a) \(5x^3-125=0\)
\(\Leftrightarrow5x^3=125\)
\(\Leftrightarrow x^3=25\)
\(\Leftrightarrow x^3=25\)
\(\Leftrightarrow x=\sqrt[3]{25}\)
c) \(6x+13x+5=0\)
\(\Leftrightarrow19x+5=0\)
\(\Leftrightarrow19x=-5\)
\(\Leftrightarrow x=\dfrac{-5}{19}\)
a, \(5x^3-125=0\)
\(\Leftrightarrow5x^3=125\)
\(\Leftrightarrow x^3=25\)
\(\Leftrightarrow x=\sqrt[3]{25}\)
c, \(6x+13x+5=0\)
\(\Leftrightarrow19x+5=0\)
\(\Leftrightarrow19x=-5\)
\(\Leftrightarrow x=\dfrac{-5}{19}\)
Câu a :
\(5x^3-125=0\)
\(\Leftrightarrow5\left(x^3-25\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5=0\\x^3-25=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}loại\\x=\sqrt[3]{25}\end{matrix}\right.\)
Vậy.........
Câu b :
\(x^2-6x-21=0\)
\(\Delta=\left(-6\right)^2-4.\left(-21\right)=120\)
Do \(\Delta>0\) nên phương trình có 2 nghiệm phân biệt .
\(x_1=\dfrac{-\left(-6\right)+\sqrt{120}}{2}=3+\sqrt{30}\)
\(x_2=\dfrac{-\left(-6\right)-\sqrt{120}}{2}=3-\sqrt{30}\)
Kết luận .
Câu c :
\(6x+13x+5=0\)
\(\Leftrightarrow19x+5=0\)
\(\Rightarrow x=-\dfrac{5}{19}\)
