Giải thích các bước giải:
a.ĐKXĐ: $x\ne 0$
Ta có:
$\dfrac{120}{x}+\dfrac{50}{60}=\dfrac{120}{x-24}$
$\to \dfrac{120}{x}\cdot \:60x\left(x-24\right)+\dfrac{50}{60}\cdot \:60x\left(x-24\right)=\dfrac{120}{x-24}\cdot \:60x\left(x-24\right)$
$\to 7200\left(x-24\right)+50x\left(x-24\right)=7200x$
$\to 50x^2+6000x-172800=7200x$
$\to 50x^2-1200x-172800=0$
$\to 50(x-72)(x+48)=0$
$\to x\in\{72,-48\}$
b.ĐKXĐ: $x\ne 0,-15$
Ta có:
$\dfrac{60}{x}-\dfrac{60}{x+15}=\dfrac23$
$\to \dfrac{60}{x}\cdot \:3x\left(x+15\right)-\dfrac{60}{x+15}\cdot \:3x\left(x+15\right)=\dfrac{2}{3}\cdot \:3x\left(x+15\right)$
$\to 180\left(x+15\right)-180x=2x\left(x+15\right)$
$\to 2700=2x^2+30x$
$\to 2x^2+30x-2700=0$
$\to 2(x-30)(x+45)=0$
$\to x\in\{30,-45\}$
c.ĐKXĐ: $x\ne 0,-5$
Ta có:
$\dfrac{270}{x}-\dfrac{270}{x+5}=\dfrac34$
$\to \dfrac{270}{x}\cdot \:4x\left(x+5\right)-\dfrac{270}{x+5}\cdot \:4x\left(x+5\right)=\dfrac{3}{4}\cdot \:4x\left(x+5\right)$
$\to 1080\left(x+5\right)-1080x=3x\left(x+5\right)$
$\to 5400=3x^2+15x$
$\to 3x^2+15x-5400=0$
$\to 3(x-40)(x+45)=0$
$\to x\in\{40, -45\}$
