A=3x/x-3 ( dấu ‘/’ này phần nhé 3x phần x-3)
a,tìm x để A>2
b, tìm x nguyên để A nguyên
A=3x/x-3 ( dấu ‘/’ này phần nhé 3x phần x-3) a,tìm x để A>2 b, tìm x nguyên để A nguyên
By Adalynn
By Adalynn
A=3x/x-3 ( dấu ‘/’ này phần nhé 3x phần x-3)
a,tìm x để A>2
b, tìm x nguyên để A nguyên
`a.`
`ĐKXĐ: x \ne 3`
`A>2`
`⇔(3x)/(x-3)>2`
`⇔(3x)/(x-3)-2>0`
`⇔(3x-2(x-3))/(x-3)>0`
`⇔(3x-2x+6)/(x-3)>0`
`⇔(x+6)/(x-3)>0`
`⇔`\(\left[ \begin{array}{l}\left \{ {{x+6<0} \atop {x-3<0}} \right. \\\left \{ {{x+6>0} \atop {x-3>0}} \right.\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}\left \{ {{x<-6} \atop {x<3}} \right. \\\left \{ {{x>-6} \atop {x>3}} \right.\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x>3\\x<-6\end{array} \right.\)
Vậy để `A>2` thì `x>3` hoặc `x< -6`
`b.`
`A=(3x)/(x-3)=(3x-9+9)/(x-3)=(3(x-3)+9)/(x-3)=(3(x-3))/(x-3)+9/(x-3)=3+9/(x-3)`
`⇒`Để `A` nguyên thì:
`9/(x-3)∈Z⇔x-3 ∈Ư(9)∈{±1;±3;±9}`
`⇒`\(\left[ \begin{array}{l}x-3=1\\x-3=-1\\x-3=3\\x-3=-3\\x-3=9\\x-3=-9\end{array} \right.\) `⇔` \(\left[ \begin{array}{l}x=4(tm)\\x=2(tm)\\x=6(tm)\\x=0(tm)\\x=12(tm)\\x=-6(tm)\end{array} \right.\)
Vậy để `A` nguyên thì `x∈{0;2;4;±6;12}`
$A = \dfrac{3x}{x – 3}$ $\quad \, ĐK: \, x \ne 3$
a) $A > 2$
$\Leftrightarrow \dfrac{3x}{x – 3} > 2$
$\Leftrightarrow \dfrac{3x}{x – 3} – 2 > 0$
$\Leftrightarrow \dfrac{x + 6}{x – 3} > 0$
$\Leftrightarrow \left[\begin{array}{l}\begin{cases}x + 6 > 0\\x – 3 > 0\end{cases}\\\begin{cases}x + 6 < 0\\x – 3 < 0\end{cases}\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}\begin{cases}x >-6\\x >3\end{cases}\\\begin{cases}x <-6\\x <3\end{cases}\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x < -6\\x > 3\end{array}\right.$
b) $A = \dfrac{3x}{x – 3} = \dfrac{3x – 9 + 9}{x – 3} = \dfrac{3(x-3) + 9}{x – 3} = 3 + \dfrac{9}{x- 3}$
$A \in \Bbb Z \Leftrightarrow \dfrac{9}{x- 3} \in \Bbb Z \Leftrightarrow x – 3 \in Ư(9) = \left\{-9;-3;-1;1;3;9\right\}$
Ta có bảng giá trị sau:
$\begin{array}{|l|r|}
\hline
x-3 & -9&-3&-1&1&3&9\\
\hline
x &-6&0&2&4&6&12\\
\hline
\end{array}$