\(\Leftrightarrow\hept{\begin{cases}x-y=20\\\frac{y-x}{xy}=\frac{1}{120}\end{cases}\Leftrightarrow}\hept{\begin{cases}x-y=20\\xy=-2400\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}y=x-20\\x\left(x-20\right)+2400=0\end{cases}}\)

Đến đây dễ rồi nhé

\(\frac{x+10}{90}+\frac{x+20}{80}+\frac{x+30}{70}+\frac{x+40}{60}+\frac{x+50}{50}=-5\)

<=> \(\frac{x+10}{90}+1+\frac{x+20}{80}+1+\frac{x+30}{70}+1+\frac{x+40}{60}+1+\frac{x+50}{50}+1=0\)

<=> \(\frac{x+100}{90}+\frac{x+100}{80}+\frac{x+100}{70}+\frac{x+100}{60}+\frac{x+100}{50}=0\)

<=> \(\left(x+100\right)\left(\frac{1}{90}+\frac{1}{80}+\frac{1}{70}+\frac{1}{60}+\frac{1}{50}\right)=0\)

<=> x + 100 = 0 

<=> x = -100

Vậy x = -100

\(ĐKXĐ:\hept{\begin{cases}x\ne0\\x\ne30\\x\ne24\end{cases}}\)

Ta có \(\frac{60}{\frac{120}{x}-4}+\frac{60}{\frac{120}{x}-5}=x\)

\(\Leftrightarrow\frac{60}{\frac{120-4x}{x}}+\frac{60}{\frac{120-5x}{x}}=x\)

\(\Leftrightarrow\frac{60x}{120-4x}+\frac{60x}{120-5x}=x\)

\(\Leftrightarrow\frac{60}{120-4x}+\frac{60}{120-5x}=1\left(Do\text{ }x\ne0\right)\)    

\(\Leftrightarrow\frac{15}{30-x}=1-\frac{12}{24-x}\)

\(\Leftrightarrow\frac{15}{30-x}=\frac{24-x-12}{24-x}\)

\(\Leftrightarrow\frac{15}{30-x}=\frac{12-x}{24-x}\)

\(\Leftrightarrow360-15x=\left(12-x\right)\left(30-x\right)\)

\(\Leftrightarrow360-15x=360-42x+x^2\)

\(\Leftrightarrow x^2-27x=0\)

\(\Leftrightarrow x\left(x-27\right)=0\)

\(\Leftrightarrow x=27\left(Tm\text{ }ĐKXĐ\right)\)

\(\frac{6}{x-5}+\frac{2}{x-8}=\frac{18}{x^2-13x+40}-1\)

điều kiện: \(x\ne5;8\)

\(\frac{6\left(x-8\right)+2\left(x-5\right)}{\left(x-5\right)\left(x-8\right)}-\frac{18}{x^2-13x+40}+1=0\)

\(\frac{6x-48+2x-10}{\left(x-5\right)\left(x-8\right)}-\frac{18}{x^2-8x-5x+40}+1=0\)

\(\frac{8x-58}{\left(x-5\right)\left(x-8\right)}-\frac{18}{x\left(x-8\right)-5\left(x-8\right)}+1=0\)

\(\frac{8x-58}{\left(x-5\right)\left(x-8\right)}-\frac{18}{\left(x-5\right)\left(x-8\right)}+\frac{\left(x-5\right)\left(x-8\right)}{\left(x-5\right)\left(x-8\right)}=0\)

\(\frac{8x-58-18+x^2-13x+40}{\left(x-5\right)\left(x-8\right)}=0\)

\(\frac{x^2-5x-36}{\left(x-5\right)\left(x-8\right)}=0\)

=> \(x^2-5x-36=0\)

\(x^2+4x-9x-36=0\)

\(x\left(x+4\right)-9\left(x+4\right)=0\)

\(\left(x-9\right)\left(x+4\right)=0\)

Vậy x - 9 = 0 hoặc x + 4 = 0

hay x = 9 (thỏa mãn điều kiện) hoặc x = -4 (thỏa mãn điều kiện)

vậy...

\(\frac{6}{x-5}+\frac{2}{x-8}=\frac{18}{x^2-13x+40}-1\)

ĐKXĐ: \(x\ne5,8\)

\(\Leftrightarrow\frac{6}{x-5}+\frac{2}{x-8}=\frac{18}{\left(x-5\right)\left(x-8\right)}-1\)

\(\Leftrightarrow6\left(x-8\right)+2\left(x-5\right)=18-\left(x-5\right)\left(x-8\right)\)

\(\Leftrightarrow8x-58=-22-x^2+13x\)

\(\Leftrightarrow8x-58+22+x^2-13x=0\)

\(\Leftrightarrow-5x-36+x^2=0\)

\(\Leftrightarrow\left(x-9\right)\left(x+4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-9=0\\x+4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=9\\x=-4\end{cases}}\)

Vậy: phương trình có tập nghiệm là: S = {9; -4}

b,x-1+2x-4+3x-9=22

6x-14=22

6x=36

x=6

a,12x-180+10x-20+39x-2340+65x-4420=780

126x-6960=780

126x=7740

x=430/7