`100/x-100/(x+10)=1/2`
`<=>(100x+1000-100x)/(x^2+10x)=1/2`
`<=>1000/(x^2+10x)=1/2`
`<=>x^2+10x=2000`
`<=>x^2+10x-2000=0`
`Delta'=25+2000=2025`
`<=>x_1=40,x_2=-50`
Vậy `S={40,-50}`
100x−100x+10=12100x-100x+10=12
⇔100x+1000−100xx2+10x=12⇔100x+1000-100xx2+10x=12
⇔1000x2+10x=12⇔1000x2+10x=12
⇔x2+10x=2000⇔x2+10x=2000
⇔x2+10x−2000=0⇔x2+10x-2000=0
Δ'=25+2000=2025Δ′=25+2000=2025
⇔x1=40,x2=−50⇔x1=40,x2=-50
-> S={40,−50}
\(\dfrac{100}{x}=\dfrac{1}{2}+\dfrac{100}{x+10}\Rightarrow\dfrac{100}{x}=\dfrac{x+210}{2x+20}\Rightarrow200x+2000=x^2+210x\)
\(x^2+10x-2000=0\Leftrightarrow\left(x+50\right)\left(x-40\right)=0\Rightarrow x=40\)
\(\dfrac{100}{x}-\dfrac{100}{x+10}=\dfrac{1}{2}\)
=>\(200\left(x+10\right)-200x=x^2+10x\)
=>\(-x^2-10x+2000=0\)
<=>\(-x^2-50x+40x+2000=0\)
<=>\(-x\left(x+50\right)+40\left(x+50\right)=0\)
\(\left(-x+40\right)\left(x+50\right)=0\)
=>\(-x+40=0=>x=40\)
hoặc \(x+50=0=>x=-50\)
100/x -100/x+10 = 1/2
=> 100.2.(x+10) - 100.2.x = x.(x+10)
<=> 200.x + 2000 - 200.x = x^2 + 10.x
<=> x^2 + 10.x -2000 = 0
Tính đenta' = 2025
=> Căn của 2025 là 45
=> x1 = 40
x2 = -50
Vậy phương trình có 2 nghiệm là x=40 hoặc x=-50