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