Question

Giải các phương trình:

a) 1,2x³ - x² - 0,2x = 0;                 b) 5x³ - x² - 5x + 1 = 0.

Answer 1

a, \(1,2x^3-x^2-0,2x=0\)
\(\Leftrightarrow12x^3-10x^2-2x=0\)
\(\Leftrightarrow6x^3-5x^2-x=0\)
\(\Leftrightarrow x\left(6x^2-5x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\6x^2-5x-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-\dfrac{1}{6}\end{matrix}\right.\)
Vậy tập nghiệm của pt đã cho là \(S=\left\{-\dfrac{1}{6};0;1\right\}\)

b, \(5x^3-x^2-5x+1=0\)
\(\Leftrightarrow x^2\left(5x-1\right)-\left(5x-1\right)=0\)
\(\Leftrightarrow\left(5x-1\right)\left(x^2-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-1=0\\5x-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\pm1\\x=\dfrac{1}{5}\end{matrix}\right.\)
Vậy tập nghiệm của phương trình đã cho là \(S=\left\{-1;\dfrac{1}{5};1\right\}\)

Answer 2

\(a,1,2x^3-x^2-0,2x=0\Leftrightarrow x\left(1,2x^2-x-0,2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\1,2x^2-x-0,2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=\dfrac{-1}{6}\end{matrix}\right.\)

\(b,5x^3-x^2-5x+1=0\Leftrightarrow x^2\left(5x-1\right)-\left(5x-1\right)=0\Leftrightarrow\left(5x-1\right)\left(x-1\right)\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=1\\x=-1\end{matrix}\right.\)

Answer 3

a, 1,2x³-x²-0,2x=0

<=>x(1,2x²-x-0,2)=0

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\5\left(1,2x^2-x-0,2\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\6x^2-5x-1=0\left(1\right)\end{matrix}\right.\)

giải PT (1) 6x²-5x-1=0

a+b+c=6-5-1=0

=>\(\left\{{}\begin{matrix}x_1=1\\x_2=\dfrac{-1}{6}\end{matrix}\right.\)

vậy PT có 2 nghiệm là x₁=1 ;x₂=\(\dfrac{-1}{6}\)

b,5x³-x²-5x+1=0

<=>x²(5x-1)-(5x-1)=0

<=>(x²-1)(5x-1)=0

\(\Leftrightarrow\left[{}\begin{matrix}x^2-1=0\\5x-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\pm1\\x=\dfrac{1}{5}\end{matrix}\right.\)

vậy PT có 2 nghiệm x₁=-1; x₂₌1; x₃=\(\dfrac{1}{5}\)