\(\dfrac{x-130}{20}\)+\(\dfrac{x-100}{25}\)+\(\dfrac{x-60}{30}\)+\(\dfrac{x-10}{35}\)=10

⇔\(\dfrac{2625\left(x-130\right)}{52500}\)+\(\dfrac{2100\left(x-100\right)}{52500}\)+\(\dfrac{1750\left(x-60\right)}{52500}\)+\(\dfrac{1500\left(x-10\right)}{52500}\)=\(\dfrac{525000}{52500}\)

⇔2625\(x\)-341250+2100\(x\)-210000+1750\(x\)-105000+1500\(x\)-15000=525000

⇔ 7975\(x\) = 1196250

⇔ \(x\) = \(\dfrac{1196250}{7975}\)

⇔\(x \) = 150

\(\Leftrightarrow3x\left(x-10\right)=60x-60\left(x-10\right)\)

\(\Leftrightarrow3x\left(x-10\right)=600\)

\(\Leftrightarrow x^2-10x-200=0\)

=>(x-20)(x+10)=0

=>x=20 hoặc x=-10

\(\dfrac{60}{x-10}-\dfrac{60}{x}=\dfrac{3}{10}\)đk : x khác 10 ; 0 

\(\Leftrightarrow600x-600\left(x-10\right)=3x\left(x-10\right)\)

\(\Leftrightarrow3x^2-30x-6000=0\Leftrightarrow x=50;x=-40\left(tm\right)\)

a: \(\Leftrightarrow\left(\dfrac{x+2001}{5}+1\right)+\left(\dfrac{x+1999}{7}+1\right)+\left(\dfrac{x+1997}{9}+1\right)+\left(\dfrac{x+1995}{11}+1\right)=0\)

=>x+2006=0

=>x=-2006

b: \(\Leftrightarrow\left(\dfrac{x-15}{100}-1\right)+\left(\dfrac{x-10}{105}-1\right)+\left(\dfrac{x-100}{5}-1\right)=\left(\dfrac{x-100}{15}-1\right)+\left(\dfrac{x-105}{10}-1\right)+\left(\dfrac{x-110}{5}-1\right)\)

=>x-105=0

=>x=105

a: =>\(\dfrac{2x-4}{2014}+\dfrac{2x-2}{2016}< \dfrac{2x-1}{2017}+\dfrac{2x-3}{2015}\)

=>\(\dfrac{2x-2018}{2014}+\dfrac{2x-2018}{2016}< \dfrac{2x-2018}{2017}+\dfrac{2x-2018}{2015}\)

=>2x-2018<0

=>x<2019

b: \(\Leftrightarrow\left(\dfrac{3-x}{100}+\dfrac{4-x}{101}\right)>\dfrac{5-x}{102}+\dfrac{6-x}{103}\)

=>\(\dfrac{x-3}{100}+\dfrac{x-4}{101}-\dfrac{x-5}{102}-\dfrac{x-6}{103}< 0\)

=>\(x+97< 0\)

=>x<-97

Lời giải:

PT $\Leftrightarrow \frac{x+25}{75}+1+\frac{x+30}{70}+1=\frac{x+35}{65}+1+\frac{x+40}{60}+1$

$\Leftrightarrow \frac{x+100}{75}+\frac{x+100}{70}=\frac{x+100}{65}+\frac{x+100}{60}$
$\Leftrightarrow (x+100)(\frac{1}{75}+\frac{1}{70}-\frac{1}{65}-\frac{1}{60})=0$

Dễ thấy $\frac{1}{75}+\frac{1}{70}-\frac{1}{65}-\frac{1}{60}<0$

$\Rightarrow x+100=0$

$\Leftrightarrow x=-100$ (tm)

-90

\(\dfrac{10-x}{100}\) + \(\dfrac{20-x}{110}\)+\(\dfrac{30-x}{120}\)=3

<=> \(\dfrac{10-x}{100}\)-1+\(\dfrac{20-x}{110}\)-1+\(\dfrac{30-x}{120}\)-1 = 0

<=> \(\dfrac{-x-90}{100}\)+\(\dfrac{-x-90}{110}\)+\(\dfrac{-x-90}{120}\)=0

<=> (-x-90) ( \(\dfrac{1}{100}\)+\(\dfrac{1}{110}\)+\(\dfrac{1}{120}\))=0

<=> (-x-90) = 0 ( do 1/100 +1/110+1/120 khác 0)

<=> -x-90 = 0

<=> -x = 90

<=> x =-90

Vậy nghiệm của pt là x=-90

\(\dfrac{200x}{100}+\dfrac{300\left(x-20\right)}{100}=\dfrac{33.500}{100}\)

=> 200x + 300(x - 20) = 16500

<=> 200x + 300x - 6000 = 16500

<=> 500x = 22500

<=> x = 45

S = {45}

Ta có: \(\dfrac{x\cdot200}{100}+\dfrac{\left(x-20\right)\cdot300}{100}=\dfrac{33\cdot500}{100}\)

\(\Leftrightarrow200x+300x-6000=16500\)

\(\Leftrightarrow500x=22500\)

hay x=45

Vậy: S={45}

x.10=(x-10).30

=>10x=30x-300

=>10x-30x+300=0

=>-20x-300=0

=>x=15

\(\Leftrightarrow\dfrac{x}{30}-\dfrac{x-10}{10}=0\)

\(\Leftrightarrow\dfrac{x-3\left(x-10\right)}{30}=0\)

\(\Leftrightarrow x-3x+30=0\)

\(\Leftrightarrow-2x+30=0\)

\(\Leftrightarrow-2x=-30\)

\(\Leftrightarrow x=15\)

\(=>x.10=30x-300\)

\(=>10x-30x-300=0\)

\(=>-20x=300\)

\(=>x=-15\)

b: \(\Leftrightarrow\dfrac{20}{x}-\dfrac{20}{x+20}=\dfrac{1}{6}\)

=>\(\dfrac{20x+400-20x}{x\left(x+20\right)}=\dfrac{1}{6}\)

=>x*(x+20)=400*6=2400

=>x^2+20x-2400=0

=>(x+60)(x-40)=0

=>x=-60 hoặc x=40

c: \(\dfrac{2x+1}{2x-1}-\dfrac{2x-1}{2x+1}=\dfrac{8}{4x^2-1}\)

=>(2x+1)^2-(2x-1)^2=8

=>4x^2+4x+1-4x^2+4x-1=8

=>8x=8

=>x=1(nhận)