Bài 1 : Tập nghiệm của phương trình
a) ( 2x + 1) ( 2 – 3x) = 0
b) 2x( x + 1) = x2 – 1
c) 3( x – 2) + x2 + 4= 0
d) 3×2 + 6x = 9 =0
Bài 1 : Tập nghiệm của phương trình
a) ( 2x + 1) ( 2 – 3x) = 0
b) 2x( x + 1) = x2 – 1
c) 3( x – 2) + x2 + 4= 0
d) 3×2 + 6x = 9 =0
Bài 1:
$a)$ `(2x +1).(2 -3x) = 0`
$⇔ \left[ \begin{array}{l}2x +1 = 0=2\\2 -3x = 0\end{array} \right. ⇔ \left[ \begin{array}{l}x=\dfrac{-1}{2}\\x=\dfrac{2}{3}\end{array} \right.$
Vậy `S = {-1/2; 2/3}`
$b)$ $2x.(x +1) = x² -1$
`⇔ 2x² +2x -x² +1 = 0`
`⇔ x² +2x +1= 0`
`⇔ (x +1)² = 0`
`⇔ x +1 = 0`
`⇔ x = -1`
Vậy `S = {-1}`
$c)$ `3.(x -2) + x² +4 = 0`
`⇔ 3x -6 +x² +4 = 0`
`⇔ x² +3x -2 = 0`
`⇔ (x +3/2)² -17/4 = 0`
`⇔ (x +3/2 -\sqrt{17}/2).(x +3/2 +\sqrt{17}/2) = 0`
$⇔ \left[ \begin{array}{l}x +\dfrac{3}{2} -\dfrac{\sqrt{17}}{2}=0\\x +\dfrac{3}{2} +\dfrac{\sqrt{17}}{2}=0\end{array} \right. ⇔ \left[ \begin{array}{l}x=\dfrac{-3 +\sqrt{17}}{2}\\x=\dfrac{-3 -\sqrt{17}}{2}\end{array} \right.$
Vậy `S = {(-3 ±\sqrt{17})/2}`
$d$) `3x² +6x -9 = 0`
`⇔ x² +2x -3 = 0`
`⇔ x.(x -1) +3.(x -1) = 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 `S = {1; -3}`
`\text{~~Holi~~}`
`a.(2x+1)(2-3x)=0`
`->`\(\left[ \begin{array}{l}2x+1=0\\2-3x=0\end{array} \right.\)
`->`\(\left[ \begin{array}{l}x=-\dfrac{1}{2}\\x=\dfrac{2}{3}\end{array} \right.\)
Vậy `S={-1/2;2/3}`
`b. 2x(x+1)=x^2-1`
`-> 2x(x+1)=(x-1)(x+1)`
`-> 2x=(x-1)`
`-> 2x-(x-1)=0`
`-> 2x-x+1=0`
`-> x+1=0`
`-> x=-1`
Vậy `S={-1}`
`c. 3(x-2)+x^2+4=0`
`-> 3x-6+x^2+4=0`
`-> x^2+3x-2=0`
`-> (x+3/2)^2-17/4=0`
`-> (x+3/2)^2-(\sqrt{17}/2)^2=0`
`->(x+3/2-\sqrt{17}/2)(x+3/2+\sqrt{17}/2)=0`
`->`\(\left[ \begin{array}{l}x+\dfrac{3}{2}-\dfrac{\sqrt{17}}{2}=0\\x+\dfrac{3}{2}+\dfrac{\sqrt{17}}{2}=0\end{array} \right.\)
`->`\(\left[ \begin{array}{l}x=-\dfrac{3-\sqrt{17}}{2}\\x=-\dfrac{3+\sqrt{17}}{2}\end{array} \right.\)
Vậy `S={-\frac{3±\sqrt{17}}{2}}`
`d. 3x^2+6x-9=0`
`-> 3(x^2+2x-3)=0`
`-> x^2+2x-3=0`
`-> x^2-x+3x-3=0`
`-> x(x-1)+3(x-1)=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 `S={1;-3}`