(x+5\right)}=\dfrac{1}{550}\)

<=> \(\dfrac{5}{x\left(x+5\right)}=\dfrac{1}{550}\)

<=> \(x^2+5x=2750\)

<=> \(x^2+5x-2750=0\)

<=> \(\left(x^2+5x+2,5^2\right)-52,5^2=0\) (bước này hơi tắt xíu nha :<)

<=> \(\left(x+2,5\right)^2-52,5^2=0\)

<=> \(\left(x+55\right)\left(x-50\right)=0\)

<=> \(\left[{}\begin{matrix}x+55=0\\x-50=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-55\\x=50\end{matrix}\right.\)

Vậy nghiệm của phương trình là x \(\in\left\{-55;50\right\}\)

\(\dfrac{1100}{x}-\dfrac{1100}{x+5}=2\)

\(\dfrac{1100\left(x+5\right)-1100x}{x\left(x+5\right)}=\dfrac{2x\left(x+5\right)}{x\left(x+5\right)}\)

\(1100x+5500-1100x=2x^2+10x\)

\(2x^2+10x-5500=0\)

Δ' \(=5^2-2\left(-5500\right)\)

Δ'\(=11025\)

\(\left[{}\begin{matrix}x=50\\x=-55\end{matrix}\right.\)

\(\left\{{}\begin{matrix}xy=1100\\y-\dfrac{1100}{x+5}=2\end{matrix}\right.< =>\left\{{}\begin{matrix}y=\dfrac{1100}{x}\left(x\ne0\right)\left(1\right)\\\dfrac{1100}{x}-\dfrac{1100}{x+5}=2\left(2\right)\end{matrix}\right.\)

* giải pt(2)\(=>\dfrac{1100x+5500-1100x}{x^2+5x}=2\)

\(=>5500=2x^2+10x=>2x^2+10x-5500=0\)

\(=>\Delta=10^2-4\left(-5500\right)2=44100>0\)

\(=>\left[{}\begin{matrix}x1=\dfrac{-10+\sqrt{44100}}{2.2}=50\left(TM\right)\left(3\right)\\x2=\dfrac{-10-\sqrt{44100}}{2.2}=-55\left(TM\right)\left(4\right)\end{matrix}\right.\)

thế(3)(4) vào(1)\(=>\left[{}\begin{matrix}y=\dfrac{1100}{50}=22\\y=\dfrac{1100}{-55}=-20\end{matrix}\right.\)

vậy...

đk : x khác 0 ; -5 

\(1100x+5500-1100x=2x\left(x+5\right)\)

\(\Leftrightarrow2x^2+10x-5500=0\Leftrightarrow x=50;x=-55\)(tm) 

= - 55 tm

c: Ta có: \(\dfrac{x+5}{x-2}=\dfrac{5\left(x+2\right)}{x^2-4}\)

\(\Leftrightarrow\dfrac{x+5}{x-2}=\dfrac{5}{x-2}\)

Suy ra: x+5=5

hay \(x=0\left(nhận\right)\)

d: Ta có: \(\dfrac{5}{x-1}=\dfrac{4}{x+2}\)

\(\Leftrightarrow5\left(x+2\right)=4\left(x-1\right)\)

\(\Leftrightarrow x=-14\left(nhận\right)\)

\(PT.\Rightarrow3x-9-\left(10-4x\right)=6x+5.\)

\(\Leftrightarrow3x-9-10+4x=6x+5.\\ \Leftrightarrow7x-19=6x+5.\\ \Leftrightarrow x=24.\)

\(\Leftrightarrow x\left(4x-3\right)-\left(x-2\right)\left(3x+2\right)=x^2-5\)

\(\Leftrightarrow4x^2-3x-3x^2-2x+6x+4=x^2-5\)

\(\Leftrightarrow x^2+x+4=x^2-5\)

=>x+4=-5

hay x=-9(nhận)

`(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

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

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

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

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

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

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

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

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

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

\(\text{Vì A }\ne0\)

\(\Rightarrow x-2014=0\)

\(\Leftrightarrow x=2014\)

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