Bài 2 : Tìm nghiệm các đa thức sau : 1.(x-5).(3 – 1/5x) 2.(x-2)^3= 1/8 14/08/2021 Bởi Bella Bài 2 : Tìm nghiệm các đa thức sau : 1.(x-5).(3 – 1/5x) 2.(x-2)^3= 1/8
`1. (x-5).(3 – 1/5x)` `(x-5).(3 – 1/5x) = 0` $\huge ⇔ \left \{ {{x – 5 = 0} \atop {3 – \dfrac{1}{5}x = 0}} \right. ⇔ \left \{ {{x = 5} \atop {x = 15}} \right.$ `2. (x-2)^3= 1/8` `(x – 2)^3 = (1/2)^3` `x – 2 = 1/2` `x = 1/2 + 2` `x = 5/2` Bình luận
`a,(x-5).(3-1/5 x)=0` `⇔` \(\left[ \begin{array}{l}x-5=0\\3-\dfrac{1}{5}x=0\end{array} \right.\) `⇔` \(\left[ \begin{array}{l}x=5\\x=15\end{array} \right.\) `b,(x-2)^3=1/8` `⇔(x-2)^3=(1/2)^3` `⇔x-2=1/2` `⇔x=5/2` Bình luận
`1. (x-5).(3 – 1/5x)`
`(x-5).(3 – 1/5x) = 0`
$\huge ⇔ \left \{ {{x – 5 = 0} \atop {3 – \dfrac{1}{5}x = 0}} \right. ⇔ \left \{ {{x = 5} \atop {x = 15}} \right.$
`2. (x-2)^3= 1/8`
`(x – 2)^3 = (1/2)^3`
`x – 2 = 1/2`
`x = 1/2 + 2`
`x = 5/2`
`a,(x-5).(3-1/5 x)=0`
`⇔` \(\left[ \begin{array}{l}x-5=0\\3-\dfrac{1}{5}x=0\end{array} \right.\) `⇔` \(\left[ \begin{array}{l}x=5\\x=15\end{array} \right.\)
`b,(x-2)^3=1/8`
`⇔(x-2)^3=(1/2)^3`
`⇔x-2=1/2`
`⇔x=5/2`