Đáp án:
Bài `10`
`a)`
` g(x) = 0`
` => (6-3x)(-2x+5) = 0`
` =>` \(\left[ \begin{array}{l}6-3x=0\\-2x+5=0\end{array} \right.\)
`=>` \(\left[ \begin{array}{l}3x=6\\-2x=-5\end{array} \right.\)
`=>` \(\left[ \begin{array}{l}x=2\\x=\dfrac{5}{2}\end{array} \right.\)
`b)` `h(x) = 0`
` => x^2 +x= 0`
` => x(x+1)=0`
` => ` \(\left[ \begin{array}{l}x=0\\x+1=0\end{array} \right.\)
` =>` \(\left[ \begin{array}{l}x=0\\x=-1\end{array} \right.\)
Bài `11`
`a)`
Sắp xểp:
`f(x) = -x^5 - 7x^4 - 2x^3 + x^2 + 4x + 9`
`g(x) = x^5 + 7x^4 + 2x^3 + 2x^2 - 3x - 9`
`b)`
` h(x) = (-x^5 - 7x^4 - 2x^3 + x^2 + 4x + 9) + ( x^5 + 7x^4 + 2x^3 + 2x^2 - 3x - 9)`
` = -x^5 - 7x^4 - 2x^3 + x^2 + 4x + 9 + x^5 + 7x^4 + 2x^3 + 2x^2 - 3x - 9`
` = ( -x^5 + x^5) - (7x^4 - 7x^4) - (2x^3 - 2x^3) + (x^2 + 2x^2) + (4x-3x) + (9-9)`
` = 0 - 0 - 0 + 3x^2 + x + 0`
` = 3x^2 +x`
`c)`
` h(x) = 0`
` => 3x^2 + x = 0`
` => x(3x +1) = 0`
` =>` \(\left[ \begin{array}{l}x=0\\3x+1=0\end{array} \right.\)
` =>` \(\left[ \begin{array}{l}x=0\\x=\dfrac{-1}{3}\end{array} \right.\)
Vậy nghiệm của đa thức là ` x=0 ; x=\frac{-1}{3}`
