Tìm `x` `a) (x – 1)^6 = (x – 1)^8` `b) (x^3 + 3) . (x^3 + 10) . (x^3 + 15) . (x^3 + 30) < 0` 28/10/2021 Bởi Hailey Tìm `x` `a) (x – 1)^6 = (x – 1)^8` `b) (x^3 + 3) . (x^3 + 10) . (x^3 + 15) . (x^3 + 30) < 0`
`\text{~~Holi~~}` `a. (x-1)^6=(x-1)^8` `-> (x-1)^6-(x-1)^8=0` `-> (x-1)^6.[1-(x-1)^2]=0` `->`\(\left[ \begin{array}{l}(x-1)^6=0\\1-(x-1)^2=0\end{array} \right.\) `->`\(\left[ \begin{array}{l}x-1=0\\(x-1)^2=1\end{array} \right.\) `->`\(\left[ \begin{array}{l}x=1\\x-1=-1\\x-1=1\end{array} \right.\) `->`\(\left[ \begin{array}{l}x=1\\x=0\\x=2\end{array} \right.\) Vậy `x_1=0,x_2=0,x_3=2` _______________________ Ở bình luận bn chủ tus cho em lm câu a thôi ạ. Bình luận
$(x-1)^6=(x-1)^8$$⇔(x-1)^8-(x-1)^6=0$$⇔(x-1)^6[(x-1)^2-1]=0$$⇔(x-1)^6(x-1-1)(x-1+1)=0$$⇔(x-1)^6(x-2)x=0$$⇔\left[\begin{array}{l}x=0\\x-1=0\\x-2=0\end{array}\right.⇔\left[\begin{array}{l}x=0\\x=1\\x=2\end{array}\right.$ Vậy $x \in \{0;1;2\}$ Bình luận
`\text{~~Holi~~}`
`a. (x-1)^6=(x-1)^8`
`-> (x-1)^6-(x-1)^8=0`
`-> (x-1)^6.[1-(x-1)^2]=0`
`->`\(\left[ \begin{array}{l}(x-1)^6=0\\1-(x-1)^2=0\end{array} \right.\)
`->`\(\left[ \begin{array}{l}x-1=0\\(x-1)^2=1\end{array} \right.\)
`->`\(\left[ \begin{array}{l}x=1\\x-1=-1\\x-1=1\end{array} \right.\)
`->`\(\left[ \begin{array}{l}x=1\\x=0\\x=2\end{array} \right.\)
Vậy `x_1=0,x_2=0,x_3=2`
_______________________
Ở bình luận bn chủ tus cho em lm câu a thôi ạ.
$(x-1)^6=(x-1)^8$
$⇔(x-1)^8-(x-1)^6=0$
$⇔(x-1)^6[(x-1)^2-1]=0$
$⇔(x-1)^6(x-1-1)(x-1+1)=0$
$⇔(x-1)^6(x-2)x=0$
$⇔\left[\begin{array}{l}x=0\\x-1=0\\x-2=0\end{array}\right.⇔\left[\begin{array}{l}x=0\\x=1\\x=2\end{array}\right.$
Vậy $x \in \{0;1;2\}$