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

\(\dfrac{100}{x}-\dfrac{100}{x+10}=\dfrac{30}{60}=0,5\left(ĐKXĐ:x\ne0;x\ne-10\right)\\ \Leftrightarrow\dfrac{100\left(x+10\right)-100x}{x\left(x+10\right)}=\dfrac{0,5x\left(x+10\right)}{x\left(x+10\right)}\\ \Leftrightarrow100x-100x+1000=0,5x^2+5x\\ \Leftrightarrow0,5x^2+5x-1000=0\\ \Leftrightarrow0,5x^2-20x+25x-1000=0\\ \Leftrightarrow0,5x.\left(x-40\right)+25.\left(x-40\right)=0\\ \Leftrightarrow\left(0,5x+25\right)\left(x-40\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}0,5x+25=0\\x-40=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-50\\x=40\end{matrix}\right.\\ Vậy:S=\left\{-50;40\right\}\)

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

\(\frac{2}{1^2}.\frac{6}{2^2}.\frac{10}{3^2}.\frac{20}{4^2}.......\frac{110}{10^2}\left(x-2\right)=-20\left(x+1\right)+60\)

\(\Rightarrow\frac{1.2}{1.1}.\frac{2.3}{2.2}.\frac{3.4}{3.3}.\frac{4.5}{4.4}......\frac{10.11}{10.10}\left(x-2\right)=-20x-20+60\)

\(\Rightarrow\frac{1.2.3.4.....10}{1.2.3.4.....10}.\frac{2.3.4.5.....11}{1.2.3.4.....10}\left(x-2\right)=-20x+40\)

\(\Rightarrow11\left(x-2\right)=-20x+40\)

\(\Rightarrow11x-22=-20x+40\)

\(\Rightarrow11x+20x=22+40\)

\(\Rightarrow31x=62\)

\(\Rightarrow x=2\)

Vậy \(x=2\)

\(\dfrac{120}{x}+\dfrac{120}{x-10}=\dfrac{3}{5}\left(dkxd:x>0,x\ne10\right)\)

\(\Leftrightarrow\dfrac{120}{x}+\dfrac{120}{x-10}-\dfrac{3}{5}=0\)

\(\Leftrightarrow\dfrac{120.5\left(x-10\right)+5.120x-3x\left(x-10\right)}{5x\left(x-10\right)}=0\)

\(\Leftrightarrow600x-6000+600x-3x^2+30x=0\)

\(\Leftrightarrow-3x^2+1230x-6000=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\approx405\\x\approx5\end{matrix}\right.\)\(\left(tmdk\right)\)

Vậy ...

\(\dfrac{1.2}{1^2}.\dfrac{2.3}{2^2}.\dfrac{3.4}{3^2}...\dfrac{9.10}{9^2}.\dfrac{10.11}{10^2}\left(x-2\right)=-20\left(x+1\right)+60\)

\(\Leftrightarrow\dfrac{1.2^2.3^2.4^2...10^2.11}{1^2.2^2.3^2....10^2}\left(x-2\right)+20\left(x+1\right)=60\)

\(\Leftrightarrow11\left(x-2\right)+20\left(x+1\right)=60\)

\(\Leftrightarrow31x=62\)

\(\Rightarrow x=2\)

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

$ĐKXĐ:x \neq -4;-5;-6;-7$

$pt⇔\dfrac{1}{x^2+4x+5x+20}+\dfrac{1}{x^2+5x+6x+30}+\dfrac{1}{x^2+6x+7x+42}=\dfrac{1}{18}$

$⇔\dfrac{1}{(x+4)(x+5)}+\dfrac{1}{(x+5)(x+6)}+\dfrac{1}{(x+6)(x+7)}=\dfrac{1}{18}$

$⇔\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+7}=\dfrac{1}{18}$

$⇔\dfrac{1}{x+4}-\dfrac{1}{x+7}=\dfrac{1}{18}$

$⇔\dfrac{3}{(x+4)(x+7)}=\dfrac{1}{18}$

$⇔x^2+11x+28=54$

$⇔x^2+11x-26=0$

$⇔x^2-2x+13x-26=0$

$⇔(x-2)(x+13)=0$

$⇔$ \(\left[{}\begin{matrix}x=2\\x=-13\end{matrix}\right.\)(t/m)

Vậy phương trình đã cho có tập nghiệm $S=(2;-13)$

x.10=(x-10).30

=>10x=30x-300

=>10x-30x+300=0

=>-20x-300=0

=>x=15

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

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

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

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

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

\(\Leftrightarrow x=15\)

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

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

\(=>-20x=300\)

\(=>x=-15\)