bài 1: a)x ∧ 4-16 ² =0 b)x ∧8+36x ∧4=0 d)5(x-2)-x ²+4=0 c)(x-5) ³-x+5=0 10/08/2021 Bởi Liliana bài 1: a)x ∧ 4-16 ² =0 b)x ∧8+36x ∧4=0 d)5(x-2)-x ²+4=0 c)(x-5) ³-x+5=0
a) $x^4-16^2=0$ $↔ x^4=16^2$ $↔ x^4=4^4$ $↔ \left[ \begin{array}{l}x=4\\x=-4\end{array} \right.$ b) $x^8+36x^4=0$ $↔ x^4(x^4+36)=0$ $↔ \left[ \begin{array}{l}x^4=0\\x^4=-36\end{array} \right.$ $→ x=0$ c) $5(x-2)-x^2+4=0$ $↔ 5(x-2)-(x-2)(x+2)=0$ $↔ (x-2)(5-x-2)=0$ $↔ \left[ \begin{array}{l}x-2=0\\3-x=0\end{array} \right.$ $↔ \left[ \begin{array}{l}x=2\\x=3\end{array} \right.$ d) $(x-5)^3-x+5=0$ $↔ (x-5)^3-(x-5)=0$ Đặt $x-5=t$, ta có: $t^3-t=0$ $↔ t(t^2-1)=0$ $↔ \left[ \begin{array}{l}t=0\\t=1\\t=-1\end{array} \right.$ Với $t=0$ ta có: $x-5=0 ↔ x=5$ Với $t=1$ ta có: $x-5=1 ↔ x=6$ Với $t=-1$ ta có: $x-5=-1 ↔ x=4$ Bình luận
`a,` `x^4-16^2=0` `⇒x^4=16^2` `⇒x^4=(±4)^4` `⇒x=±4` `b,` `x^8+36x^4=0` `⇒x^4(x^4+36)=0` \(⇒\left[ \begin{array}{l}x^4=0\\x^4+36=0\end{array} \right.⇒\left[ \begin{array}{l}x=0\\x^4=-36\text{(vô lý)}\end{array} \right.\) `c,` `(x-5)^3-x+5=0` `⇒(x-5)^3-(x-5)=0` `⇒(x-5)[(x-5)^2-1]=0` \(⇒\left[ \begin{array}{l}x-5=0\\(x-5)^2-1=0\end{array} \right.⇒\left[ \begin{array}{l}x=5\\\left[ \begin{array}{l}x-5=1\\x-5=-1\end{array} \right.⇒\left[ \begin{array}{l}x=6\\x=-4\end{array} \right.\end{array} \right.\) `d,` `5(x-2)-x^2+4=0` `5(x-2)-(x+2)(x-2)=0` `(x-2)(5-x-2)=0` \(⇒\left[ \begin{array}{l}x-2=0\\5-x-2\end{array} \right.⇒\left[ \begin{array}{l}x=2\\x=3\end{array} \right.\) Bình luận
a) $x^4-16^2=0$
$↔ x^4=16^2$
$↔ x^4=4^4$
$↔ \left[ \begin{array}{l}x=4\\x=-4\end{array} \right.$
b) $x^8+36x^4=0$
$↔ x^4(x^4+36)=0$
$↔ \left[ \begin{array}{l}x^4=0\\x^4=-36\end{array} \right.$
$→ x=0$
c) $5(x-2)-x^2+4=0$
$↔ 5(x-2)-(x-2)(x+2)=0$
$↔ (x-2)(5-x-2)=0$
$↔ \left[ \begin{array}{l}x-2=0\\3-x=0\end{array} \right.$
$↔ \left[ \begin{array}{l}x=2\\x=3\end{array} \right.$
d) $(x-5)^3-x+5=0$
$↔ (x-5)^3-(x-5)=0$
Đặt $x-5=t$, ta có:
$t^3-t=0$
$↔ t(t^2-1)=0$
$↔ \left[ \begin{array}{l}t=0\\t=1\\t=-1\end{array} \right.$
Với $t=0$ ta có: $x-5=0 ↔ x=5$
Với $t=1$ ta có: $x-5=1 ↔ x=6$
Với $t=-1$ ta có: $x-5=-1 ↔ x=4$
`a,` `x^4-16^2=0`
`⇒x^4=16^2`
`⇒x^4=(±4)^4`
`⇒x=±4`
`b,` `x^8+36x^4=0`
`⇒x^4(x^4+36)=0`
\(⇒\left[ \begin{array}{l}x^4=0\\x^4+36=0\end{array} \right.⇒\left[ \begin{array}{l}x=0\\x^4=-36\text{(vô lý)}\end{array} \right.\)
`c,` `(x-5)^3-x+5=0`
`⇒(x-5)^3-(x-5)=0`
`⇒(x-5)[(x-5)^2-1]=0`
\(⇒\left[ \begin{array}{l}x-5=0\\(x-5)^2-1=0\end{array} \right.⇒\left[ \begin{array}{l}x=5\\\left[ \begin{array}{l}x-5=1\\x-5=-1\end{array} \right.⇒\left[ \begin{array}{l}x=6\\x=-4\end{array} \right.\end{array} \right.\)
`d,` `5(x-2)-x^2+4=0`
`5(x-2)-(x+2)(x-2)=0`
`(x-2)(5-x-2)=0`
\(⇒\left[ \begin{array}{l}x-2=0\\5-x-2\end{array} \right.⇒\left[ \begin{array}{l}x=2\\x=3\end{array} \right.\)