a) $\frac{120}{x}$ -1= $\frac{120-x}{x+10}$ + $\frac{2}{5}$ b)(2x-4)()$x^{2}$ +1)>0 giải đủ nhé

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1. `a) 120/x-1=(120-x)/(x+10)+2/5` ĐK: `x \ne 0,-10`

   `<=>(120.5(x+10)-5x(x+10))/(5x(x+10))=((120-x)5x+2x(x+10))/(5x(x+10))`

   `=> 600(x+10)-5x^2-50x=600x-5x^2+2x^2+20x`

   `<=> 600x+6000-5x^2-50x=-3x^2+620x`

   `<=> 2x^2+70x-6000=0`

   `<=> x^2+35x-3000=0`

   `<=> x^2-40x+75x-3000=0`

   `<=> (x-40)(x+75)=0`

   `<=>`\(\left[ \begin{array}{l}x=40\\x=-75\end{array} \right.\) ™

   Vậy `S={40;-75}`

   `b) (2x-4)(x^2+1)>0`

   Do `x^2+1>0` với `AAx`

   `-> 2x-4>0`

   `-> 2x>4`

   `-> x>2`

   Vậy `x>2`

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2. `a, 120/x – 1 = (120-x)/(x+10) + 2/5`

   ĐKXĐ : `x \ne 0 , x \ne – 10`

   `⇔ (120.5(x+10))/(5x(x+10)) – 1.5x(x+10) = (120-x.5x)/(5x(x+10)) + (2x(x+10))/(5x(x+10))`

   `⇒ 600(x+10) – 5x(x+10) = 5x(120-x) + 2x(x+10)`

   `⇔ -5x^2 + 550x + 6000 = -3x^2 + 620x`

   `⇔ -5x^2 – 70x + 6000 = -3x^2`

   `⇔ -2x^2 – 70x + 6000 = 0`

   `⇔ -2(x-40)(x+75) = 0`

   `⇔`\(\left[ \begin{array}{l}x-40=0\\x+75=0\end{array} \right.\) 

   `⇔`\(\left[ \begin{array}{l}x=40(TM)\\x=-75(TM)\end{array} \right.\) 

   Vậy `S ={40,75}`

   `b, (2x-4)(x^2+1) > 0`

   `⇔ x^3 – 2x^2 + x – 2 > 0`

   `⇔ (x-2)(x^2+1) > 0`

   Vì `x^2 + 1 > 0`

   `⇔ x – 2 > 0`

   `⇔ x > 2`

   Vậy `S = {x|x > 2}`

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