Đáp án:
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Bài `7.`
`a,`
`f (x) = x^2 - mx + 3`
Vì `f (x)` có nghiệm là `x=3`
`-> f (3)=0`
`-> 3^2 - m . 3 + 3=0`
`-> 9 - 3m + 3=0`
`-> (9 + 3)-3m = 0`
`-> 12-3m=0`
`-> 3m=12-0`
`-> 3m=12`
`->m=12 : 3`
`-> m=4`
Vậy `m=4` để `f (x)` có nghiệm là `x=3`
`b,`
`f (x) = x^2 - mx + 3`
Thay `m=4` vào ta được :
`-> f (x) = x^2 - 4x + 3`
`-> f (x) =x^2 - x - 3x + 3`
`-> f (x) = (x^2 - x) - (3x-3)`
`-> f (x) = x (x-1) -3 (x-1)`
`-> f (x) = (x-1) (x-3)`
Cho `f (x)=0`
`-> (x-1) (x-3)=0`
`->` \(\left[ \begin{array}{l}x-1=0\\x-3=0\end{array} \right.\)
`->` \(\left[ \begin{array}{l}x=1\\x=3\end{array} \right.\)
Vậy `x=1,x=3` là 2 nghiệm của `f (x)` khi `m=4`
Bài `8.`
`A (x) = x+3`
Cho `A (x)=0`
`->x+3=0`
`->x=0-3`
`->x=-3`
Vậy ....
`B (x) = 3x + 6`
Cho `B (x)=0`
`-> 3x+6=0`
`->3x=0-6`
`->3x=-6`
`->x=-6:3`
`->x=-2`
Vậy ...
`C (x) = 4x^2-1`
Cho `C (x)=0`
`-> 4x^2-1=0`
`->4x^2=1`
`->x^2=1/4`
`->` \(\left[ \begin{array}{l}x^2=(\dfrac{1}{2})^2\\x^2=(\dfrac{-1}{2})^2\end{array} \right.\)
`->` \(\left[ \begin{array}{l}x=\dfrac{1}{2}\\x=\dfrac{-1}{2}\end{array} \right.\)
Vậy ...
`D (x) = (x-2) (2x^2-8)`
Cho `D (x)=0`
`-> (x-2) (2x^2-8)=0`
`->` \(\left[ \begin{array}{l}x-2=0\\2x^2-8=0\end{array} \right.\)
`->` \(\left[ \begin{array}{l}x=2\\2x^2=8\end{array} \right.\)
`->` \(\left[ \begin{array}{l}x=2\\x^2=4\end{array} \right.\)
`->` \(\left[ \begin{array}{l}x=2\\x^2=2^2\\x^2=(-2)^2\end{array} \right.\)
`->` \(\left[ \begin{array}{l}x=2\\x=2\\x=-2\end{array} \right.\)
Vậy ...
`E (x) = 3x^3 -6x`
Cho `E (x)=0`
`->3x^3-6x=0`
`-> x (3x^2-6)=0`
`->` \(\left[ \begin{array}{l}x=0\\3x^2-6=0\end{array} \right.\)
`->` \(\left[ \begin{array}{l}x=0\\3x^2=6\end{array} \right.\)
`->` \(\left[ \begin{array}{l}x=0\\x^2=2\end{array} \right.\)
`->` \(\left[ \begin{array}{l}x=0\\x^2=(\sqrt{2})^2\\x^2=(-\sqrt{2})^2\end{array} \right.\)
`->` \(\left[ \begin{array}{l}x=0\\x=\sqrt{2}\\x=-\sqrt{2}\end{array} \right.\)
Vậy ..
`F (x) = 9x^3 - 25x`
Cho `F (x)=0`
`-> 9x^3 - 25x=0`
`-> x (9x^2 - 25)=0`
`->` \(\left[ \begin{array}{l}x=0\\9x^2-25=0\end{array} \right.\)
`->` \(\left[ \begin{array}{l}x=0\\9x^2=25\end{array} \right.\)
`->` \(\left[ \begin{array}{l}x=0\\x^2=\dfrac{25}{9}\end{array} \right.\)
`->` \(\left[ \begin{array}{l}x=0\\x^2=(\dfrac{5}{3})^2\\x^2=(\dfrac{-5}{3})^2\end{array} \right.\)
`->` \(\left[ \begin{array}{l}x=0\\x=\dfrac{5}{3}\\x=\dfrac{-5}{3}\end{array} \right.\)
Vậy ...
`G (x) = x^3 + 2x`
Cho `G (x)=0`
`->x^3+2x=0`
`-> x (x^2+2)=0`
`->` \(\left[ \begin{array}{l}x=0\\x^2+2=0\end{array} \right.\)
`->` \(\left[ \begin{array}{l}x=0\\x^2=-2 \text{(Vô lí)}\end{array} \right.\)
`->` \(\left[ \begin{array}{l}x=0\\x=∅\end{array} \right.\)
Vậy ...
`H (x) = -9x^2 + 4`
Cho `H (x)=0`
`-> -9x^2+4=0`
`-> -9x^2=-4`
`-> x^2=4/9`
`->` \(\left[ \begin{array}{l}x^2=(\dfrac{2}{3})^2\\x^2=(\dfrac{-2}{3})^2\end{array} \right.\)
`->` \(\left[ \begin{array}{l}x=\dfrac{2}{3}\\x=\dfrac{-2}{3}\end{array} \right.\)
Vậy ...
`P (x) = 2 (-x+3) - 4 (x-2)`
Cho `P (x)=0`
`-> 2 (-x+3) - 4 (x-2)=0`
`-> -2x + 6 - 4x +8=0`
`-> (-2x-4x)+(6+8)=0`
`-> -6x + 14=0`
`-> -6x=-14`
`->x=7/3`
Vậy ...
`Q (x) = x^2 -6x + 5`
Cho `Q (x)=0`
`-> x^2-6x+5=0`
`-> x^2 - x - 5x + 5=0`
`-> (x^2-x) - (5x-5)=0`
`-> x (x-1) - 5 (x-1)=0`
`-> (x-1) (x-5)=0`
`->` \(\left[ \begin{array}{l}x-1=0\\x-5=0\end{array} \right.\)
`->` \(\left[ \begin{array}{l}x=1\\x=5\end{array} \right.\)
Vậy ...
