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)

\(ĐKXĐ:\left\{{}\begin{matrix}x\ne-1\\x\ne4\end{matrix}\right.\)

\(\dfrac{14x}{x+1}< \dfrac{9x-30}{x-4}\\ \Leftrightarrow14x\left(x-4\right)< \left(9x-30\right)\left(x+1\right)\\ \Leftrightarrow14x^2-56x< 9x^2-21x-30\\ \Leftrightarrow5x^2-35x+30< 0\\ \Leftrightarrow1< x< 6\)

\(x-\dfrac{\dfrac{x}{3}-\dfrac{3+x}{2}}{4}=\dfrac{x-\dfrac{15-7x}{3}}{4}-2x+3\)

\(<=>4x-\dfrac{x}{3}+\dfrac{3+x}{2}=x-\dfrac{15-7x}{3}-8x+12\)

`<=>24x-2x+3(3+x)=6x-2(15-7x)-48x+72`

`<=>24x-2x+9+3x=6x-30+14x-48x+72`

`<=>53x=33`

`<=>x=33/53`

a) \(5x - 30 = 0\)

\(5x = 0 + 30\)     

\(5x = 30\)

\(x = 30:5\)

\(x = 6\)      

Vậy phương trình có nghiệm \(x = 6\).

b) \(4 - 3x = 11\)

\( - 3x = 11 - 4\)

\( - 3x =  7\)

\(x = \left( { 7} \right):\left( { - 3} \right)\)

\(x = \dfrac{-7}{3}\)

Vậy phương trình có nghiệm \(x = \dfrac{7}{3}\).

c) \(3x + x + 20 = 0\)               

\(4x + 20 = 0\)

\(4x = 0 - 20\)

\(4x =  - 20\)

\(x = \left( { - 20} \right):4\)

\(x =  - 5\)   

Vậy phương trình có nghiệm \(x =  - 5\).

d) \(\dfrac{1}{3}x + \dfrac{1}{2} = x + 2\)

\(\dfrac{1}{3}x - x = 2 - \dfrac{1}{2}\)

\(\dfrac{{ - 2}}{3}x = \dfrac{3}{2}\)

\(x = \dfrac{3}{2}:\left( {\dfrac{{ - 2}}{3}} \right)\)

\(x = \dfrac{{ - 9}}{4}\)

Vậy phương trình có nghiệm \(x = \dfrac{{ - 9}}{4}\).

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)

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

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

\(\Leftrightarrow\dfrac{16}{x+4}+\dfrac{16}{x-4}=\dfrac{5}{3}\)

=>\(\dfrac{16x-64+16x+64}{x^2-16}=\dfrac{5}{3}\)

=>5(x^2-16)=3*32x=96x

=>5x^2-96x-80=0

=>x=20 hoặc x=-4/5

ĐKXĐ : \(x\notin\left\{0;-1;-2;-3;-4\right\}\)

Ta có \(\dfrac{1}{x}+\dfrac{1}{x+1}+\dfrac{1}{x+2}+\dfrac{1}{x+3}+\dfrac{1}{x+4}=0\)

\(\Leftrightarrow\dfrac{2x+4}{x.\left(x+4\right)}+\dfrac{2x+4}{\left(x+1\right).\left(x+3\right)}+\dfrac{1}{x+2}=0\)

\(\Leftrightarrow\dfrac{2x+4}{\left(x+2\right)^2-4}+\dfrac{2x+4}{\left(x+2\right)^2-1}+\dfrac{1}{x+2}=0\) (*)

Đặt x + 2 = a \(\left(a\ne0\right)\) 

(*) \(\Leftrightarrow\dfrac{2a}{a^2-4}+\dfrac{2a}{a^2-1}+\dfrac{1}{a}=0\)

\(\Leftrightarrow\dfrac{2}{a-\dfrac{4}{a}}+\dfrac{2}{a-\dfrac{1}{a}}+\dfrac{1}{a}=0\) (**)

Đặt \(\dfrac{1}{a}=b\left(b\ne0\right)\) \(\Rightarrow ab=1\)

Ta được (**) \(\Leftrightarrow\dfrac{2}{a-4b}+\dfrac{2}{a-b}+b=0\)

\(\Leftrightarrow\dfrac{2b}{1-4b^2}+\dfrac{2b}{1-b^2}+b=0\)

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

\(\Rightarrow4-10b^2=-4b^4+5b^2-1\)

\(\Leftrightarrow4b^4-15b^2+5=0\) (***)

Đặt b² = t > 0

Ta có (***) <=> \(4t^2-15t+5=0\Leftrightarrow t=\dfrac{15\pm\sqrt{145}}{8}\) (tm)

\(\Leftrightarrow b=\pm\sqrt{\dfrac{15\pm\sqrt{145}}{8}}\) 

mà x + 2 = a ; ab = 1 

nên \(x=\pm\sqrt{\dfrac{8}{15\pm\sqrt{145}}}-2\)

Thử lại ta có phương trình có 4 nghiệm như trên