\(\Leftrightarrow x^2-10x+25-4x^2-20x=0\)
\(\Leftrightarrow-3x^2-30x+25=0\)
\(\Leftrightarrow3x^2+30x-25=0\)
\(\text{Δ}=30^2-4\cdot3\cdot\left(-25\right)=900+300=1200>0\)
Do đó: Phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{-30-20\sqrt{3}}{6}=\dfrac{-15-10\sqrt{3}}{3}\\x_2=\dfrac{-15+10\sqrt{3}}{3}\end{matrix}\right.\)
a) x2 + 10x + 25 - 4x2 - 20x = 0
<=> 3x2 + 10x - 25 = 0
<=> (x + 5)(3x - 5) = 0 <=> 0RB\(\left\{{}\begin{matrix}-5\\\dfrac{5}{3}\\\end{matrix}\right.\)
Vậy S = {−5;\(\dfrac{5}{3}\)}
\(\Leftrightarrow x^2-10x+25-4x^2-20x=0\)
\(\Leftrightarrow-3x^2-30x+25=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-15+10\sqrt{3}}{3}\\x=\dfrac{-15-10\sqrt{3}}{3}\end{matrix}\right.\)
<=>\(x^2-10x+25-4x^2-20x=0\)
<=>\(-3x^2-30x+25=0\)
<=>\(3x^2+30x-25=0\)
<=>\(\left(x-\dfrac{-15+10\sqrt{3}}{3}\right)\left(x-\dfrac{-15-10\sqrt{3}}{3}\right)=0\)
<=>\(\left[{}\begin{matrix}x=\dfrac{10\sqrt{3}-15}{3}\\x=\dfrac{10\sqrt{3}+15}{3}\end{matrix}\right.\)
⇔x2−10x+25−4x2−20x=0⇔x2−10x+25−4x2−20x=0
⇔−3x2−30x+25=0⇔−3x2−30x+25=0
⇔3x2+30x−25=0⇔3x2+30x−25=0
Δ=302−4⋅3⋅(−25)=900+300=1200>0Δ=302−4⋅3⋅(−25)=900+300=1200>0
Do đó: Phương trình có hai nghiệm phân biệt là:
