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

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

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

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

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

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

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

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

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

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

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

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

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

Lời giải:
$12=2^2.3$

$5=1.5$

$8=2^3$

$\Rightarrow \text{BCNN(12,5,8)}=2^3.3.5=120$

$x\in \text{BC(12,5,8)}$ nên $x\in \left\{120; 240; 360;...\right\}$

Mà $60\leq x\leq 240$ nên $x\in \left\{120; 240\right\}$

\(1+\frac{-1}{60}+\frac{19}{120}

a: =>x(5-2)=120

=>3x=120

=>x=40

b: =>x(13-10)=240

=>3x=240

=>x=80

c; =>10x=250

=>x=25

d: =>3x=36

=>x=12

\(1+\frac{-1}{60}+\frac{19}{120}< \frac{x}{36}< \frac{58}{90}+\frac{59}{72}+\frac{-1}{60}\)

=> \(\frac{137}{120}< \frac{x}{36}< \frac{521}{360}\)

=> \(\frac{411}{360}< \frac{10x}{360}< \frac{521}{360}\)

=> 411 < 10x < 521

=> x \(\in\){ 42,43,44,...,52}

\(\frac{240}{x-12}-\frac{240}{x}=\frac{5}{3}\)

\(\Leftrightarrow\frac{240}{x-12}-\frac{240}{x}=\frac{5}{3}\left(x\ne12,x\ne0\right)\)

\(\Leftrightarrow\frac{240}{x-12}-\frac{240}{x}-\frac{5}{3}=0\)

\(\Leftrightarrow\frac{720x-720x+8640-5x^2+60x}{3x\left(x-12\right)}=0\)

\(\Leftrightarrow8640-5x^2+60x=0\)

\(\Leftrightarrow5\left(1728-x^2+12x\right)=0\)

\(\Leftrightarrow5\left(-x^2+12x+1728\right)=0\)

\(\Leftrightarrow5\left(-x^2+48x-36x+1728\right)=0\)

\(\Leftrightarrow5\left[-x\left(x-48\right)-36\left(x-48\right)\right]=0\)

\(\Leftrightarrow5\left[-\left(x-48\right)\right].\left(x+36\right)=0\)

\(\Leftrightarrow-5\left(x-48\right).\left(x+36\right)=0\)

\(\Leftrightarrow\left(x-48\right)\left(x+36\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-48=0\\x+36=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=48\\x=-36\end{matrix}\right.,x\ne12,x\ne0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=48\\x=-36\end{matrix}\right.\)