a) a) (x+2)(x – 3) = 0
b) (x – 5)(7 – x) = 0
c) (x – 1)(x+5)(-3x + 8) = 0
d) (x – 1)(3x+1) = 0
e) (x – 1)(x+2)(x-3) = 0
f) x(x2 – 1) = 0
giúp em với
a) a) (x+2)(x – 3) = 0
b) (x – 5)(7 – x) = 0
c) (x – 1)(x+5)(-3x + 8) = 0
d) (x – 1)(3x+1) = 0
e) (x – 1)(x+2)(x-3) = 0
f) x(x2 – 1) = 0
giúp em với
`a) (x+2)(x – 3) = 0`
`<=>x=-2` hoặc `x=3`
`b) (x – 5)(7 – x) = 0`
`<=>x=5` hoặc `x=7`
`c) (x – 1)(x+5)(-3x + 8) = 0`
`<=>x={1;-5;8/3}`
`d) (x – 1)(3x+1) = 0`
`x={1;-1/3}`
`e) (x – 1)(x+2)(x-3) = 0`
`<=>x={1;-2;3}`
`f) x(x2 – 1) = 0`
`<=>x=0` hoặc `x=1/2`
*Bạn tham khảo nhé!!!!
a. (x + 2)(x – 3) = 0
⇔\(\left[ \begin{array}{l}x+2=0\\x-3=0\end{array} \right.\) ⇒ \(\left[ \begin{array}{l}x=-2\\x=3\end{array} \right.\)
b. (x – 5)(7 – x) = 0
⇔\(\left[ \begin{array}{l}x-5=0\\7-x=0\end{array} \right.\) ⇒\(\left[ \begin{array}{l}x=5\\x=7\end{array} \right.\)
c. (x – 1)(x + 5)(-3x + 8) = 0
⇔\(\left[ \begin{array}{l}x-1=0\\x+5=0\\-3x+8=0\end{array} \right.\) ⇒ \(\left[ \begin{array}{l}x=1\\x=-5\\x=\frac{8}{3}\end{array} \right.\)
d. (x – 1)(3x + 1) = 0
⇔\(\left[ \begin{array}{l}x-1=0\\3x+1=0\end{array} \right.\) ⇒\(\left[ \begin{array}{l}x=1\\x=\frac{-1}{3}\end{array} \right.\)
e. (x – 1)(x + 2)(x – 3) = 0
⇔\(\left[ \begin{array}{l}x-1=0\\x+2=0\\x-3=0\end{array} \right.\) ⇒\(\left[ \begin{array}{l}x=1\\x=-2\\x=3\end{array} \right.\)
f. x(x² – 1) = 0
⇔x(x + 1)(x – 1) = 0
⇔\(\left[ \begin{array}{l}x=0\\x+1=0\\x-1=0\end{array} \right.\) ⇒\(\left[ \begin{array}{l}x=0\\x=-1\\x=1\end{array} \right.\)