Question

giải phương trình:

j) 3x² +7x+2=0

k) x² -4x+1=0

l) 2x² - 6x +1=0

m) 3x² +4x - 4=0

n) x⁶ - x² =0

o) x³ -12 =13x

p) -x⁵ +4x⁴ = -12x³

q) x³ =4x

Answer 1

\(j,3x^2+7x+2=0\)

\(\Leftrightarrow3x^2+6x+x+2=0\)

\(\Leftrightarrow3x\left(x+2\right)+\left(x+2\right)=0\)

\(\Leftrightarrow\left(3x+1\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+1=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\frac{1}{3}\\x=-2\end{matrix}\right.\)

Vậy...............................

Answer 2

\(m,3x^2+4x-4=0\)

\(\Leftrightarrow3x^2+6x-2x-4=0\)

\(\Leftrightarrow3x\left(x+2\right)-2\left(x+2\right)=0\)

\(\Leftrightarrow\left(3x-2\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\x+2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{2}{3}\\x=-2\end{matrix}\right.\)