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

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)

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

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

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

=>(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 

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

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

\(PT\Leftrightarrow\left(\dfrac{x-70}{130}-1\right)+\left(\dfrac{x-25}{175}-1\right)+\left(\dfrac{x-50}{150}-1\right)+\left(\dfrac{x-275}{25}+3\right)=0\)

\(\Leftrightarrow\left(x-200\right)\left(\dfrac{1}{130}+\dfrac{1}{175}+\dfrac{1}{150}+\dfrac{1}{25}\right)=0\Leftrightarrow x=200\).

Vậy...

-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

`[158-x]/31+[185-x]/29+[208-x]/27+[227-x]/25=10`

`<=>[158-x]/31-1+[185-x]/29-2+[208-x]/27-3+[227-x]/25-4=0`

`<=>[127-x]/21+[127-x]/29+[127-x]/27+[127-x]/25=0`

`<=>(127-x)(1/21+1/29+1/27+1/25)=0`

`<=>127-x=0`

`<=>x=127`

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

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

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

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

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

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

\(\Rightarrow31x=62\)

\(\Rightarrow x=2\)

Vậy \(x=2\)

x.10=(x-10).30

=>10x=30x-300

=>10x-30x+300=0

=>-20x-300=0

=>x=15

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

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

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

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

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

\(\Leftrightarrow x=15\)

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

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

\(=>-20x=300\)

\(=>x=-15\)

\(\dfrac{x^2-26}{10}+\dfrac{x^2-25}{11}\ge\dfrac{x^2-24}{12}+\dfrac{x^2-23}{13}\)

\(\Leftrightarrow\left(\dfrac{x^2-26}{10}-1\right)+\left(\dfrac{x^2-25}{11}-1\right)\ge\left(\dfrac{x^2-24}{12}-1\right)+\left(\dfrac{x^2-23}{13}-1\right)\)

\(\Leftrightarrow\dfrac{x^2-36}{10}+\dfrac{x^2-36}{11}\ge\dfrac{x^2-36}{12}+\dfrac{x^2-36}{13}\)

\(\Leftrightarrow\dfrac{x^2-36}{10}+\dfrac{x^2-36}{11}-\dfrac{x^2-36}{12}-\dfrac{x^2-36}{13}\ge0\)

\(\Leftrightarrow\left(x^2-36\right)\left(\dfrac{1}{10}+\dfrac{1}{11}-\dfrac{1}{12}-\dfrac{1}{13}\right)\ge0\)

Vì \(\dfrac{1}{10}+\dfrac{1}{11}-\dfrac{1}{12}-\dfrac{1}{13}>0\Rightarrow x^2-36\ge0\Leftrightarrow\left[{}\begin{matrix}x\le-6\\x\ge6\end{matrix}\right.\)

Bất phương trình đó tương đương với:

 \(\left(\dfrac{x^2-26}{10}-1\right)+\left(\dfrac{x^2-25}{11}-1\right)\ge\left(\dfrac{x^2-24}{12}-1\right)+\left(\dfrac{x^2-23}{13}-1\right)\)

⇔ \(\dfrac{x^2-36}{10}+\dfrac{x^2-36}{11}\ge\dfrac{x^2-36}{12}+\dfrac{x^2-36}{13}\)

⇔ \(\dfrac{x^2-36}{10}+\dfrac{x^2-36}{11}-\dfrac{x^2-36}{12}-\dfrac{x^2-36}{13}\ge0\)

⇔ \(\left(x^2-36\right)\left(\dfrac{1}{10}+\dfrac{1}{11}-\dfrac{1}{12}-\dfrac{1}{13}\right)\ge0\)

+)Vì \(\dfrac{1}{10}>\dfrac{1}{11}>\dfrac{1}{12}>\dfrac{1}{13}\) nên \(\dfrac{1}{10}+\dfrac{1}{11}-\dfrac{1}{12}-\dfrac{1}{13}>0\) 

⇔ \(x^2-36\ge0\)

⇔ \(x^2\ge36\)

⇔ \(\sqrt{x^2}\ge6\)

⇔ \(\left|x\right|\ge6\)

⇔ \(\left[{}\begin{matrix}x\ge6\\x\le-6\end{matrix}\right.\)

➤ Vậy \(\left[{}\begin{matrix}x\ge6\\x\le-6\end{matrix}\right.\)